phy 1

1. UNITSMeasurements of physical quantities take place by means of a comparison with a standard. For example: a meter stick, a weight of 1 kilogram, etc.

The base units that will be used in this course are:

meter (m): One meter is equal to the path length traveled by light in vacuum during a time interval of 1/299,792,458 of a second.

kilogram (kg): One kilogram is the mass of a Platinum-Iridium cylinder kept at the International Bureau of Weights and Measures in Paris.

second (s): One second is the time occupied by 9,192,631,770 vibrations of the light (of a specified wavelength) emitted by a Cesium-133 atom.

A unit that is being used as a base unit must be both accessible and invariable. The original meter bar kept in Paris was not very accessible (this is still true for the kilogram). In addition, the length of the standard bar is temperature dependent. The definition of the meter in terms of the number of wavelengths of a particular atomic transition in Krypton-86 made the meter more accessible; the transition is characteristic for Krypton-86, and is the same for each Krypton-86 atom. However, Doppler shifts due to the thermal motion of the atoms can slightly change the wavelength, and produce a small uncertainty in the definition of the meter. The current definition of the meter in terms of the speed of light is not affected by thermal motion: the speed of light in vacuum is constant, independent of temperature and velocity of observer and/or source.

Example: how to measure the distance d from the earth to the moon ?

APOLLO astronauts placed a mirror on the moon. It can be used to measure the distance between the earth and the moon very accurately. The reflection of a laser beam aimed at this mirror reaches the earth after 2.495 s. The distance can then be calculated:

The definition of the standard mass makes it very inaccessible. In principle, the weight of individual nuclei can be used as a standard; nuclear weight does not depend on location, temperature, pressure, etc. However, counting the number of nuclei in a standard (or assembling a fixed number of nuclei) is an almost impossible task.

Note:

1. Improved definitions of base units must be defined such that it matches the previous definition as closely as possible (no need to change all meter sticks in 1983).

2. Speed of light is now defined as 299,792,458 m/s.

All physical quantities are expressed in terms of base units. For example, the velocity is usually given in units of m/s. All other units are derived units and may be expressed as a combination of base units. For example (see Appendix A):

Other examples are:

density: kg/m3

area: m2

When dealing with very small or large numbers, it is convenient to use prefixes (see Table 1.1, and Table 1-2 on page 3 in Halliday, Resnick, and Walker).

Factor Prefix Symbol

1018 exa- E 1015 peta- P 1012 tera T 109 giga- G 106 mega- M 103 kilo- k 10-3 milli- m 10-6 micro- u 10-9 nano- n 10-12 pico- p 10-15 femto- f 10-18 atto- a

Table 1.1. SI Prefixes

Example: 2.35 10-9 s = 2.35 ns

In many cases, the data are not supplied in the correct units, and one needs to convert units (see Appendix F in Halliday, Resnick, and Walker).

Examples: 1" = 2.54 x 10-2 m

55 miles = 88 x 103 m

1 h = 3600 s

55 miles/h = 88 x 103/3600 m/s = 24.4 m/s

2. MOTION IN A STRAIGHT LINEIn mechanics we are interested in trying to understand the motion of objects. In this chapter, the motion of objects in 1 dimension will be discussed. Motion in 1 dimension is motion along a straight line.

2.1. Position

The position of an object along a straight line can be uniquely identified by its distance from a (user chosen) origin. (see Figure 2.1). Note: the position is fully specified by 1 coordinate (that is why this a 1 dimensional problem).

Figure 2.1. One-dimensional position.

Figure 2.2. x vs. t graphs for various velocities.

For a given problem, the origin can be chosen at whatever point is convenient. For example, the position of the object at time t = 0 is often chosen as the origin. The position of the object will in general be a function of time: x(t). Figure 2.2. shows the position as a function of time for an object at rest, and for objects moving to the left and to the right.

The slope of the curve in the position versus time graph depends on the velocity of the object. See for example Figure 2.3. After 10 seconds, the cheetah has covered a distance of 310 meter, the human 100 meter, and the pig 50 meter. Obviously, the cheetah has the highest velocity. A similar conclusion is obtained when we consider the time required to cover a fixed distance. The cheetah covers 300 meter in 10 s, the human in 30 s, and the pig requires 60 s. It is clear that a steeper slope of the curve in the x vs. t graph corresponds to a higher velocity.

Figure 2.3. x vs. t graphs for various creatures.

2.2. Velocity

An object that changes its position has a non-zero velocity. The average velocity of an object during a specified time interval is defined as:

If the object moves to the right, the average velocity is positive. An object moving to the left has a negative average velocity. It is clear from the definition of the average velocity that depends only on the position of the object at time t = t1 and at time t = t2. This is nicely illustrated in sample problem 2-1 and 2-2.

Sample Problem 2-1

You drive a beat-up pickup truck down a straight road for 5.2 mi at 43 mi/h, at which point you run out of fuel. You walk 1.2 mi farther, to the nearest gas station, in 27 min (= 0.450 h). What is your average velocity from the time you started your truck to the time that you arrived at the station ?

The pickup truck initially covers a distance of 5.2 miles with a velocity of 43 miles/hour. This takes 7.3 minutes. After the pickup truck runs out of gas, it takes you 27 minutes to walk to the nearest gas station which is 1.2 miles down the road. When you arrive at the gas station, you have covered (5.2 + 1.2) = 6.4 miles, during a period of (7.3 + 27) = 34.3 minutes. Your average velocity up to this point is:

Sample Problem 2-2

Suppose you next carry the fuel back to the truck, making the round-trip in 35 min. What is your average velocity for the full journey, from the start of your driving to you arrival back at the truck with the fuel ?

It takes you another 35 minutes to walk back to your car. When you reach your truck, you are again 5.2 miles from the origin, and have been traveling for (34.4 + 35) = 69.4 minutes. At that point your average velocity is:

After this episode, you return back home. You cover the 5.2 miles again in 7.3 minutes (velocity equals 43 miles/hour). When you arrives home, you are 0 miles from your origin, and obviously your average velocity is:

The average velocity of the pickup truck which was left in the garage is also 0 miles/hour. Since the average velocity of an object depends only on its initial and final location and time, and not on the motion of the object in between, it is in general not a useful parameter. A more useful quantity is the instantaneous velocity of an object at a given instant. The instantaneous velocity is the value that the average velocity approaches as the time interval over which it is measured approaches zero:

For example: see sample problem 2-5.

The velocity of the object at t = 3.5 s can now be calculated:

2.3. Acceleration

The velocity of an object is defined in terms of the change of position of that object over time. A quantity used to describe the change of the velocity of an object over time is the acceleration a. The average acceleration over a time interval between t1 and t2 is defined as:

Note the similarity between the definition of the average velocity and the definition of the average acceleration. The instantaneous acceleration a is defined as:

From the definition of the acceleration, it is clear that the acceleration has the following units:

A positive acceleration is in general interpreted as meaning an increase in velocity. However, this is not correct. From the definition of the acceleration, we can conclude that the acceleration is positive if

This is obviously true if the velocities are positive, and the velocity is increasing with time. However, it is also true for negative velocities if the velocity becomes less negative over time.

2.4. Constant Acceleration

Objects falling under the influence of gravity are one example of objects moving with constant acceleration. A constant acceleration means that the acceleration does not depend on time:

Integrating this equation, the velocity of the object can be obtained:

where v0 is the velocity of the object at time t = 0. From the velocity, the position of the object as function of time can be calculated:

where x0 is the position of the object at time t = 0.

Note 1: verify these relations by integrating the formulas for the position and the velocity.

Note 2: the equations of motion are the basis for most problems (see sample problem 7).

Sample Problem 2-8

Spotting a police car, you brake a Porsche from 75 km/h to 45 km/h over a distance of 88m. a) What is the acceleration, assumed to be constant ? b) What is the elapsed time ? c) If you continue to slow down with the acceleration calculated in (a) above, how much time would elapse in bringing the car to rest from 75 km/h ? d) In (c) above, what distance would be covered ? e) Suppose that, on a second trial with the acceleration calculated in (a) above and a different initial velocity, you bring your car to rest after traversing 200 m. What was the total braking time ?

a) Our starting points are the equations of motion:

(1)

(2)

The following information is provided:

* v(t = 0) = v0 = 75 km/h = 20.8 m/s

* v(t1) = 45 km/h = 12.5 m/s

* x(t = 0) = x0 = 0 m (Note: origin defined as position of Porsche at t = 0 s)

* x(t1) = 88 m

* a = constant

From eq.(1) we obtain:

(3)

Substitute (3) in (2):

(4)

From eq.(4) we can obtain the acceleration a:

(5)

b) Substitute eq.(5) into eq.(3):

(6)

c) The car is at rest at time t2:

(7)

Substituting the acceleration calculated using eq.(5) into eq.(3):

(8)

d) Substitute t2 (from eq.(8)) and a (from eq.(5)) into eq.(2):

(9)

e) The following information is provided:

* v(t3) = 0 m/s (Note: Porsche at rest at t = t3)

* x(t = 0) = x0 = 0 m (Note: origin defined as position of Porsche at t = 0)

* x(t3) = 200 m

* a = constant = - 1.6 m/s2

Eq.(1) tells us:

(10)

Substitute eq.(10) into eq.(2):

(11)

The time t3 can now easily be calculated:

(12)

2.5. Gravitational Acceleration

A special case of constant acceleration is free fall (falling in vacuum). In problems of free fall, the direction of free fall is defined along the y-axis, and the positive position along the y-axis corresponds to upward motion. The acceleration due to gravity (g) equals 9.8 m/s2 (along the negative y-axis). The equations of motion for free fall are very similar to those discussed previously for constant acceleration:

where y0 and v0 are the position and the velocity of the object at time t = 0.

Example

A pitcher tosses a baseball straight up, with an initial speed of 25 m/s. (a) How long does it take to reach its highest point ? (b) How high does the ball rise above its release point ? (c) How long will it take for the ball to reach a point 25 m above its release point.

Figure 2.4. Vertical position of baseball as function of time.

a) Our starting points are the equations of motion:

The initial conditions are:

* v(t = 0) = v0 = 25 m/s (upwards movement)

* y(t = 0) = y0 = 0 m (Note: origin defined as position of ball at t = 0)

* g = 9.8 m/s2

The highest point is obtained at time t = t1. At that point, the velocity is zero:

The ball reaches its highest point after 2.6 s (see Figure 2.4).

b) The position of the ball at t1 = 2.6 s can be easily calculated:

c) The quation for y(t) can be easily rewritten as:

where y is the height of the ball at time t. This Equation can be easily solved for t:

Using the initial conditions specified in (a) this equation can be used to calculate the time at which the ball reaches a height of 25 m (y = 25 m):

t = 1.4 s

t = 3.7 s

Figure 2.5. Velocity of the baseball as function of time.

The velocities of the ball at these times are (see also Figure 2.5):

v(t = 1.4 s) = + 11.3 m/s

v(t = 3.7 s) = - 11.3 m/s

At t = 1.4 s, the ball is at y = 25 m with positive velocity (upwards motion). At t = 2.6 s, the ball reaches its highest point (v = 0). After t = 2.6 s, the ball starts falling down (negative velocity). At t= 3.7 s the ball is located again at y = 25 m, but now moves downwards.

3. VECTORSThe motion of a particle in one dimension is simple. Its velocity is either positive or negative: positive velocity corresponds to a motion to the right while negative velocity corresponds to a motion to the left. To describe the motion of an object in 3 dimensions, we need to specify not only the magnitude of its velocity, but also its direction: velocity is a vector. A quantity not involving a direction is a scalar. Examples of scalars are temperature, pressure and time. In this Chapter we will discuss the various vector operations that will be used in this course.

3.1. Vector Addition

One important vector operation that we will frequently encounter is vector addition. There are two methods for vector addition: the graphical method and the analytical method. We will start discussing vector addition by using the graphical method.

3.1.1. The graphical method

Assume two vectors and are defined. If is added to , a third vector is created (see Figure 3.1).

Figure 3.1. Commutative Law of Vector Addition.

There are however two ways of combining the vectors and (see Figure 3.1). Inspection of the resulting vector shows that vector addition satisfies thecommutative law (order of addition does not influence the final result):

Vector addition also satisfies the associative law (the result of vector addition is independent of the order in which the vectors are added, see Figure 3.2):

Figure 3.2. Associative Law of Vector Addition.

Figure 3.3. Vector and -

The opposite of vector is a vector with the same magnitude as but pointing in the opposite direction (see Figure 3.3):

+ (- ) = 0

Subtracting from is the same as adding the opposite of to (see Figure 3.4):

= - = + (- )

Figure 3.4. Vector Subtraction.

Figure 3.4 also shows that + = .

In actual calculations the graphical method is not practical, and the vector algebra is performed on its components (this is the analytical method).

3.1.2. The analytical method

To demonstrate the use of the analytical method of vector addition, we limit ourselves to 2 dimensions. Define a coordinate system with an x-axis and y-

axis (see Figure 3.5). We can always find 2 vectors, and , whose vector

sum equals . These two vectors, and , are called the components of , and by definition satisfy the following relation:

Suppose that [theta] is the angle between the vector and the x-axis. The 2 components of are defined such that their direction is along the x-axis and y-axis. The length of each of the components can now be easily calculated:

and the vector can be written as:

Figure 3.5. Decomposition of vector .

Note: in contrast to Halliday, Resnick, and Walker we use and to indicate the unit vectors along the x- and y-axis, respectively (Halliday, Resnick, and Walker use i, j, and k which is harder to write). The decomposition of vector into 2 components is not unambiguous. It depends on the choice of the coordinate system (see Figure 3.6).

Although the decomposition of a vector depends on the coordinate system chosen, relations between vectors are not affected by the choice of the coordinate system (for example, if two vectors are perpendicular in one coordinate system, they are perpendicular in every coordinate system). The physics (and the relation between physical quantities) is also not affected by the choice of the coordinate system, and we usually choose the coordinate system such that our problems can be solved most easily. This was already demonstrated in Chapter 2, where the origin of the coordinate system was usually defined as the position of the object at time t = 0. In two or three dimensions, we can usually choose the coordinate system such that one of the vectors coincides with one of the coordinate axes. For example, if the coordinate system is defined such that the direction of is along the x-axis, the components of are:

ax = a

ay = 0

az = 0

Figure 3.6. Decomposition of vector in different coordinate systems.

Vector algebra using the analytical method is based on the following rule:

"Two vectors are equal to each other if their corresponding components are equal"

Applying this rule to vector addition:

Comparing the equations for and , we can conclude that since is equal to , their components are related as follows:

These relations describe vector addition using the analytical technique (and similar relations hold for vector addition in three dimensions).

We have just described how to find the components of a vector if its magnitude and direction are provided. If the components of a vector are provided, we can also calculate its direction and magnitude. Suppose the components of vector

along the x-axis and y-axis are ax and ay, respectively. The length of the vector can be easily calculated:

The angle [theta] between the vector and the positive x-axis can be obtained from the following relation (see also Figure 3.5):

Note: one should exercise great care in using the previous relation for [theta]. Atan(ay/ax) has 2 solutions; any calculator will return the solution between - [pi]/2 and [pi]/2. The correct solution will depend on the sign of ax:

if ax > 0: - [pi]/2 < [theta] < [pi]/2

if ax < 0: [pi]/2 < [theta] < 3[pi]/2

Figure 3.7. Example of angle ambiguity.

Example: The two vectors shown in Figure 3.7 can be written as:

It is clear that for both vectors: atan([theta]) = 1. For vector , the solution is [theta] = 45deg., while for vector , the solution is [theta] = 135deg.. Note that ax = 1 > 0, and bx = -1 < 0.

3.2. Multiplying Vectors - Multiplying a Vector by a Scalar

The product of a vector and a scalar s is a new vector, whose direction is the same as that of if s is positive or opposite to that direction if s is negative (see Figure 3.8). The magnitude of the new vector is the magnitude of multiplied by the absolute value of s. This procedure can be summarized as follows:

Figure 3.8. Multiplying a vector by a scalar.

Since two vectors are only equal only if their corresponding components are equal, we obtain the following relation between the components of and the components of :

The following relations are summarizing the relations between the magnitude and direction of the vectors and :

3.3. Multiplying Vectors - Scalar Product

Two vectors and are shown in Figure 3.9. The angle between these two vectors is [phi]. The scalar product of and (represented by . ) is defined as:

In a coordinate system in which the x, y and z-axes are mutual perpendicular, the following relations hold for the scalar product between the various unit vectors:

Suppose that the vectors and are defined as follows:

Figure 3.9. Scalar Product of vectors and .

The scalar product of and can now be rewritten in terms of the scalar product between the unit vectors along the x, y and z-axes:

Note that in deriving this equation, we have applied the following rule:

An alternative derivation of the expression of the scalar product in terms of the components of the two vectors can be easily derived as follows (see Figure 3.10). The components of and are given by:

Figure 3.10. Alternative Derivation of Scalar Product.

ax = a cos(a)

ay = a sin(a)

bx = b cos([beta])

by = b sin([beta])

The scalar product can now be obtained as follows:

What is so useful about the scalar product ? If two vectors are perpendicular, their relative angle equals 90deg., and . = 0. The scalar product therefore provides an easy test to determine whether two vectors are perpendicular.

Suppose and are defined as follows:

These two vectors are perpendicular since:

In a similar manner we can easily determine whether the angle between the two vector is less than 90deg. (in which case . > 0) or more than 90deg. (in which case . < 0)

3.4. Multiplying Vectors - Vector Product

The vector product of two vectors and , written as x , is a third vector with the following properties:

the magnitude of is given by:

where [phi] is the smallest angle between and . Note: the angle between and is [phi] or 360deg. - [phi]. However, since sin([phi]) = - sin(2[pi] - [phi]), the vector product is different for these two angles.

The direction of is perpendicular to the plane defined by and , and the direction (up or down) is determined using the right-hand rule.

It is clear from the definition of the vector product that the order of the components is important. It can be shown, by applying the right hand rule, that the following relation holds:

x = - x

The following expression can be used to calculate x if the components of and are provided:

4. MOTION IN A PLANE

4.1. Position

In Chapter 2 we discussed the motion of an object in one dimension. Its position was unambiguously defined by its distance (positive or negative) from a user defined origin. The motion of this object could be described in terms of scalars. The discussion about motion in two or three dimensions is more complicated. To answer the question "where am I ?" in two dimensions, one needs to specify two coordinates. In three dimensions one needs to specify three coordinates. To specify the position of an object the concept of the position vector needs to be introduced. The position vector is defined as a vector that starts at the (user defined) origin and ends at the current position of the object (see Figure 4.1). In general, the position vector will be time dependent (t). Using the techniques developed in Chapter 3, we can write the position vector in terms of its components:

Figure 4.1. Definition of Position Vector.

Note: In Chapter 2 we got used to plotting the position of the object, its velocity and its acceleration as function of time. In two or three dimensions, this is much more difficult, and most graphs will show for example the trajectory of the object (without providing direct information concerning the time).

4.2. Velocity

The velocity of an object in two or three dimensions is defined analogously to its definition in Chapter 2:

This equation shows that the velocity of an object in two or three dimensions is also a vector. Again, the velocity vector can be decomposed into its three components:

The components of (t) can be calculated from the corresponding components of the position vector (t):

We conclude:

4.3. Acceleration

The acceleration of an object in three dimensions is defined analogously to its definition in Chapter 2:

This equation shows that the acceleration of an object in two or three dimensions is also a vector, which can be decomposed into three components:

The components of (t) can be calculated from the corresponding components of the position vector (t) and the velocity vector (t):

We conclude:

Assuming a constant acceleration in the x, y and z direction, we can write down the following equations of motion:

vx(t) = vx0 + ax t

vy(t) = vy0 + ay t

vz(t) = vz0 + az t

The magnitude of the velocity as function of time can be calculated:

If we look at a very small time interval, the change in the velocity vector will be small. In this limit:

The magnitude of the velocity for small time intervals can be written as:

From this formula we can conclude:

Sample Problem 4-2 - 4-4

A rabbit runs across a parking lot on which a set of coordinate axes has, strangely enough, been drawn. The path is such that the components of the rabbit's position with respect to the origin of the coordinate frame are given as function of time by

The units of the numerical coefficients in these equations are such that, if you substitute t in seconds, x and y are in meters.

The position vector of the rabbit at time t can be expressed as:

From the equations of motion for x(t) and y(t) we can calculate the velocity and acceleration:

The acceleration of the rabbit is constant (independent of time).

The scalar product of the velocity and the acceleration will tell us something about the change of velocity of the rabbit:

From this equation we conclude that for t < 14.6 s (negative scalar product) the rabbit will slow down, while for t > 14.6 s (positive scalar product) the rabbits

speed will increase. To check this prediction, we calculate the magnitude of the velocity of the rabbit:

The velocity of the rabbit has a minimum at t = 14.6 s.

4.4. Projectile Motion

We will start considering the motion of a projectile in 2 dimensions. The coordinate system that will be used to describe the motion of the projectile consist of an x-axis (horizontal direction) and a y-axis (vertical direction). Assuming that we are dealing with constant acceleration, we can obtain the velocity and position of the projectile using the procedure outlined in Chapter 2:

where x0 and y0 are the x and y position of the object at t = 0 s, and vx0 and vy0 are the x and y components of the velocity of the object at time t = 0. Note that axonly affects vx and not vy, and ay only affects vy.

In describing the motion of the projectile, we will assume that there is no acceleration in the x-direction, while the acceleration in the y-direction is equal to the free-fall acceleration:

ax = 0

ay = - g = - 9.8 m/s2

In this case, the equations of motion for the projectile are:

The trajectory of the projectile is completely determined by the equations of motion x(t) and y(t). The coordinate system in which we will analyze the trajectory of the projectile is chosen such that x0 = y0 = 0. In this case:

The time t can be eliminated from these two equations:

Substituting this expression for t into the equation of motion for y, the following relation between x and y can be obtained:

We can conclude that the trajectory of the projectile is described by a parabola.

Note: Often, the total velocity v0 of the object at time t = 0 s and the angle [theta] between the direction of the projectile and the positive x-axis is provided. From this information the components of the velocity at time t = 0 s can be calculated:

Figure 4.2. Projectile Motion.

Example:

Suppose a projectile is launched with an initial velocity v0 and an angle [theta] with respect to the x-axis (see Figure 4.2). What is its range R ?

We start with defining the coordinate system to be used in this problem:

x(t = 0) = x0 = 0

y(t = 0) = y0 = 0

The position of the projectile at any given time t can be obtained from the following expressions:

At impact, y(t) = 0. The time of impact can therefore be obtained by requiring that y(t) = 0 and solving for t:

This equation has two solutions:

The first solution corresponds to the time that the projectile was launched, while the second solution gives us the time that the projectile hits the ground again. The x-coordinate at that time can be obtained by substituting the expression for t into the expression for x(t):

The maximum range is obtained when sin(2[theta]) = 1, which corresponds to [theta] = 45deg.. The velocity of the projectile on impact can be calculated using the equations for vx(t) and vy(t):

Comparing the velocity on impact with the velocity at t = 0, we observe that the velocity component parallel to the x-axis is unchanged, while the component along the y-axis changed sign.

If we look at the equation of the range R, we observe that the for each value of R (less than Rmax) there are two possible launch angles: 45deg. + [Delta][theta] and 45deg. - [Delta][theta] (sin(2[theta]) is symmetric around [theta] = 45deg.). The time of flight for the two cases are however different: a larger launch angle corresponds to a longer time of flight (time of flight is proportional to sin([theta])).

Note: in all our calculations we have neglected air resistance.

Sample Problem 4-7

A movie stunt man is to run across a rooftop and then horizontally off it, to land on the roof of the next building (see Figure 4-16 in Halliday, Resnick and Walker). Before he attempts the jump, he wisely asks you to determine whether it is possible. Can he make the jump if his maximum rooftop speed is 4.5 m/s ?

The coordinate system is chosen such that the origin is defined as the position of the stunt man at the moment he starts his jump from the roof (this is also defined as time t = 0). In this case, the following initial conditions apply:

x0 = 0 m

y0 = 0 m

vx0 = + 4.5 m/s

vy0 = 0 m/s

ax = 0 m/s2

ay = - g

The equations of motion describing the trajectory of the stunt man can now be written as:

The time at which the stunt man has dropped 4.8 m can be calculated from the equation for y(t):

The horizontal distance traveled by the stunt man during this time interval can be calculated from the equation of motion for x(t):

However, to reach the next building, the stunt man should have moved 6.2 m horizontally. Unless the stunt man wants to commit suicide, he should not jump.

Example

An antitank gun is located on the edge of a plateau that is 60 m above the surrounding plain. The gun crew sights an enemy tank stationary on the plain at a horizontal distance of 2.2 km from the gun. At the same moment, the tank crew sees the gun and starts to move directly away from it with an acceleration of 0.90 m/s2. If the antitank gun fires a shell with a muzzle speed of 240 m/s at an elevation of 10deg. above the horizontal, how long should the gun crew wait before firing if they are to hit the tank.

Our starting point are the equations of motion of the shell.

(1)

(2)

Our coordinate system is defined such that the shell is launched at t = 0 s and its location at that instant is specified by x = 0 m and y = h. Therefore, x0 = 0 m and y0= h. In order to determine the trajectory of the shell we first determine its time of flight between launch and impact. This time t1 can be obtained from eq.(2) by requiring that on impact y(t1) = 0 m. Thus

(3)

The solutions for t1 are

(4)

Since the shell will hit the ground after being fired (at t = 0) we only need to consider the positive solution for t1. The range R of the shell can be obtained by substituting t1 into eq.(1):

(5)

The problem provides the following information concerning the firing of the projectile:

h = 60 m

v0 = 240 m/s

[theta] = 10deg.

Using this information we can calculate v0x and v0y:

v0x = v0 cos([theta]) = 236 m/s

v0y = v0 sin([theta]) = 42 m/s

Substituting these values in eq.(4) and eq.(5) we obtain:

t1 = 9.8 s

R = 2320 m

The distance between the impact point and the original position of the tank is 120 m. The tank starts from rest (v = 0 m/s) and has an acceleration of 0.9 m/s2. The time t2 it takes for the tank to travel to the impact point can be found by solving the following equation:

(6)

This shows that

(7)

If the shell is fired at the same time that the tank starts to move, the tank will not reach the impact point until (16.3 - 9.8) s = 6.5 s after the shell has landed. This means that the gun crew has to wait 6.5 s before firing the antitank gun if they are to hit the tank.

4.5. Circular Motion

Suppose the motion of an object as function of time can be described by the following relations:

(8)

The equations in (8) describe a periodic motion: the position of the object at time t and at time t + T are identical. The period of the periodic motion is obviously T. The path of this object will be circular. This can be easily verified by calculating the distance of this object to the origin:

(9)

Clearly, the distance of the object to the origin is constant (independent of time). See Figure 4.3. The velocity of the object can be easily calculated using the relations in eq.(8):

(10a)

(10b)

The direction and magnitude of the velocity vector are:

Figure 4.3. Definition of trajectory.

(11)

(12)

Figure 4.4. Direction of velocity and acceleration.

Equation (11) shows that the direction of the velocity is perpendicular to the position vector. In a very similar manner the acceleration of the object can be calculated:

(13a)

(13b)

The direction and magnitude of the acceleration vector are:

(14)

(15)

Equation (14) shows that the direction of the acceleration is in the radial direction (opposite to the position vector). See also Figure 4.4.

Note: Equation (12) shows that the magnitude of the velocity of the object is constant (independent of time). Equation (11) shows that only its direction changes with time.

Note: Equation (15) shows that the acceleration is non-zero. It is clear that this non-zero acceleration does not change the magnitude of the velocity, but it

changes its direction. The acceleration will be non-zero if either the magnitude or the direction of the velocity changes.

4.6. Relative Motion in 1D

The description of the motion of an object depends on the motion of the observer. A man standing along a highway observes cars moving with a velocity of + 55 miles/hour. If our observer would be traveling in one of these cars with a velocity of + 55 miles/hour he would see the other cars moving with a velocity of 0 miles/hour. This demonstrates that the observed velocity of an object depends on the motion of the observer. We will start our discussion with relative motion in one dimension.

Figure 4.5. Relative Motion in One Dimension.

Suppose a moving car is observed by an observer located at the origin of frame A, and by an observer located at the origin of frame B (see Figure 4.5). The snapshot shown in Figure 4.5 shows that at that instant, the distance between observer A and observer B equals xBA. The position of the car measured by observer A, xCA, and the position of the car measured by observer B, xCB, are related as follows:

xCA = xBA + xCB

If we differentiate this equation with respect to time, the following relation is obtained for the velocity of the car measured by observer A and by observer B:

vCA = vBA + vCB

where vCA is the velocity of the car measured by observer A, vCB is the velocity of the car measured by observer B, and vBA is the velocity of observer B measured by observer A. If observer A and observer B do not move with

respect to each other (vBA = 0 m/s), the velocity of the car measured by observer A is equal to the velocity of the car measured by observer B.

In a similar fashion, we can obtain a relation between the acceleration of the car measured by observer A and the acceleration measured by observer B:

aCA = aBA + aCB

If the observer in frame B is moving with a constant velocity with respect to the observer in frame A (vBA = constant and aBA = 0 m/s2), the acceleration measured in frame A is equal to the acceleration measured in frame B:

aCA = aCB

Sample Problem 4-12

Alex, parked by the side of an east-west road, is watching car P, which is moving in a westerly direction. Barbara, driving east at a speed vBA = 52 km/h, watches the same car. Take the easterly direction as positive.

a) If Alex measures a speed of 78 km/h for car P, what velocity will Barbara measure ?

This problem involves three cars that are located somewhere along a highway running from east to west. Our coordinate system is chosen such that a positive velocity corresponds to motion towards the east while a negative velocity corresponds to motion towards the west. Alex is sitting in a car parked along the side of the highway (frame A) and Barbara is driving east with a velocity vBA (measured by Alex) equal to + 52 km/h. Both observe a car P. The velocity of P, VPA (measured by Alex), is - 78 km/h (see Figure 4.6). The velocity of car P measured by Barbara can be calculated as follows:

vPA = vBA + vPB

vPA = - 78 km/h

vBA = 52 km/h

vPB = vPA - vBA = (- 78 - 52) km/h = - 130 km/h

The velocity of Alex measured by Barbara can be calculated in a similar manner, and the resulting velocity is - 52 km/h. The corresponding velocity diagram is shown in Figure 4.7.

Figure 4.6. Velocity Diagram Measured Relative to Alex.

Figure 4.7. Velocity Diagram Measured Relative to Barbara.

b) If ALEX sees car P brake to a halt in 10 s, what acceleration (assumed constant) will he measure for it ?

Alex observes that car P brakes to a halt in 10 s. Alex can calculate the acceleration of car P:

v(t) = v0 + a t

The initial conditions are:

v0 = v(t = 0) = - 78 km/h = - 21.7 m/s

The final conditions are:

v(t = 10 s) = 0 m/s

Assuming a constant acceleration, a can be calculated as follows:

c) What acceleration would Barbara measure for the braking car ?

Barbara also observe that car P slows down during this 10 s interval. At time t = 0 she observes car P moving with a velocity equal to v0 = - 130 km/h (- 36.1 m/s). After 10 s, she observes car P moving with a velocity equal to v(t = 10 s) = - 52 km/h (- 14.4 m/s). Barbara can also calculate the acceleration of car P:

Both Alex and Barbara measure the same acceleration of car P. This is what we expected since Barbara moves with a constant acceleration with respect to Alex.

4.7. Relative Motion in 2D

The description of the motion of an object in two or three dimensions depends on the choice of the coordinate system. Figure 4.8 shows two reference frames in two dimensions. The vectors rPA and rPB are the position vectors of object P in reference frame A and in reference frame B, respectively. Vector rBA is the position of observer B (located at the origin of reference frame B) in frame A. The position vector of the object P in reference frame B can be obtained from the position vector in reference frame A:

Figure 4.8. Reference Frames in Two Dimensions.

The velocity and acceleration of the object P in reference frame B can be obtained by differentiating the position vector of P in reference frame B with respect to time:

If the observer in reference frame B is moving with a constant velocity with respect to an observer in reference frame A, the acceleration of the object P measured by observer A will be the same as the acceleration measured by observer B.

5. FORCE AND MOTION - 1When an object all of a sudden changes its velocity and/or direction, we can always find an interaction between that object and its surroundings that is responsible for this change. We state that the surroundings exert a force on the object studied. Under the influence of a force, an object will accelerate. The force laws allow us to calculate the force acting on a body from the properties of the body and its environment. The laws of motion are subsequently used to calculate the acceleration of the object under influence of the force.

In this course we will be discussing the laws of motion obtained by Newton. This is called Newtonian mechanics. It should be realized that Newtonian mechanics does not always provide correct answers. If the speed of the objects involved is an appreciable fraction of the speed of light, we must replace Newtonian mechanics by Einstein's special theory of relativity. For problems on the scale of atomic structure we must replace Newtonian mechanics by quantum mechanics.

5.1. Newtons First Law

All around us we observe that all moving objects will come eventually to rest, unless we apply a force to them. We need to keep pedaling if we want to keep a bicycle going with constant speed, we need to have our engine running if we want to keep driving with a speed of 55 miles/hour. In all these cases, friction will ultimately stop any moving object, unless the friction force is canceled by the force supplied by our legs, our engine, etc. If we reduce friction, the moving object will take longer to slow down, and the force needed to overcome the friction force will be less. In the limit of no friction, our object will keep moving with a constant velocity, and no force need to be applied. This conclusion is summarized in Newton's first law:

" Consider a body on which no net force acts. If the body is at rest, it will remain at rest. If the body is moving with constant velocity, it will continue to do so. "

Newton's first law is really a statement about reference frames in that it defines the kinds of reference frames in which the laws of Newtonian mechanics hold. Reference frames in which Newton's first law applies are called inertial reference frames.

One way to test whether a reference frame is an inertial reference frame, is to put a test body at rest and arrange things so that no net force acts in it. If the reference frame is an inertial frame, the body will remain at rest; if the body does not remain at rest, the reference frame is not an inertial frame. If you put a bowling ball at rest on a merry-go-round, no identifiable forces act on the ball, but it does not remain at rest. Rotating reference frames are not inertial reference frames. Strictly speaking, the earth is therefore also not an inertial frame, however, only if we consider large scale motion such as wind and ocean current do we need to take into account the non inertial character of the rotating earth.

5.2. Force

If we exert the same force on several objects with different mass, we will observe different accelerations. For example, one can throw a baseball significantly further (and faster) than a ball of similar size made of lead. The unit of force is the Newton (N), and a force of 1 N is defined as the force that when applied to an object with a mass of 1 kg, produces an acceleration of 1 m/s2. If we apply a force equal to 2 N, the corresponding acceleration is 2 m/s2.

Experiments have shown that the force is a vector. This can be shown by demonstrating that the force has a magnitude and a direction. Suppose, we apply a force of 3 N to our standard object (mass 1 kg). The force is applied such that the resulting acceleration of 3 m/s2 is upwards (positive y-direction). In addition, we apply a force of 4 N in the horizontal direction (this force is applied such that the standard object will accelerate with an acceleration of 4 m/s2 in the direction of the positive x-axis if this is the only force applied). The situation is illustrated in Figure 5.1. If both forces are acting on the standard mass simultaneously, the acceleration of the object is measured to be 5 m/s2, and the direction of the acceleration coincides with the direction of the vector sum of the two forces. The total force is 5 N and is equal to the magnitude of the sum vector of the two forces (if we assume that the direction of the force is equal to the direction of the acceleration). We conclude that indeed the force is

a vector and that both the force and the corresponding acceleration have the same direction.

Figure 5.1. The Acceleration of the standard body under the influence of two forces.

The acceleration produced by a certain force depends on the mass of the object. The acceleration of an object with twice the mass of the standard mass under the influence of a certain force is half that of the acceleration of the standard mass due to the same force. The following list summarizes what we have learned so far about forces:

1. Force is a vector.

2. A force acting on an object produces an acceleration. The direction of the acceleration is the same as the direction of the force applied.

3. For a given force, the resulting acceleration of a body with a mass twice that of the standard mass, is half that of the acceleration of the standard mass under the influence of the same force.

The conclusions are summarized in Newton's second law:'

where [Sigma]F is the vector sum of all forces acting on an object with mass m, and is the resulting acceleration (note: the sum includes only external forces). If we decompose both the force and the acceleration into their individual components along the x, y and z-axis, we obtain the following relations:

Newton's second law includes a formal statement of Newton's first law: if there is no net force acting on an object ([Sigma]F = 0 N) the acceleration is zero (and the velocity of the object is constant).

Sample Problem 5-1

A student pushes a loaded sled whose mass is 240 kg for a distance of 2.3 m over the frictionless surface of a frozen lake. He exerts a horizontal force equal to 130 N. If the sled starts from rest, what is its final velocity ?

Figure 5.2. Coordinate System Sample Problem 5-1.

This is a one-dimensional problem. The coordinate system is defined such that the origin coincides with the position of the sled at time t = 0 s, and the force is applied in the positive direction (see Figure 5.2). Since the force is constant, the resulting acceleration a is also constant, and can be calculated by applying Newton's second law:

The constant acceleration is only applied over a distance d (= 2.3 m). In the coordinate system chosen, the equation of motion can be written as follows:

From this equation, the time at which the sled has covered a distance d can be calculated:

and the velocity of the sled at that time is equal to

Sample Problem 5-2

In a two dimensional tug war, Alex, Betty and Charles pull on ropes that are tied to an automobile tire. The ropes make angles as shown in Figure 5.3, which is a view from above. Alex pulls with a force FA = 220 N and Charles with a force FC = 170 N. With what force must Betty pull in order to keep the tire stationary ?

Since the tire is stationary, the net force on the tire must be zero. This also means that the net force along the x and y direction must be zero:

Substituting the known values for FA, FC and [theta] in the first equation, we can calculate [phi]:

Substituting this value for [phi] into the second equation we can calculate FB:

Figure 5.3. Force Diagram Sample Problem 5-2.

5.3. Newtons Third Law

If a hammer exerts a force on a nail, the nail exerts an equal but oppositely directed force on the hammer. This is true in general, and is described by Newton's third law:'

" Suppose a body A exerts a force (FBA) on body B. Experiments show that in that case body B exerts a force (FAB) on body A. These two forces are equal in magnitude and oppositely directed:

"

Note: Since the two members of an action-reaction pair always act on different bodies, they can not cancel each other.

5.4. Mass and Weight

The mass of a body and the weight of a body are totally different properties. The mass m of a body is a scalar; its SI unit the kilogram. The mass of a body can be determined by comparing it to the standard kilogram. Mass is an intrinsic property of a body; it is the same on the earth's surface, in an orbiting satellite, on Mars, or in interstellar space. The weight of a body is a vector; its SI unit in the Newton. The weight of a body with mass m is defined as:

where g is the free-fall acceleration at the location of the body. Since the free-fall acceleration varies from point to point, the weight of an object depends on its location, and is therefore not an intrinsic property of a body.

5.4.1. Measuring Mass

The mass of a body can be determined via comparison with the standard mass. The equal-arm balance is designed for this purpose (see Figure 5.4). The equal-arm balance is balanced if the force on the left equals the force on the right:

Figure 5.4. Equal-Arm Balance.

These two forces are the gravitational forces acting on m1 and m2 and can be calculated easily:

If the free-fall acceleration is constant at the location of the balance, we can conclude that if the arms are balanced:

m1 = m2

Therefore, the equal-arm balance determines the relative mass of two objects by comparing their weight.

5.4.2. Measuring Weight

The measurement of the weight of an object can be carried out using a spring scale (see Figure 5.5). The spring scale uses a spring to measure the weight of the object. There is a one-to-one relation between the stretch of the spring and the applied force (responsible for the stretch). In general, spring scales are calibrated and show the mass of the object. However, it should be stressed that the mass of the object is obtained from the measured weight, and in this process it is assumed that the free-fall acceleration equals 9.8 m/s2. Therefore, a spring scale will only indicate the correct mass if it is used at a location at which the free-fall acceleration is equal to that at the calibration site (note: a spring scale will incorrectly determine the mass of an object if it is used on the moon, or in an accelerating elevator).

Figure 5.5. Spring Scale.

5.5. Applications

Sample Problem 5-7

Figure 5.6 shows a block of mass m = 15 kg hanging from three cords. What are the tensions in these cords ?

The mass m experiences a gravitational force equal to mg. Since the mass is at rest, cord C must provide an opposing force equal to mg. Applying Newton's third law, we conclude that cord C exerts a force on the knot whose magnitude is equal to mg (and pointed in the direction shown in Figure 5.6). Since the system is at rest, the net force on the knot must be equal to zero:

This vector equation can be rewritten in terms of its components along the x-axis and y-axis, using the following information:

Figure 5.6. Sample Problem 5-7.

Using these expressions we can write down the equations for the x and y-components of the net force:

The first expression can be used to express TA in terms of TB:

Substituting this expression into the equation for [Sigma]Fy we obtain:

from which we can calculate TB:

Knowing TB, we can now calculate TA:

In the case of sample problem 5-6, the tensions in the cords are:

TA = 100 N

TB = 140 N

TC = 150 N

Problem

Figure 5.7 shows a block with mass m on a frictionless plane, tilted by an angle [theta]. What is the acceleration of the block ?

Figure 5.7. Mass m on an inclined plane.

In order to determine the acceleration of the block we have to determine the total force acting on the block along the inclined plane. Two forces act on the block: the gravitational force exerted by the earth on the block, and a force, called the normal force exerted by the plane on the block (see Figure 5.8). This force must be present since in its absence mass m will experience free fall (instead of sliding motion). Since the normal force is normal to the inclined plane it does not have a component along it. The component of the gravitational force along the inclined plane is given by

The acceleration produced by this force can be determined from Newton's second law

Figure 5.8. Forces acting on mass m.

Sample Problem 5-8

Figure 5.9 shows a block with mass m held by a cord on a frictionless plane, tilted by an angle [theta]. What is the tension in the cord ? What force does the plane exert on the block ?


This problem can be solved easily if the coordinate system is chosen carefully. The best choice of coordinate system is shown in Figure 5.10. Since the block is at rest, the net force on it must be zero:

Due to the choice of the coordinate system, both N and T only have components along the y-axis and x-axis, respectively:

The mass will stay at rest if all components of the net force are zero:

From these equations we can obtain N and T:

Figure 5.10. Coordinate System # 1 used in Sample Problem 5-8.

Figure 5.11. Coordinate System # 2 used in Sample Problem 5-8.

The standard choice of coordinate system with the x-axis coinciding with the horizontal direction and the y-axis coinciding with the vertical direction (see Figure 5.11) would have made the problem significantly more difficult. In this coordinate system, N and T have component along both the x and y direction:

In this case, N and T can be obtained by solving the following equations:

Of course, the solutions for N and T are identical to those derived previously, but the derivation is harder.

Sample Problem 5-10

Two blocks are connected by a cord that passes over a (weightless) pulley (see Figure 5.12). Find the tension in the cord and the (common) acceleration.

The blocks are moving with a constant acceleration. Since the cord is assumed to rigid, the acceleration of mass m has to be equal to the acceleration of mass M. However, since the pulley reverses the direction of motion, the direction of the acceleration of mass m is opposed to the direction of the acceleration of mass M. For each of the masses we can write down the following force equations:

The first equation can be used to express T in terms of a:

Substituting this expression for T into the second equation, we obtain:

Figure 5.12. Setup Sample Problem 5-10.

The acceleration a can now be calculated:

Note that a is positive when M > m, and a is negative when M < m. The acceleration is zero if m = M. This of course agrees with what our expectations. The tension in the cord can now be calculated:

Problem

A block of mass m1 on a smooth inclined plane of angle [theta] is connected by a cord over a small frictionless pulley to a second block of mass m2 hanging vertically (see Figure 5.13). The mass of the cord and the pulley can be neglected.

a) What is the acceleration of each block ?

b) What is the tension in the cord ?

Figure 5.13. Inclined plane and pulley.

In order to determine the acceleration and the tension, we have to identify all forces acting on both masses. The following forces act on m1 (see Figure 5.14):

* The gravitational force W1 = m1 g. This force is pointing downwards in the vertical direction.

* The normal force N. This force is exerted by the surface of the inclined plane on the mass and is pointing in a direction perpendicular to the inclined it.

* Tension T. The cord exerts this force on the mass. Its direction is parallel to the inclined plane.

In general, the net force acting on m1 will be non-zero and m1 will have a non-zero acceleration. The acceleration will be along the x-axis (see Figure 5.14) and is defined to be positive if the acceleration is in the same direction as the tension T. The components of the net force acting on m1 are given by

(1)

(2)

Figure 5.14. Forces acting on m1.


The following forces act on m2 (see Figure 5.15):

* The gravitational force W2 = m2 g. This force is pointing downwards along the vertical.

* The tension T. The cord exerts this force on the mass. This force is pointing upwards along the vertical. The tension in the cord is the same at each point, and the magnitude of this force is therefore equal to the one acting on m1 although it points in a different direction.

The net force on m2 will be non-zero and the mass will accelerate. Since m1 and m2 are connected via a cord, they will have the same acceleration. If the direction of the acceleration of m1 is along T, the direction of the acceleration on m2 will be along W2 (see Figure 5.15). None of the forces acting on m2 has a component along the x-axis and we will therefore only consider the net force along the y-axis:

(3)

Equations (1) and (3) are two equations with two unknown (T and a), and can be solved. Equation (3) can be rewritten as

(4)

Substituting equation (4) for T in equation (1), we can determine a:

(5)

Substituting equation (5) into equation (4) we obtain the tension T:

(6)

6. FORCE AND MOTION - II

Figure 6.1. Static Friction.

6.1. Static and Kinetic Friction

Suppose that a horizontal force F is applied to a block resting on a rough surface (see Figure 6.1). As long as the applied force F is less than a certain maximum force (Fmax), the block will not move. This means that the net force on the block in the horizontal direction is zero. Therefore, besides the applied force F, there must be a second force f acting on the block. The force f must have a strength equal to F, and it must be pointing in the opposite direction. This force f is called the friction force, and because the block does not move, we are dealing with static friction. Experiments have shown that the force of static friction is largely independent of the area of contact and proportional to the normal force N acting between the block and the surface. The static friction force is

f <= us N

where us is the coefficient of static friction (which is dimensionless). The coefficient of static friction is approximately constant (independent of N). The maximum force that can be applied without moving the block is

Fmax = us N

Once the block has been set in motion, the force F needed to keep it in motion with a constant velocity is usually less than the critical force needed to get the motion started. In this situation we are dealing with kinetic friction and the friction force fk is given by

fk = uk N

where uk is the coefficient of kinetic friction. The kinetic friction force is independent of the applied force, but always points in the opposite direction. The equation for fk is not a vector equation since fk and N do not point in the same direction.

Note: The friction between car tires and the road is static friction. This friction is crucial when you try to stop a car. Since the maximum static friction force is larger than the kinetic friction force, a car can be stopped fastest if we prevent the wheels from locking up. When the wheels lock up, the friction force is changed to kinetic friction (the tires and the ground are moving with respect to each other) thereby reducing the acceleration and increasing the time and length required to bring the car to a halt.

Sample Problem 6-1

Figure 6.2 shows a mass m on an inclined slope. At a certain angle [theta] the mass begins to slide down the slope. Calculate the coefficient of static friction.

Figure 6.2. Coordinate System used in Sample Problem 1.

Figure 2 shows the coordinate system used in this problem. Note that with this choice of coordinate system, the normal force N and the friction force f have each only one component; N is directed along the y-axis and f is directed along the x-axis. Since this is the maximum angle at which the object will remain at rest, the friction force has reached it maximum value:

Since the object is at rest, the net force on the object equals zero:

In terms of the components of the net force along the x-axis and the y-axis:

The coefficient of static friction can be easily obtained from these two equations:

Note The friction force between car tires and the road is reduced when the car travels uphill or downhill. It is harder to drive uphill or downhill when the roads are slick than it is to drive on leveled surface.

Figure 6.3. Free-Body Diagram for Sled.

Sample Problem 6-3

A woman pulls a loaded sled (mass m) along a horizontal surface at constant speed. The coefficient of kinetic friction between the runners and the snow is uk and the angle between the rope and the horizontal axis is [phi] (see Figure 6.3). What is the tension in the rope ?

Since the sled is moving with a constant velocity, the net force on the sled must be zero. Decomposing the net force into its components along the x-axis and the y-axis, we obtain the following equations of force:

The second equation can be used to eliminate N:

Substituting this expression in the first equation we obtain:

The tension T can now be calculated:

The normal force N is always perpendicular to the surface. In the previous two sample problems, the normal force N was proportional to the weight of the object. However, this is not always true. For example, suppose I am pressing an eraser against the blackboard. I ask myself, what is the minimum force that I need to apply in order to prevent the eraser from slipping ? This situation is shown schematically in Figure 6.4. Since the eraser is at rest, the net force acting on it must be zero (and therefore, the components of the net force in both the x and the y-direction must be equal to zero):

Figure 6.4. Eraser on the Black Board.

The second equation tells us that the static friction force fs must be equal to W. This implies the following for the normal force N:

However, the normal force N is equal to the applied force F. In order to prevent the eraser from slipping, the force F will need to exceed a minimum threshold:

This relation shows that if the mass of the eraser is increased, the applied force needed to prevent the eraser from slipping will increase (the minimum force is proportional to the mass). This example also illustrates a situation in which the normal force is not related to the mass of the object.

Figure 6.5. Problem 25P.

Problem 25P

In Figure 6.5, A, and B are blocks with weights of 44 N and 22N, respectively. (a) Determine the minimum weight of block C that must be placed on A to keep it from sliding if us between A and the table is 0.20. (b) Block C is suddenly lifted off A. What is the acceleration of block A, if uk between A and the table is 0.15 ?

A weight (block C) is placed on top of block A and prevents it from sliding. The acceleration of the system is therefore 0 m/s2. Consequently, the net force on each block is equal to 0 N. In order to determine the minimum weight of block C required to accomplish this, we start evaluating the net force on each block. The forces acting on block B, the weight WB and the tension T, are schematically indicated in Figure 6.6. The net force acting on block B is directed along the y-axis and has a magnitude equal to

Figure 6.6. Forces acting on block B.

Since the net force acting on block B must be zero we conclude that

T = WB

The forces acting on block A and block C are indicated in Figure 6.7. The net force acting in the y-direction is zero and thus

N = WA + WC

Since the system remains at rest, the net force acting on block A and C along the x-direction must also be zero. This means that the static friction force fs must be equal to the tension T. Experiments show that fs has a maximum value which is determined by the normal force N and the static friction coefficient us

fs <= us N = us (WA + WC)

The minimum weight of block C that will prevent the system from slipping can be found by requiring that

us (WA + WC) >= fs = T = WB

and thus

Figure 6.7. Forces acting on block A and C.

When block C is removed the static friction force is changed (since the normal force is changed). The maximum static friction force is now us WA = 8.8 N which is less than the weight of block B. Obviously block A will slip and both blocks will accelerate. At this point the friction force acting on block A is the kinetic friction force fk whose magnitude is equal to

fk = uk N = uk WA

The net force acting on block A is given by

The net force acting on block B is given by

Eliminating the tension T from these last two equations we obtain for the acceleration a


Problem 36P

Two masses, m1 = 1.65 kg and m2 = 3.30 kg, attached by a massless rod parallel to the inclined plane on which they both slide (see Figure 6.8), travel along the plane with m1 trailing m2. The angle of incline is 30deg.. The coefficient of kinetic friction between m1 and the incline is u1 = 0.226; that between m2 and the incline is u2 = 0.113. Compute (a) the tension in the rod and (b) the common acceleration of the two masses. (c) How would the answers to (a) and (b) change if m2 trailedm1 ?

The forces acting on mass m1 are schematically shown in Figure 6.9. The x and y-components of the net force acting on m1 are given by


In the coordinate system chosen, there is no acceleration along the y-axis. The normal force N1 must therefore be equal to m1 g cos([theta]). This fixes the kinetic friction force

f1k = u1k N1 = u1k m1 g cos([theta])

Mass m1 will accelerate down hill with an acceleration a. The acceleration a is related to the x-component of the net force acting on mass m1

The forces acting on mass m2 are schematically shown in Figure 6.10. The friction force f2k acting on mass m2 can be determined easily (see calculation of f1k):

f2k = u2k N2 = u2k m2 g cos([theta])

The x-component of the net force acting on mass m2 is given by

and is related to the acceleration a of mass m2


We now have two equations with two unknown (a and T). Eliminating the tension T from these two equations we obtain the following expression for a

Substituting the values of the parameters given we find that a = 3.62 m/s2. The tension T in the rod can now be determined easily

which is equal to 1.06 N. If mass m1 and mass m2 are reversed, we will still obtain the same acceleration, but the tension in the rod will be negative (which means that the rod is being compressed).

6.2. Drag Force

The friction force we have discussed so far acts when two surfaces touch. The force that tends to reduce the velocity of objects moving through air is very similar to the friction force; this force is called the drag force. The drag force D acting on an object moving through air is given by

where A is the effective cross-sectional area of the body, [rho] is the density of air and v is the speed of the object. C is a dimensionless drag coefficient that depends on the shape of the object and whose value generally lies in the range between 0.5 and 1.0. The direction of the drag force is opposite to the direction of the velocity.

Figure 6.11. Drag Force.

Because of the drag force, a falling body will eventually fall with a constant velocity, the so called terminal velocity vt. When the object is moving with its terminal velocity vt the net force on it must be zero (no change in velocity means no acceleration). This occurs when D = mg, and the terminal velocity vt has to satisfy the following relation:

and vt is calculated to be

The equation for vt shows that the terminal velocity of an object increases with a decreasing effective area.

The terminal velocity of an object is the final velocity it obtains during free fall. The object will obtain this velocity independent of whether its initial velocity is larger or smaller than the terminal velocity (see Figure 6.11).

6.3. Uniform Circular Motion

In chapter 4 we have seen that when a particle moves in a circle, it experiences an acceleration a, directed towards the center of the circle, with a magnitude equal to

where v is the velocity of the particle, and r is the radius of the circle. The acceleration a is called the centripetal acceleration. To account for the centripetal acceleration, a centripetal force must be acting on this object. This force must be directed towards the center of the circle, and can be calculated from Newton's second law:

An example of uniform circular motion is the motion of the moon around the earth. Suppose the period of this motion is T. What does this tell us about the distance r between the earth and the moon ? During one period, the moon covers a total distance equal to 2[pi]r. The velocity of the moon, vm, can be calculated:

The corresponding centripetal force is

Here we assumed that mm is the mass of the moon. The centripetal force is supplied by the gravitational attraction between the earth and the moon. In Chapter 15 we will see that the strength of the gravitational interaction can be calculated as follows:

where G is the gravitational constant and me is the mass of the earth. For a constant circular motion, the gravitational force must provide the required centripetal force:

The distance between the earth and the moon can therefore be calculated:

The constant of gravity is known to be G = 6.67 x 10-11 m3/(s kg) and the mass of the earth is known to be me = 5.98 x 1024 kg. The measured period of the moon is 27.3 days (2.3 x 106 s). The distance between the moon and the earth can therefore be calculated:

r = 3.82 x 108 m

which agrees nicely with the distance obtained using other techniques (for example the measurement of the time it takes for light to travel from the earth to the moon and back).

Figure 6.12. Forces acting on a car while rounding an unbanked curve.

6.4. Rounding a Curve

Friction is critical if we want to round a curve while driving a car or bicycle. This can be easily understood if we consider the forces that act on a car while it is making a turn. Suppose that the car in question make a turn with radius R and velocity v. Figure 6.12 shows the forces acting on the car. There is no motion in the vertical direction and the net force in this direction must therefore be equal to zero. This requires that the normal force N is equal to the weight of the car:

N = m g

When the car rounds the curve it carries out uniform circular motion. The corresponding centripetal acceleration of this motion is given by

In order for the car to carry out this circular motion there must exist a radial force with a strength equal to

This force can only be supplied by the static friction force and therefore we require that

The static friction force fs has a maximum value equal to us N and this limits the velocity and the radius of curvature of the curve that the car can take:

We conclude that the car will be able to make a turn with radius R and velocity v if the coefficient of static friction between the tires and the road is

Figure 6.13. Forces acting on a car while rounding an banked curve.

If the road is frictionless (us = 0) because of a cover of ice, the car will not be able to round any curve at all. In order to avoid problems like this, curves on highways are usually banked. The effect of banking the curves can be easily understood. Figure 6.13 shows the forces acting on an automobile when it is rounding a curve on a banked highway. We assume that there is no friction between the tires and the road. The normal force N has components both along the radial and the vertical axes. Since there is no motion along the vertical direction, the net force along the vertical axis must be zero. This requires that

and fixes the normal force N

The radial component of the normal force is given by

This component of the normal force can produce the radial acceleration required to allow the car to round the curve, even in the absence of friction. If the car has to round a curve with radius R and with velocity v, we require that

or

This last equation shows that the banking angle of a highway curve is designed for a specific velocity and radius of curvature.

Sample Problem 6-9

A conical pendulum whirls around in a horizontal circle at constant speed v at the end of a cord whose length is L. The cord makes an angle [theta] with the vertical. What is the period of the pendulum ?

The pendulum is shown schematically in figure 6.14. Since the pendulum is carrying out a uniform circular motion, the acceleration of the pendulum has to point toward the center of the circle (direction along the position vector r) and the magnitude of the acceleration equals v2/r, where v is the velocity of the pendulum and r is the radius of the circle. The net force in the horizontal plane should therefore be always directed towards the center of the circle and have strength determined by Newton's second law.

The coordinate system chosen is such that the origin coincides with the center of the circle describing the motion of the pendulum. Since the horizontal component of the force is always directed towards the center we will be using an r-axis (rather than an x-axis). The y-axis coincides with the vertical direction (see Figure 6.14). Since the y-coordinate of the bob is constant, the acceleration in y-direction must be zero. The net force in this direction must therefore be zero:

This expression allows us to calculate the tension T:

The net force in the radial direction can now be determined:


The centripetal acceleration a can now be calculated:

From the radius R of the trajectory and the centripetal acceleration a, the velocity of the object can be calculated:

The period T can be calculated from the known velocity v and radius R:

For L = 1.7 m and [theta] = 37deg. the period T = 2.3 s.

Sample Problem 6-10

A Cadillac with mass m moves at a constant speed v on a curved (unbanked) roadway whose radius of curvature is R. What is the minimum coefficient of static friction us between the tires and the roadway ?


The situation is schematically shown in Figure 6.15. Since there is no acceleration in the y-direction, the net force in this direction must be zero:

The centripetal force Fc is given by:

In this situation, the centripetal force is provided by the static friction force. If no slipping occurs, the maximum static friction force must exceed the required centripetal force:

The minimum coefficient of static friction can be obtained from this equation:

If the velocity of the car is 72 km/hr (20 m/s) and the radius of curvature R = 190 m, the minimum value of the coefficient of static friction is 0.21. Note that

the mass of the car does not enter in the calculation, and the friction coefficient is therefore the same for all objects moving with the same velocity. The minimum coefficient of static friction scales with the square of the velocity; a reduction of the velocity by a factor if two, will reduce the minimum friction coefficient by a factor of four.

Problem 58E

A stunt man drives a car over the top of a hill, the cross section of which can be approximated by a circle of radius 250 m. What is the greatest speed at which he can drive without the car leaving the road at the top of the hill ?

The car will not leave the road at the top of the hill if the net radial force acting on it can supply the required centripetal acceleration. The only radial forces acting on the car are the gravitational force and the normal force (see Figure 6.16). The net radial force Fr acting on the car is equal to

Figure 6.16. Forces acting on the car.

Fr = W - N

Since the normal force N is always directed along the positive y-axis, the radial force will never exceed the weight W of the car. This therefore also limits the centripetal force and therefore the speed of the car.

The maximum velocity of the car can now be found easily

Suppose the car is driving with a velocity less than 178 km/h. The normal force N can now be calculated and will be a function of the velocity v. If the car carries out a uniform circular motion than we know that a net radial force must be acting on it and that its magnitude is equal to mv2/R. The net radial force acting on the car is equal to W - N. We conclude that

or

Problem 60P

A small object is placed 10 cm from the center of a phonograph turntable. It is observed to remain on the table when it rotates at 33 1/3 revolutions per minute but slides off when it rotates at 45 revolutions per minute. Between what limits must the coefficient of static friction between the object and the surface of the turntable lie ?

The object is located a distance R away from the rotation axis. During one revolution the object covers a distance 2[pi]R. If one revolution is completed during a time T, the linear velocity of the object can be obtained using the following equation:

In order for the object to carry out such a uniform circular motion it must provide a radial force with magnitude equal to

The only radial force acting on the object is the static friction force. The friction force fs has a maximum value given by

If the object remains on the table, the static friction coefficient needs to satisfy the following relation:

In this problem the distance to the rotation axis is 0.1 m. The block remains on the table when the table rotates at 33 1/3 rev/min. This corresponds to 1 revolution per 1.80 s, and a linear velocity of 0.35 m/s. This implies that the coefficient of static friction must be at least 0.12. When the table rotates at 45 rev/min the block leaves the table. This implies the coefficient of static friction is less than 0.22

7. WORK AND KINETIC ENERGYIn this chapter we will introduce the concepts of work and kinetic energy. These tools will significantly simplify the manner in which certain problems can be solved.

Figure 7.1. A force F acting on a body. The resulting displacement is indicated by the vector d.

7.1. Work: constant force

Suppose a constant force F acts on a body while the object moves over a distance d. Both the force F and the displacement d are vectors who are not necessarily pointing in the same direction (see Figure 7.1). The work done by the force F on the object as it undergoes a displacement d is defined as

The work done by the force F is zero if:

* d = 0: displacement equal to zero

* [phi] = 90deg.: force perpendicular to displacement

Figure 7.2. Positive or Negative Work.

The work done by the force F can be positive or negative, depending on [phi]. For example, suppose we have an object moving with constant velocity. At time t = 0 s, a force F is applied. If F is the only force acting on the body, the object will either increase or decrease its speed depending on whether or not the velocity v and the force F are pointing in the same direction (see Figure 7.2). If (F * v) > 0, the speed of the object will increase and the work done by the force on the object is positive. If (F * v) < 0, the speed of the object will decrease and the work done by the force on the object is negative. If (F * v) = 0 we are dealing with centripetal motion and the speed of the object remains constant. Note that for the friction force (F * v) < 0 (always) and the speed of the object is always reduced !

Per definition, work is a scalar. The unit of work is the Joule (J). From the definition of the work it is clear that:

1 J = 1 N m = 1 kg m2/s2

Figure 7.3. Forces acting on the safe.

Sample Problem 7-2

A safe with mass m is pushed across a tiled floor with constant velocity for a distance d. The coefficient of friction between the bottom of the safe and the floor is uk. Identify all the forces acting on the safe and calculate the work done by each of them. What is the total work done ?

Figure 7.3 shows all the forces that act on the safe. Since the safe is moving with constant velocity, its acceleration is zero, and the net force acting on it is zero

The components of the net force along the x-axis and along the y-axis must therefore also be zero

The second equation shows that N = W = m g. The force that is applied to the safe can now be calculated

The work done on the safe by each of the four forces can now be calculated:

The total work done on the safe is therefore

which could be expected since the net force on the safe is zero.

Example Problem 1

A crate with mass m is pulled up a slope (angle of inclination is [theta]) with constant velocity. Calculate the amount of work done by the force after the crate has moved to a height h (see Figure 7.4).

Figure 7.4. Example Problem 1.

The coordinate system that will be used is shown in Figure 4. Since the crate is moving with a constant velocity, the net force in the x and y direction must be zero. The net force in the x direction is given by

and the force F required to move the crate with constant velocity is hereby fixed:

This force acts over a distance d. The value of d is fixed by the angle [theta] and the height h:

(see Figure 7.4). The work done by the force on the crate is given by

The work done on the crate by the gravitational force is given by

The work done on the crate by the normal force N is zero since N is perpendicular to d. We conclude that the total work done on the crate is given by

which was expected sine the net force on the crate is zero.

Figure 7.5. Crate moved in vertical direction.

If the same crate had been lifted by a height h in the vertical direction (see Figure 7.5), the force F required to produce a constant velocity would be equal to

F = m g

This force acts over a distance h, and the work done by this force on the object is

WF = m g h

which is equal to the work done by the force on the inclined slope. Although the work done by each force is the same, the strength of the required force is very different in each of the two cases.

Example Problem 2

A 3.57 kg block is drawn at constant speed 4.06 m along a horizontal floor by a rope exerting a 7.68 N force at angle of 15deg. above the horizontal. Compute (a) the work done by the rope on the block, and (b) the coefficient of kinetic friction between block and floor.


A total of four forces act on the mass m: the gravitational force W, the normal force N, the friction force fk and the applied force F. These four forces are shown schematically in Figure 7.6. Since the velocity of the mass is constant, its acceleration is equal to zero. The x and y-components of the net force acting on the mass are given by

Since the net force acting on the mass must be zero, the last equation can be used to determine the normal force N:

The kinetic friction force fk is given by

However, since the net component of the force along the x-axis must also be zero, the kinetic friction force fk is also related to the applies force in the following manner

Combining these last two expressions we can determine the coefficient of kinetic friction:

The work done by the rope on the mass m can be calculated rather easily:

The work done by the friction force is given by

The work done by the normal force N and the weight W is zero since the force and displacement are perpendicular. The total work done on the mass is therefore given by

This is not unexpected since the net force acting on the mass is zero.

7.2. Work: variable force

In the previous discussion we have assumed that the force acting on the object is constant (not dependent on position and/or time). However, in many cases this is not a correct assumption. By reducing the size of the displacement (for example by reducing the time interval) we can obtain an interval over which the force is almost constant. The work done over this small interval (dW) can be calculated

The total work done by the force F is the sum of all dW

Example: The Spring

An example of a varying force is the force exerted by a spring that is stretched or compressed. Suppose we define our coordinate system such that its origin coincides with the end point of a spring in its relaxed state (see Figure 7.7). The spring is stretched if x > 0 and compressed if x < 0. The force exerted by the spring will attempt to return the spring to its relaxed state:

if x < 0: F > 0

if x > 0: F < 0

It is found experimentally that for many springs the force is proportional to x:

F = - k x

Figure 7.7. Relaxed, Stretched and Compressed Springs.

where k is the spring constant (which is positive and independent of x). The SI units for the spring constant is N/m. The larger the spring constant, the stiffer the spring. The work done by the spring on an object attached to its end can be calculated if we know the initial position xi and final position xf of the object:

If the spring is initially in its relaxed state (xi = 0) we find that the work done by the spring is

Figure 7.8. Pendulum in x-y plane

7.3. Work in 2D

Consider the pendulum shown in Figure 7.8. The pendulum is moved from position 1 to position 2 by a constant force F, pointing in the horizontal direction (see Figure 7.8). The mass of the pendulum is m. What is the work done by the sum of the applied force and the gravitational force to move the pendulum from position 1 to position 2 ?

Method 1 - Difficult

The vector sum of the applied force and the gravitational force is shown in Figure 7.9. The angle between the applied force F and the vector sum Ft is a. Figure 7.9 shows that the following equations relate F to Ft and Fg to Ft:

Figure 7.9. Vector sum Ft of Fg and F.

In order to calculate the work done by the total force on the pendulum, we need to know the angle between the total force and the direction of motion. Figure 7.10 shows that if the angle between the pendulum and the y-axis is [theta] , the angle between the total force and the direction of motion is [theta] + a. The distance dr is a function of d[theta]:

For a very small distance dr, the angle between dr and Ft will not change. The work done by Ft on the pendulum is given by

The total work done by Ft can be obtained by integrating the equation for dW over all angles between [theta] = 0deg. and [theta] = [theta]max. The maximum angle can be easily expressed in terms of r and h:

Figure 7.10. Angle between sum force and direction.

The total work done is

Using one of the trigonometric identities (Appendix, page A15) we can rewrite this expression as

Using the equations shown above for Ft cos(a), Ft sin(a), r cos([theta]max) and r sin([theta]max) we can rewrite this expression and obtain for W:

Method 2 - Easy

The total work done on the pendulum by the applied force F and the gravitational force Fg could have been obtained much easier if the following relation had been used:

The total work W is the sum of the work done by the applied force F and the work done by the gravitational force Fg. These two quantities can be calculated easily:

And the total work is

which is identical to the result obtained using method 1.

7.4. Kinetic Energy

The observation that an object is moving with a certain velocity indicates that at some time in the past work must have been done on it. Suppose our object

has mass m and is moving with velocity v. Its current velocity is the result of a force F. For a given force F we can obtain the acceleration of our object:

Assuming that the object was at rest at time t = 0 we can obtain the velocity at any later time:

Therefore the time at which the mass reaches a velocity v can be calculated:

If at that time the force is turned off, the mass will keep moving with a constant velocity equal to v. In order to calculate the work done by the force F on the mass, we need to know the total distance over which this force acted. This distance d can be found easily from the equations of motion:

The work done by the force F on the mass is given by

The work is independent of the strength of the force F and depends only on the mass of the object and its velocity. Since this work is related to the motion of the object, it is called its kinetic energy K:

If the kinetic energy of a particle changes from some initial value Ki to some final value Kf the amount of work done on the particle is given by

W = Kf - Ki

This indicates that the change in the kinetic energy of a particle is equal to the total work done on that particle by all the forces that act on it.

Alternative Derivation

Consider a particle with mass m moving along the x-axis and acted on by a net force F(x) that points along the x-axis. The work done by the force F on the mass m as the particle moves from its initial position xi to its final position xf is

From the definition of a we can conclude

Substituting this expression into the integral we obtain

Example Problem 3

An object with mass m is at rest at time t = 0. It falls under the influence of gravity through a distance h (see Figure 7.11). What is its velocity at that point ?

Since the object is initially at rest, its initial kinetic energy is zero:

Ki = 0 J

The force acting on the object is the gravitational force

Fg = m g

Figure 7.11. Falling Object.

The work done by the gravitational force on the object is simply

W = Fg h = m g h

The kinetic energy of the object after falling a distance h can be calculated:

W = m . g . h = Kf - Ki = Kf

and its velocity at that point is

Figure 7.12. Projectile motion.

Example Problem 4

A baseball is thrown up in the air with an initial velocity v0 (see Figure 7.12). What is the highest point it reaches ?

The initial kinetic energy of the baseball is

At its highest point the velocity of the baseball is zero, and therefore its kinetic energy is equal to zero. The work done on the baseball by the gravitational force can be obtained:

W = Kf - Ki = - Ki

In this case the direction of the displacement of the ball is opposite to the direction of the gravitational force. Suppose the baseball reaches a height h. At that point the work done on the baseball is

W = - m g h

The maximum height h can now be calculated:

7.5. Power

In every day life, the amount of work an apparatus can do is not always important. In general it is more important to know the time within which a certain amount of work can be done. For example: the explosive effect of dynamite is based on its capability to release large amounts of energy in a very short time. The same amount of work could have been done using a small space heater (and having it run for a long time) but the space heater would cause no explosion. The quantity of interest is power. The power tells us something about the rate of doing work. If an amount of work W is carried out in a time interval [Delta]t, the average power for this time-interval is

The instantaneous power can be written as

The SI unit of power is J/s or W (Watt). For example, our usage of electricity is always expressed in units of kilowatt . hour. This is equivalent to

7.3.1. kW . h = (103 W) (3600 s) = 3.6 x 106 J = 3.6 MJ

We can also express the power delivered to a body in terms of the force that acts on the body and its velocity. Thus for a particle moving in one dimension we obtain

In the more general case of motion in 3 dimensions the power P can be expressed as

8. THE CONSERVATION OF ENERGY8.1. Conservation laws

In this chapter we will discuss conservation of energy. The conservation laws in physics can be expressed in very simple form:

" Consider a system of particles, completely isolated from outside influence. As the particles move about and interact with each other, there are certain properties of the system that do not change "

In short we can express this as

X = constant

in which X is the conserved property.

8.2. Conservation of mechanical energy

A mass hanging from the ceiling will have a kinetic energy equal to zero. If the cord breaks, the mass will rapidly increase its kinetic energy. This kinetic energy was somehow stored in the mass when it was hanging from the ceiling: the energy was hidden, but has the potential to reappear as kinetic energy. The stored energy is called potential energy. Conservation of energy tells us that the total energy of the system is conserved, and in this case, the sum of kinetic and potential energy must be constant. This means that every change in the kinetic energy of a system must be accompanied by an equal but opposite change in the potential energy:

[Delta]U + [Delta]K = 0

and

E = U + K = constant

The work-energy theorem discussed in Chapter 7 relates the amount of work W to the change in the kinetic energy of the system

W = [Delta]K

The change in the potential energy of the system can now be related to the amount of work done on the system

[Delta]U = - [Delta]K = - W

which will be the definition of the potential energy. The unit of potential energy is the Joule (J).

The potential energy U can be obtained from the applied force F

and

where U(x0) is the potential energy of the system at its chosen reference configuration. It turns out that only changes in the potential energy are important, andwe are free to assign the arbitrary value of zero to the potential energy of the system when it is in its reference configuration.

Sometimes, the potential energy function U(x) is known. The force responsible for this potential can then be obtained

We will continue with discussing some example of conservation of energy.

8.2.1. The spring force

The force exerted by a spring on a mass m can be calculated using Hooke's law

F(x) = - k x

where k is the spring constant, and x is the amount by which the spring is stretched (x > 0) or compressed (x < 0). When a moving object runs into a relaxed spring it will slow down, come to rest momentarily, before accelerating in a direction opposite to its original direction (see Figure 8.1). While the object is slowing down, it will compress the spring. As the spring is compressed, the kinetic energy of the block is gradually transferred to the spring where it is stored as potential energy. The potential energy of the spring in its relaxed position is defined to be zero. The potential energy of the spring in any other state can be obtained from Hooke's law

Suppose the total energy of the ball-spring system is E. Conservation of energy tells us

Note that the amount of work done by the spring on the block after it returns to its original position is zero.

Figure 8.1. Conversion of kinetic energy into potential energy and vice-versa.

Sample Problem 8-4

A spring of a spring gun is compressed a distance d from its relaxed state. A ball if mass m is put in the barrel. With what speed will the ball leave the barrel once the gun is fired ?

Suppose Ei is the mechanical energy of the system when the spring is compressed. Since the system is initially at rest, the total energy is just the potential energy of the compressed spring:

At the moment that the ball leaves the barrel, the spring is in its relaxed position, and its potential energy is zero. The total energy at that point is therefore just the kinetic energy of the moving mass:

Conservation of energy requires that Ei = Ef. This means

We can now calculate the velocity of the ball

Example Problem 1

Suppose the ball in Figure 8.1 has an initial velocity v0 and a mass m. If the spring constant is k, what is the maximum compression of the spring ?

In the initial situation, the spring is in its relaxed position (U = 0). The total energy of the ball-spring system is given by

The maximum compression of the spring will occur when the ball is at rest. At this point the kinetic energy of the system is zero (K = 0) and the total energy of the system is given by

Conservation of energy tells us that Ei = Ef, and thus

and

8.2.2. Gravitational force

A ball moving upwards in the gravitational field of the earth will lose its kinetic energy and come momentarily to rest at its highest point. The ball than reverses its direction, steadily regaining its kinetic energy that was lost on the way up. When the ball arrives at its starting point it will have a kinetic energy equal to its initial kinetic energy. The work done by the gravitational force on the ball is negative during the upwards motion while it is positive on the way down. The work done when the ball returns to its original position is zero.

The potential energy due to the gravitational force can be calculated

where the potential energy at y = 0 is defined to be zero. Conservation of energy for the earth-ball system now shows

This equation holds also for a ball moving in two or three dimensions. Since Fg is perpendicular to the horizontal direction, the work done by this force on the ball is zero for a displacement in the x and/or the z-direction. In the calculation of the change in the gravitational potential energy of an object, only the displacement in the vertical direction needs to be considered.

Sample Problem 8-3

A child with mass m is released from rest at the top of a curved water slide, a height h above the level of a pool. What is the velocity of the child when she is projected into the pool ? Assume that the slide is frictionless.

The initial energy consist only out of potential energy (since child is at rest the kinetic energy is zero)

Ei = m g h

where we have taken the potential energy at pool level to be zero. At the bottom of the slide, the potential energy is zero, and the final energy consist only out of kinetic energy

Conservation of energy requires that

Ei = Ef

Thus

or

8.2.3. Friction force

A block of mass m projected onto a rough surface will be brought to rest by the kinetic friction force. There is no way to get back the original kinetic energy of the block after the friction force has brought it to rest. The directed long-scale motion of the block has been transformed into kinetic energy of the randomly directed moving atoms that make up the block and the plane. We can not associate a potential energy with the friction force.

8.3. Conservative and non-conservative forces

If a potential energy can be associated with a force, we call the force conservative. Examples of conservative forces are the spring force and the gravitational force. If a potential energy can not be associated with a force, we call that force non-conservative. An example is the friction force. Alternative tests of the conservative nature of a force are:

1. A force is conservative if the work it does on a particle that moves through a round trip is zero; otherwise, the force is non conservative. The requirement of zero work for a round trip is not met by the friction force.

2. A force is conservative if the work done by it on a particle that moves between two points is the same for all paths connecting these points; otherwise, it is non-conservative.

Test 1 and test 2 are equivalent. For example, assume that the work done for the round trip from A to B and back to A (see Figure 8.2) is zero. This means that

Figure 8.2. Particle on a round trip from A to B back to A, and from A to B via 2 different routes.

WAB,1 + WBA,2 = 0

or

WAB,1 = - WBA,2

The work done by the force on each segment reverses sign if we revert the direction

WAB,2 = - WBA,2

This relation than can be used to show that

WAB,1 = WAB,2

which is exactly what test 2 states (the work done by the force on the object depends only on the initial and final position of the object and not on the path taken).

Figure 8.3 shows two possible trajectories to get from A to B. What is the work done on the object by the gravitational force for trajectory 1 and for trajectory 2 ? The work done if the mass is moved along route 1 is equal to

The alternate route (route 2) consist out of a motion in the horizontal direction followed by one in the vertical direction. For any motion in the horizontal plane, the gravitational force is perpendicular to the displacement. The work done by the gravitational force is therefore zero. For the motion along the vertical, the gravitational force is opposed to the motion. The work done by the gravitational force is

Figure 8.3. Two possible trajectories to get from A to B.

The total work done by the gravitational force on the object when it is moved from A to B via route 2 is therefore

which is equal to W1.

8.4. Potential energy curve

A plot of the potential energy as function of the x-coordinate tells us a lot about the motion of the object (see for example Figure 8.12 in Halliday, Resnick and Walker). By differentiating U(x) we can obtain the force acting on the object

In the absence of friction the conservation of mechanical energy holds and

U(x) + K = E

Since the kinetic energy can not be negative, the particle can only be in those regions for which E - U is zero or positive. The points at which E - U = K = 0 are called the turning points. The potential energy curve (Figure 8.12 in Halliday, Resnick and Walker) shows several local maxima and minima. The force at each of these maxima and minima is zero. A point is a position of stable equilibrium if the potential energy has a minimum at that point (in this case, small displacements in either direction will result in a force that pushes the particle back towards the position of stable equilibrium). Points of unstable equilibrium appear as maxima in the potential energy curve (if the particle is displaced slightly from the position of unstable equilibrium, the forces acting on it will tend to push the particle even further away).

8.5. Non-conservative forces

If we look at a block-spring system, oscillating on a rough surface, we will see that the amplitude of the motion decreases continuously. Because of the frictional force, the mechanical energy is no longer conserved. If we look at a system on which several conservative forces act, in addition to the friction force. The total work done on the system is

which is equal to the change in the kinetic energy of the system (work-energy theorem). Each conservative force can be identified with a potential energy and

We can now rewrite the expression for the change in the kinetic energy of the system

The work done by the friction force is equal to the change in the mechanical energy of the system.

8.6. Conservation of energy

In the presence of non-conservative forces, mechanical energy is converted into internal energy Uint (or thermal energy):

[Delta]Uint = - Wf

With this definition of the internal energy, the work-energy theorem can be rewritten as

which is the law of conservation of energy. In words

" Energy may be transformed from one kind into another in an isolated system but it can not be created or destroyed; the total energy of a system always remains constant. "

Sample Problem 8-8

A ball bearing whose mass is m is fired vertically downward from a height h with an initial velocity v0 (see Figure 8.4). It buries itself in the sand at a depth d. What average upward resistive force f does the sand exert on the ball as it comes to rest ?

Figure 8.4. Sample Problem 5.

The work done by the friction force f is given by

The initial mechanical energy of the system is given by

The final mechanical energy of the system consist only out of the potential energy (Kf = 0)

Ef = Uf = m g (- d) = - m g d

The change in mechanical energy is

which must be equal to the work done on the bearing by the frictional force

The friction force f can now be calculated

Example Problem 2

A block whose mass is m is fired up an inclined plane (see Figure 8.5) with an initial velocity v0. It travels a distance d up the plane, comes momentarily to rest, and then slides back down to the bottom of the plane. What is the magnitude of the kinetic friction force that acts on the block while it is moving ? What will the velocity be when the block returns at its original position.

The work done by the friction force is equal to the change in the mechanical energy of the system. The potential energy at the origin is taken to be zero. Therefore, the initial mechanical energy of the system is just the kinetic energy of the block


The final mechanical energy (at maximum height) is just the potential energy of the block at height h:

Ef = m g h = m g d sin([theta])

The change in mechanical energy is

The work done on the block by the friction force is

Wf = - f d

and must be equal to [Delta]E. Thus

The friction force f can now be obtained

When the block returns to the origin, the friction force has again done work on the block. The total work done by the friction force on the block is now

Wf = - 2 f d

This must be equal to the change in mechanical energy of the system. When the block returns at the origin, there is no change in its potential energy. The change in the mechanical energy of the system is due to a change in the velocity of the block:

The final velocity of the block can now be calculated

9. SYSTEMS OF PARTICLES9.1. Center of mass

The motion of a rotating ax thrown between two jugglers looks rather complicated, and very different from the standard projectile motion discussed in Chapter 4. Experiments have shown that one point of the ax follows a trajectory described by the standard equations of motion of a projectile. This special point is called the center of mass of the ax.

The position of the center of mass of a system of two particles with mass m1 and m2, located at position x1 and x2, respectively, is defined as

Since we are free to define our coordinate system in whatever way is convenient, we can define the origin of our coordinate system to coincide with the left most object (see Figure 9.1). The position of the center of mass is now

Figure 9.1. Position of the center of mass in 1 dimension.

This equation shows that the center of mass lies between the two masses, closest to the heavier mass. In general, for a system with more than two particles, the position of the center of mass will satisfy the following relation

The definition of the center of mass in one dimension can be easily generalized to three dimensions

or in vector notation

For a rigid body, the summation will be replaced by an integral

Suppose we are dealing with a number of objects. Figure 9.2 shows a system consisting of 4 masses, m1, m2, m3 and m4, located at x1, x2, x3 and x4, respectively. The position of the center of mass of m1 and m2 is given by

The position of the center of mass of m3 and m4 is given by

The position of the center of mass of the whole system is given by

This can be rewritten as

Using the center of mass of m1 and m2 and of m3 and m4 we can express the center of mass of the whole system as follows

Figure 9.2. Location of 4 masses.

This shows that the center of mass of a system can be calculated from the position of the center of mass of all objects that make up the system. For example, the position of the center of mass of a system consisting out of several spheres can be calculated by assuming that the mass of each sphere is concentrated in the center of that sphere (its center of mass).

Note:

The center of mass of an object always lies on a point/line/plane of symmetry (for homogeneous objects).

The center of mass of an object does not need to lie within the body of that object (for example: the center of a donut is its center of mass even though there is no mass at that point).

Sample Problem 9-3

Figure 9.3a shows a circular metal plate of radius 2R from which a disk if radius R has been removed. Let us call it object X. Locate the center of mass of object X.

Symmetry arguments immediately tell us that the center of mass of object X is located on the x-axis. Suppose the hole in object X is filled with a disk of radius R. The new object (object C, Figure 9.3b) is symmetric around the origin of our coordinate system, and that point is therefore the center of mass of object C. However, object C consist out of object X and a disk with radius R centered on the x-axis at x = - R (this disk is called object D). The center of mass of this system (consisting out of object X and object D) can be easily calculated:

This equation can be rewritten as

For a homogeneous disk (with density [rho]) the masses of object X and D can be calculated


The position of the center of mass of object X is given by

Example Problem 9-1

Figure 9.4 shows a one-dimensional rod. The density of the rod is position dependent : [rho](x) = a - bx + cx2. Determine the location of the center-of-mass of the rod.

Figure 9.4. Position Dependent Density.

The mass of a fraction of the rod (length dx) is given by

dm = [rho](x) dx

The position of the center of mass of the rod can be determined as follows

After evaluating the integral we obtain

The total mass of the rod can be obtained easily

9.2. Motion of the Center of Mass

The definition of the center of mass of a system of particles can be rewritten as

where M is the total mass of the system. Differentiating this equation with respect to time shows

where vcm is the velocity of the center of mass and vi is the velocity of mass mi. The acceleration of the center of mass can be obtained by once again differentiating this expression with respect to time

where acm is the acceleration of the center of mass and ai is the acceleration of mass mi. Using Newton's second law we can identify mi ai with the force acting on mass mi. This shows that

This equation shows that the motion of the center of mass is only determined by the external forces. Forces exerted by one part of the system on other parts of the system are called internal forces. According to Newton's third law, the sum of all internal forces cancel out (for each interaction there are two forces acting on two parts: they are equal in magnitude but pointing in an opposite

direction and cancel if we take the vector sum of all internal forces). See Figure 9.5.

Figure 9.5. Internal and External Forces acting on a System of Particles.

The previous equations show that the center of mass of a system of particles acts like a particle of mass M, and reacts like a particle when the system is exposed to external forces. They also show that when the net external force acting on the system is zero, the velocity of the center of mass will be constant.

Example Problem 9-2

The center of mass of an exploding rocket will follow the trajectory of a projectile. although its individual pieces can follow a quit complex trajectory. The forces of the explosion are internal to the system, and the only external force acting on the system is the gravitational force.

Example Problem 9-3

A ball of mass m and radius R is placed inside a spherical shell of the same mass m and inner radius 2R (see Figure 9.6a). The ball is released and moves back and forth before coming to rest at the bottom of the shell (see Figure 9.6b). What is the displacement of the system ?

Figure 9.6. Example Problem 9-3.

The only external forces acting on the system are the gravitational force and the normal force. Both act in the y-direction. The x-component of the total external force acting on the system is zero. The x-component of the acceleration of the center of mass is therefore equal to zero. The velocity of the center of mass in

the x-direction is initially equal to zero, and will therefore remain zero. We conclude that the position of the center of mass along the x-axis will not change. In the initial configuration (Figure 9.6a) the x-position of the center of mass is given by

After the ball comes to rest, the x-coordinate of the center of mass of the system coincides with the x-coordinate of the center of the sphere (symmetry axis). Originally the center of the sphere was located at x = 0, and we conclude that the system is displaced over a distance R/2, to the left.

9.3. Linear Momentum

The linear momentum p of an object with mass m and velocity v is defined as

From this definition it is clear that the unit of momentum is (kg m/s) or (N s). Since this momentum is related to the linear motion of the object, it is called linear momentum. In Chapter 11 we will be discussing angular momentum which is the momentum related to the angular motion of the object.

Under certain circumstances the linear momentum of a system is conserved. The linear momentum of a particle is related to the net force acting on that object:

The rate of change of linear momentum of a particle is equal to the net force acting on the object, and is pointed in the direction of the force. If the net force acting on an object is zero, its linear momentum is constant (conservation of linear momentum).

The total linear momentum p of a system of particles is defined as the vector sum of the individual linear momenta

This expression can be rewritten as

where M is the total mass of the system. We conclude that

" The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass. "

If we differentiate linear momentum of the center-of-mass with respect to time we obtain

This expression shows that if the net external force acting on a system of particles is zero (Fext = 0 N), the linear momentum of the system is conserved.

Example Problem 9-4

A stream of bullets with mass m is fired horizontally with speed v into a large wooden block with mass M that is initially at rest on a horizontal table. If the block is free to slide across the table (without friction), what speed will it acquire after it has absorbed n bullets ?


Consider the closed system shown in Figure 9.7. This is an isolated system; no particles leave or enter the system. The rate of change of its linear momentum is therefore equal to the net external force. In this system, all external forces (normal and gravitational force) act in the y-direction, and we can conclude that the linear momentum in the x-direction is conserved. The system shown in Figure 9.7 consist initially out of n bullets, each moving with speed v, and the wooden block which is at rest. The total linear momentum in the x-direction is therefore

After the n bullets strike the wooden block, its mass is increased to (M + n m) and its velocity is V. The x-component of the linear momentum at that point is therefore

Since the linear momentum along the x-axis is conserved we conclude

or

The final velocity V of the wooden block will always be less than the velocity of the bullets (independent of how many bullets we fire).

Note: we did not have to consider what happened when the bullets hit the block since these forces are internal forces.

Sample Problem 9-10

Two blocks with mass m1 and mass m2 are connected by a spring and are free to slide on a frictionless horizontal surface. The blocks are pulled apart and then released from rest. What fraction of the total kinetic energy will each block have at any later time ?


Figure 9.8 shows a schematic of the system. The velocities of mass m1 and mass m2 are defined to be positive when they are directed towards the right in Figure 9.8 (in Figure 9.8 the velocity of m2 is negative).

Consider the system consisting of the two masses and the spring. This is a closed system. The only external forces acting on the system are the gravitational force and the normal force. Both these forces are directed vertically. The net force along the x-axis is zero, and therefore, linear momentum is conserved along the x-axis.

Initially, both masses are at rest, and the total linear momentum along the x-axis is zero. Suppose at a later time mass m1 has a velocity equal to v1 and mass m2 has a velocity equal to v2. The total linear momentum at that time is then given by

Since the linear momentum along the x-axis is conserved, pf must be equal to 0. The velocity v2 of mass m2 can now be expressed in terms of m1 and v1:

This shows that the velocity of mass m1 and of mass m2 always have opposite sign. The kinetic energy of mass m1 and mass m2 can now be calculated

The total kinetic energy of the system is

If f1 is the fraction of the total kinetic energy that is carried by mass m1 we obtain the following equation for f1:

Problem 43P

A vessel at rest explodes, breaking into three pieces. Two pieces, having equal mass, fly off perpendicular to one another with the same speed of 30 m/s. The third piece has three times the mass of each of the other pieces. What is the direction and magnitude of its velocity immediately after the explosion ?

The vessel is an isolated system on which no external forces are acting. This implies that the total linear momentum of the system is conserved. Since the vessel is initially at rest, the initial linear momentum of the system is zero. Since the total linear momentum is conserved, the final linear momentum of the system must also be zero. Figure 9.9 shows schematically the direction of the three fragments in which the vessel explodes. The problem states that m1 = m2 and that m3 = 3 m1. Assuming that the total mass of the system is conserved we conclude that


or

The problem also states that v1 = v2 = 30 m/s. Conservation of linear momentum along the x-axis and along the y-axis requires

These two equations can be rewritten in the following manner

These two equation can be combined to give

tan ([theta]) = 1

or

[theta] = 45deg.

The velocity of the third fragment can now be obtained easily

Problem 48P

A 1400 kg cannon, which fires a 70 kg shell with a muzzle speed of 556 m/s, is set at an elevation of 39deg. above the horizontal. The cannon is mounted on frictionless rails, so that it recoils freely. (a) What is the speed of the shell with respect to the earth ? (b) At what angle with the ground is the shell projected ?

The mass of the cannon is M, and the mass of the shell is m. The firing angle is [theta] and the muzzle speed is v0. The velocity of the cannon and the shell in with respect to the earth is vc and vs, respectively. The angle of projection of the shell with respect to the earth is a.

The external forces acting on the shell and cannon are the gravitational force and the normal force. These forces are directed along the y-axis. Since there is no external force acting on the shell along the x-axis, the linear momentum of the system along the x-axis is conserved. The total linear momentum of the system along the x-axis (the horizontal axis) is given by (see figure 9.10)

Figure 9.10. Velocity diagram of shell and canon.

This implies that

The muzzle speed provided in this problem is measured with respect to the muzzle. Since the cannon is not at rest, the speed of the shell with respect to the earth will be different than the speed of the shell with respect to the muzzle. The relation between these velocities and the firing angles is schematically shown in Figure 9.11. The figure clearly shows the following relations between the various velocities and firing angles:

Figure 9.11. Velocity diagram of shell.

The first equation can be used to eliminate vc

Using the last two equations, this expression can be rewritten as

or

We conclude that a = 40.4deg.. The velocity of the shell with respect to the earth is given by

9.4. The rocket

The motion of a rocket is a nice example of a system with a variable mass in which nevertheless conservation of linear momentum can be applied. Suppose a rocket is flying through deep space (no friction force and no gravitational force). It is burning fuel. Suppose at some time t the mass of the rocket is M. During a time interval [Delta]t, the mass of the rocket changes by [Delta]M:

M(t + [Delta]t) = M(t) + [Delta]M

Since the rocket is burning fuel, [Delta]M is negative. The mass of the exhaust products is - [Delta]M. The result of the burning of fuel is a change in the velocity of the rocket:

v(t + [Delta]t) = v(t) + [Delta]v

If we consider our system to consist of the rocket and the exhaust generated during the time interval [Delta]t, we are dealing with a closed system. Since there are no external forces acting on the system, the total linear momentum of the system is conserved. The initial linear momentum of the system (at time t) is given by

pi = M(t) v(t)

The final linear momentum of the system is given by

pf = (M(t) + [Delta]M) (v(t) + [Delta]v) + (- [Delta]M) U

where U is the velocity of the exhaust. Conservation of linear momentum therefore requires that

M(t) v(t) = (M(t) + [Delta]M) (v(t) + [Delta]v ) + (- [Delta]M) U

The exhaust velocity of the rocket depends on the design of the rocket engine. Suppose that for the engine used the velocity of the exhaust relative to the engine is measured to be U0. In the frame of reference in which the rocket is moving, the exhaust velocity is a function of both U0 and the velocity of the rocket

U - U0 = v(t) + [Delta]v

Using this expression we can rewrite the expression for conservation of linear momentum as follows

M(t) v(t) = (M(t) + [Delta]M) (v(t) + [Delta]v) + (- [Delta]M) (v(t) + [Delta]v + U0)

or

M(t) v(t) = M(t) (v(t) + [Delta]v) - [Delta]M U0

We conclude

[Delta]M U0 = M(t) [Delta]v

Dividing each side by [Delta]t gives

Now:

dM/dt = - R where R is the rate of fuel consumption. U0 = - u where u is the (positive) velocity of the exhaust gasses relative

to the rocket. dv/dt is the acceleration of the rocket.

After making these substitution we obtain the "first rocket equation"

R u = M a

The mass used in the "first rocket equation" is of course time dependent (related to R). In order to find the velocity of the rocket (after burning some fuel) we return to the differential equation previously discussed

or

Integrating both sides gives

We conclude

which is the "second rocket equation".

10. COLLISIONS10.1. Introduction

In a collision, strong mutual forces act between a few particles for a short time. These internal forces are significantly larger than any external forces during the time of the collision. The laws of conservation of linear momentum and energy, applied to the "before" and "after" situations, often allows us to predict the outcome of a collision. A great deal can be learned about the interactions between the colliding particles from the observed collision products.

Note:

external forces are small (and are ignored) during the collision particles before and after the collision can be different (for example:

nuclear reactions)

the collision force does not need to be a contact force

10.2. Impulse

Suppose a force F acts during a collision. The result of the collision force will be a change in the momentum of the particles involved. The amount of change depends not only on the average value of the force, but also on the time period during which it acts. The change in momentum dp is related to the collision force F as

The total change in the of momentum during the collision is given by

The right hand side of this equation is a measure of both the strength and the duration of the collision force. It is called the collision impulse J:

The unit of the impulse is N . s. From the definition of the impulse J we see that the relation between the impulse and the change of momentum is given by

If the average collision force Fav acts during a time period equal to [Delta]t, the impulse J is equal to

Fav [Delta]t = J

Sample Problem 10-1

A baseball of mass m in horizontal flight with speed vi is struck by a batter. After leaving the bat, the ball travels in the opposite direction with a speed vf. What impulse J acts on the ball while it was in contact with the bat ?

If the impact time is [Delta]t, what is the average force that acts on the baseball ?

What is the average acceleration of the baseball during this period ?

10.3. Collisions in One-Dimension

Consider the collision shown in Figure 10.1. If there are no external forces acting on this system (consisting of the two masses) the total momentum of the system is conserved. The first class of collisions we will discuss are the elastic collisions. Collisions are called elastic collisions if the total kinetic energy of the system is conserved. Applying conservation of linear momentum to the collision shown in Figure 10.1 gives

Conservation of the total kinetic energy gives

We now have two equations with two unknown (v1f and v2f) which can be solved. The first equation can be rewritten as

The second equation can be rewritten as

The final velocity of mass m1 can now be calculated by dividing the last two expressions

Figure 10.1. Collision in One-Dimension.

This gives

The final velocity of m1 can now be obtained

The final velocity of mass m2 can also be obtained

It is clear that the velocity of m2 is always positive. The velocity of m1 can be either positive or negative, depending on the masses of the two objects: v1f is negative if m2 > m1, and positive if m2 < m1.

In Chapter 9 we have shown that the motion of the center of mass is unaffected by the collision. The velocity of the center of masscenter of mass velocity can be calculated easily

It can be easily verified that the velocity of the center of mass after the collision is the same as it was before the collision (as it should be of course since there are no external forces acting on the system).

Some Special Cases

Equal Mass: m1 = m2.

The previously derived equations show that in this case

v1f = 0 m/s

v2f = v1i

In head-on collisions, particles of equal mass simply exchange velocities.

Massive Target: m2 >> m1.

Using the previously derived expressions for v1f and v2f we obtain

The projectile simply bounces back and the final velocity of the target will be a very small fraction of the initial velocity of the projectile.

Massive Projectile: m1 >> m2.

Using the previously derived expressions for v1f and v2f we obtain

The velocity of the projectile is almost unchanged while the target will move with twice the initial velocity of the projectile.

10.5. Motion of the Center of Mass

The collision force acting between the target and the projectile is an internal force of the system under consideration consists of these two objects. The motion of the center of mass of a number of objects is solely determined by the external forces acting on the system (see Chapter 9).

This equation shows that if no external forces act on the system, the velocity of its center of mass is constant.

Sample Problem 10-4

In a nuclear reactor, newly-produced fast neutrons must be slowed down before they can participate effectively in the chain-reaction process. By what fraction is the kinetic energy of a neutron (mass m1) reduced in a head-on collision with a nucleus of mass m2 (initially at rest) ?

Suppose v1i is the initial velocity of the neutron. Its final velocity, v1f, can be obtained using one of the previously derived equations:

The initial kinetic energy of the neutron is given by

The final kinetic energy of the neutron is given by

The fraction of the kinetic energy of the neutron lost in the collision is given by

This is equal to

Since the mass units in the equation for f cancel, we can use atomic mass units:

mass Pb: 206 amu => f = 0.019 = 1.9 %

mass C: 12 amu => f = 0.28 = 28 %

mass H: 1 amu => f = 1.00 = 100 %

We conclude that any compound that contains large amounts of H will be a good moderator.

Example Problem 10-1

A glider whose mass is m2 rests on an air track. A second glider, whose mass is m1, approaches the target glider with a velocity v1i and collides elastically with it. The target glider rebounds elastically from the end of the track and meets the projectile glider a second time (see Figure 10.2). At what distance from the end of the air track will the second collision occur ?

The velocity of m1 and m2 after the first collision are given by

Suppose the second collision occurs a distance x from the end of the track (see Figure 10.2). At that point, mass m1 has traveled a distance (d - x) after the first collision and mass m2 has traveled a distance (d + x). Both masses must cover these distances of course in the same time


or


Substituting the expressions for v1f and v2f in the expression for x, we obtain

A special case occurs when m1 = m2. In this case, x = d. Note: v1f = 0 m/s and v2f = v1i.

10.6. Collisions in One-Dimension: Inelastic

If no external forces act on a system, its momentum is conserved. However, kinetic energy is not always conserved. An example of an inelastic collision is a collision in which the particles stick together (after the collisions). This type of a collision is a completely inelastic collision (of course, even in a completely inelastic collision, the total energy is conserved. The lost kinetic energy is converted into another form of energy: for example, thermal energy, energy of deformation etc.)


Suppose a mass m1 is moving with an initial velocity vi and collides with a mass m2, which is initially at rest (see Figure 10.3). The two masses stick together. What is the final velocity of the system, and what is the change in the kinetic energy of the system ?

Figure 10.3. Completely Inelastic Collision.

The initial momentum of the system is

pi = m1 vi

The final momentum of the system is

pf = (m1 + m2) vf

Conservation of linear momentum implies

m1 vi = (m1 + m2) vf

or

The final velocity of the combined system will always be less than that of the incoming object.

The initial kinetic energy of the system is

The final kinetic energy of the system is

Note: not all the kinetic energy can be lost, even in a completely inelastic collision, since the motion of the center of mass must still be present. Only if our reference frame is chosen such that the center-of-mass velocity is zero, will the final kinetic energy in a completely inelastic collision be zero.

Sample Problem 10-5: The ballistic pendulum.

Suppose a bullet of mass m1 hits a large block of wood of mass m2. As a result, the block plus bullet swings upwards (maximum height is h). What is the velocity of the bullet ?

Suppose the velocity of the bullet is vi. The initial momentum of the system is

pi = m1 vi

The final velocity of the block plus bullet is vf. The final momentum is

pf = (m1 + m2) vf

Conservation of linear momentum implies

m1 . vi = (m1 + m2) vf

or

The initial kinetic energy of the block plus bullet is

Since it is assumed that the block plus bullet swing frictionless, the total mechanical energy of the block plus bullet must be conserved. At its highest point, the mechanical energy of the block plus bullet is equal to

Conservation of mechanical energy implies that

or

The velocity of the bullet can now be calculated

10.7. Collisions in Two-Dimensions

Suppose a mass m1, with initial velocity v1i, undergoes a collision with a mass m2 (which is initially at rest). The particles fly of at angles [theta]1 and [theta]2, as shown in Figure 10.4. Since no external forces act on the collision system, linear momentum is conserved (in both x and y direction):

and

If the collision is elastic, kinetic energy also needs to be conserved:

Figure 10.4. A Collision in Two Dimensions.

The variables are:

mass: m1 and m2

velocity: v1i, v1f and v2f

angle: [theta]1 and [theta]2

If 4 of these variables are defined, the remaining 3 can be calculated by applying conservation of energy and linear momentum.


A beam of nuclei with mass m1 and velocity v1 is incident on a target nucleus with mass m2 which us initially at rest. The velocity and scattering angles of both reaction products is measured. Determine the masses of the reaction products and the change in kinetic energy.

The collision is schematically shown in Figure 10.5. Conservation of linear momentum along the x-axis requires

where p1, p3 and p4 are the momenta of particle 1, particle 3 and particle 4, respectively. Conservation of linear momentum along the y-axis requires

The last equation can be rewritten as


Substituting this in the first equation we obtain

This immediately shows that

or

This relation can be used to find p4

and m4

The change in the kinetic energy of the system can now be calculated


A 20 kg body is moving in the direction of the positive x-axis with a speed of 200 m/s when, owing to an internal explosion, it breaks into three parts. One part, whose mass is 10 kg, moves away from the point of explosion with a speed of 100 m/s along the positive y-axis. A second fragment, with a mass of 4 kg, moves along the negative x-axis with a speed of 500 m/s.

a). What is the speed of the third (6 kg) fragment ?

b) How much energy was released in the explosion (ignore gravity) ?

The problem is schematically illustrated in Figure 10.6. Since the force of the explosion is an internal force, the total momentum of the system will be conserved. We therefore can apply this conservation law both along the x-axis and along the y-axis. Conservation of linear momentum along the x-axis requires

Conservation of linear momentum along the y-axis requires

These equations can be rewritten as

Squaring both equations and adding them gives

Solving this equation gives v3 = 1014 m/s. The energy supplied by the explosion is now easy to calculate:


11. ROTATION11.1. Rotational variables

In this chapter we will be dealing with the rotation of a rigid body about a fixed axis. Every point of the body moves in a circle, whose center lies on the axis of rotation, and every point experiences the same angular displacement during a particular time interval.

Figure 11.1. Relation between s and [theta].

Suppose the z-axis of our coordinate system coincides with the axis of rotation of the rigid body. The x-axis and the y-axis are taken to be perpendicular to the z-axis. Each part of the rigid body moves in a circle around the z-axis. Suppose a given point A on the body covers a linear distance s during the rotation (see Figure 11.1). During one complete revolution point A covers a distance equal to 2[pi]r. In that case, the angle of rotation is equal to 2[pi] radians. For the situation shown in Figure 11.1, the angle of rotation can be easily calculated:

In describing the rotation of a rigid body we have to choose a reference line with respect to which the angle of rotation is being measured. In figure 11.1 the reference line connects the origin of the coordinate system and point A. The angle of rotation is the angle between the reference line and the x-axis (as is shown in Figure 11.1).

If the angle of rotation [theta] is time dependent, it makes sense to introduce the concept of angular velocity and angular acceleration. The angular velocity [omega] is defined as

The unit of the angular velocity is rad/s. The angular velocity can be positive (counterclockwise rotation) or negative (clockwise rotation). The angular accelerationa is defined as

The unit of the angular acceleration is rad/s2.

In order to describe rotation around a point (rather than a fixed axis) the concept of an angular velocity vector is introduced. The magnitude of the angular velocity vector is equal to the absolute value of the angular velocity for rotation around a fixed axis (as defined above). The direction of the velocity vector is parallel to the rotation axis and the right-hand rule needs to be used to determine whether the vector points upwards or downwards.

Problem 7P

A wheel rotates with an angular acceleration a given by

where t is the time and a and b are constants. If the wheel has an initial angular velocity [omega]0, write the equations for (a) the angular velocity and (b) the angle turned as function of time.

To solve this problem, we start with looking at the relation between the angular acceleration and the angular velocity

This relation can be rewritten as

Substituting the given angular acceleration we obtain for the angular velocity

The angle of rotation is related to the angular velocity

Substituting the derived expression for [omega](t) the angle of rotation can be calculated

and therefore

11.2. Constant angular acceleration

If the angular acceleration a is constant (time independent) the following equations can be used to calculate [omega] and [theta] at any time t:

Note that these equations are very similar to the equations for linear motion.

Problem 19P

A wheel starting from rest, rotates with a constant angular acceleration of 2.0 rad/s2. During a certain 3.0 s interval it turns through 90 rad. (a) How long had the wheel been turning before the start of the 3.0 s interval ? (b). What was the angular velocity of the wheel at the start of the 3.0 s interval ?

Time t = 0 s is defined as the moment at which the wheel is at rest. Therefore, [omega]0 = 0 rad/s. The rotation angle at any later time is measured with respect to the position of the body at time t = 0 s: [theta]0 = 0 rad. The equations of rotation are now given by

The angle of rotation during a 3.0 s interval will depend on time:

In our problem, the rotation [Delta][theta] during a period [Delta]t is given. The time that the wheel has been turning before the time period [Delta]t can be easily calculated

The angular velocity of the wheel at the beginning of this period is

11.3. Relation between linear and angular variables

An example of the relation between angular and linear variables has already been discussed. Figure 1 illustrates how the distance s, covered by point A, is related to the radius of the circle and the angle of rotation

The velocity of point A can be obtained by differentiating this equation with respect to time

To derive this equation we have assumed that for rotations around a fixed axis the distance r from point A to the rotation axis is constant (independent of time) which is true for a rigid body. The acceleration of point A can be determined as follows

The acceleration at is the tangential component of the linear acceleration, related to the change in the magnitude of the velocity of point A. However, we have seen that an object carrying out a circular motion also experiences a radial acceleration. The magnitude of the radial component, ar, is

Using the previously derived expression for v in terms of [omega] and r, we can rewrite the radial component of the acceleration as follows

Figure 11.2 shows the direction of both the radial and the tangential components of the acceleration of point A. The radial component is always present as long as [omega] is not equal to zero; the tangential component is only present if the angular acceleration is not zero.

Figure 11.2. Components of the acceleration of point A.

We can conclude that when a rigid body is rotating around a fixed axis, every part of the body has the same angular velocity [omega] and the same angular acceleration a, but points that are located at different distances from the rotation axis have different linear velocities and different linear accelerations.

11.4. Kinetic energy of rotation

The total kinetic energy of a rotating object can be found by summing the kinetic energy of each individual particle:

To derive this equation we have used the fact that the angular velocity is the same for each particle of the rigid body. The quantity in parenthesis tells us

how the mass of the rotating body is distributed around the axis of rotation. This quantity is called the moment of inertia (or rotational inertia)

The unit for I is kg m2. Using this definition, we can write the kinetic energy of the rotating object as

Note: in many previous problems we have assumed to be dealing with massless pulleys. This assumption assures that by applying conservation of mechanical energy we do not have to consider the kinetic energy related to the rotation of the pulley.

11.5. Calculation of rotational inertia

To calculate the moment of inertia of a rigid body we have to integrate over the whole body

If the moment of inertia about an axis that passes through the center of mass is known, the moment of inertia about any other axis, parallel to it, can be found by applying the parallel-axis theorem

where Icm is the moment of inertia about an axis passing through the center of mass, M is the total mass of the body, and h is the perpendicular distance between the two parallel axes.

Sample Problem 11-8

Determine the moment of inertia of a uniform rod of mass m and length L about an axis at right angle with the rod, though its center of mass (see Figure 11.3).

The mass per unit length of the rod is m/L. The mass dm of an element of the rod with length dx is

The contribution of this mass to the total moment of inertia of the rod is

The total moment of inertia of the rod can be determined by integrating over all parts of the rod:

The moment of inertia of the rod around its end point (see Figure 11.4) can now be calculated using the parallel axes theorem


Figure 11.4. Sample Problem 11.8.

Example: Moment of Inertia of Disk

Figure 11.5. Moment of inertia of a disk.

A uniform disk has a radius R and a total mass M. The density of the disk is given by

To calculate the moment of inertia of the whole disk, we first look at a small section of the disk (see Figure 5). The area of the ring located at a distance r from the center and having a width dr is

The mass of this ring is

The contribution of this ring to the total moment of inertia of the disk is given by

The total moment of inertia can now be found by summing over all rings:

Substituting the calculated density we obtain

11.6. Torque

Suppose a force F is applied to point A (see Figure 11.6). Point A is part of a rigid body with an axis of rotation going through the origin. Suppose the angle between the force F and the position vector r is [phi]. The force F can be decomposed into two components: one parallel to the position vector and one perpendicular to the position vector. It is obvious that the component parallel to the position vector can not cause a rotation of the rigid body.

The magnitude of the component of the force perpendicular to the position vector is given by

Figure 11.6. Torque

The tangential component of the applied force F will produce a rotation of the object; the actual angular velocity will depend not only on the applied force but also on the distance between the axis of rotation and point A. To describe the effect of the force, the concept of torque is introduced.

The tangential component of F, Ft, will produce a tangential acceleration a t

The torque [tau] can be rewritten as

The tangential acceleration at is related to the angular acceleration a

We conclude that

This is just Newton's second law for rotation.

Sample Problem 11-11

Figure 11.7 shows a uniform disk with mass M and radius R. The disk is mounted on a fixed axle. A block with mass m hangs from a light cord that is wrapped around the rim of the disk. Find the acceleration of the falling block, the angular acceleration of the disk, and the tension of the cord.


The forces acting on the mass and the disk are shown in Figure 11.8. Since the mass m is moving downward, the gravitational force m . g must exceed the tension T in the cord. The linear acceleration of the mass m is defined to be positive if it points down:

The uniform disk rotates as a result of the presence of mass m. The torque exerted by the tension T of the cord on the disk is

The resulting angular acceleration a of the disk can be obtained from the torque

or


The moment of inertia of the disk is given by

The angular acceleration a is equal to

However, the linear acceleration of the cord is a, and therefore the linear acceleration of the rim of the disk must also be a. The linear acceleration of the rim and the angular acceleration a are related as follows

Combining this expression with the previous expression we can conclude that

If we combine this expression with a previously derived expression for a

we can calculate a and T:

We see that the acceleration of the falling block is always less than the gravitational acceleration, but approaches g when the mass of the disk becomes much smaller than the mass m. The angular acceleration a can be obtained from

11.7. Work

Figure 11.9. Work done by a force.

Suppose a particle with mass m is connected to the end of a rod (with negligible mass). Under the influence of a force F, the system rotates through an angle [Delta][theta] (see Figure 11.9). The work done by the force F is determined by the tangential component of F

Here r d[theta] is the length of the arc traversed by the particle. The total work done by the force during a finite rotation (from [theta]i to [theta]f) is given by

This equation is very similar to what we have derived for linear motion:

The total work done can now be calculated

or


This relation shows that the work done by the torque acting on a rigid body is equal to the change in rotational kinetic energy of that body.

12. ROLLING, TORQUE AND ANGULAR MOMENTUM12.1. Rolling Motion

Figure 12.1. Rotational Motion of Wheel

A wheel rolling over a surface has both a linear and a rotational velocity. Suppose the angular velocity of the wheel is [omega]. The corresponding linear velocity of any point on the rim of the wheel is given by

where R is the radius of the wheel (see Figure 12.1). When the wheel is in contact with the ground, its bottom part is at rest with respect to the ground. This implies that besides a rotational motion the wheel experiences a linear motion with a velocity equal to + vcm (see Figure 12.2). We conclude that the top of the wheel moves twice as fast as the center and the bottom of the wheel does not move at all.

Figure 12.2. Motion of wheel is sum of rotational and translational motion.

An alternative way of looking at the motion of a wheel is by regarding it as a pure rotation (with the same angular velocity [omega]) about an instantaneous stationary axis through the bottom of the wheel (point P, Figure 12.3).

Figure 12.3. Motion of wheel around axis through P.

12.2. Kinetic Energy

The kinetic energy of the wheel shown in Figure 12.3 can be calculated easily using the formulas derived in Chapter 11

where IP is the rotational inertia around the axis through P, and [omega] is the rotational velocity of the wheel. The rotational inertia around an axis through P, IP, is related to the rotational inertia around an axis through the center of mass, Icm

The kinetic energy of the wheel can now be rewritten as

where the first term is the kinetic energy associated with the rotation of the wheel about an axis through its center of mass and the second term is associated with the translational motion of the wheel.


Figure 12.4 shows a disk with mass M and rotational inertia I on an inclined plane. The mass is released from a height h. What is its final velocity at the bottom of the plane ?

The disk is released from rest. Its total mechanical energy at that point is equal to its potential energy

When the disk reaches the bottom of the plane, all of its potential energy is converted into kinetic energy. The kinetic energy of the disk will consist out of rotational and translational kinetic energy:

The moment of inertia of the disk is given by

where R is the radius of the disk. The kinetic energy of the disk can now be rewritten as

Figure 12.4. Mass on inclined plane.

Conservation of mechanical energy implies that Ei = Ef, or

This shows that the velocity of the disk is given by

Consider now two different disks with identical mass M but different moments of inertia. In this case the final kinetic energy can be written as

Conservation of energy now requires that

or

We conclude that in this case, the disk with the smallest moment of inertia has the largest final velocity.


Problem 15P

A small solid marble of mass m and radius r rolls without slipping along a loop-the-loop track shown in Figure 12.5, having been released from rest somewhere along the straight section of the track. From what minimum height above the bottom of the track must the marble be released in order not to leave the track at the top of the loop.

The marble will not leave the track at the top of the loop if the centripetal force exceeds the gravitational force at that point:

or

The kinetic energy of the marble at the top consists out of rotational and translational energy

where we assumed that the marble is rolling over the track (no slipping). The moment of inertia of the marble is given by

Using this expression we obtain for the kinetic energy

The marble will reach the top if

The total mechanical energy of the marble at the top of the loop-the-loop is equal to

The initial energy of the marble is just its potential energy at a height h

Conservation of energy now implies that

or

Example Problem 12-2: the yo-yo

Figure 12.6. The yo-yo.

Figure 12.6 shows a schematic drawing of a yo-yo. What is its linear acceleration ?

There are two forces acting on the yo-yo: an upward force equal to the tension in the cord, and the gravitational force. The acceleration of the system depends on these two forces:

The rotational motion of the yo-yo is determined by the torque exerted by the tension T (the torque due to the gravitational force is zero)

The rotational acceleration a is related to the linear acceleration a:

We can now write down the following equations for the tension T

The linear acceleration a can now be calculated

Thus, the yo-yo rolls down the string with a constant acceleration. The acceleration can be made smaller by increasing the rotational inertia and by decreasing the radius of the axle.

12.3. Torque

Figure 12.7. Motion of particle P in the x-y plane.

A particle with mass m moves in the x-y plane (see Figure 12.7). A single force F acts on the particle and the angle between the force and the position vector is [phi]. Per definition, the torque exerted by this force on the mass, with respect to the origin of our coordinate system, is given by

and

where r[invtee] is called the arm of the force F with respect to the origin. According to the definition of the vector product, the vector [tau] lies parallel to the z-axis, and its direction (either up or down) can be determined using the right-hand rule. Torque defined in this way has meaning only with respect to a specified origin. The direction of the torque is always at right angles to the plane formed by the vectors r and F. The torque is zero if r = 0 m, F = 0 N or r is parallel or anti-parallel to F.

12.4. Angular Momentum

The angular momentum L of particle P in Figure 12.7, with respect to the origin, is defined as

This definition implies that if the particle is moving directly away from the origin, or directly towards it, the angular momentum associated with this

motion is zero. A particle will have a different angular momentum if the origin is chosen at a different location. A particle moving in a circle will have an angular momentum (with respect to the center of the circle) equal to

Again we notice the similarity between the definition of linear momentum and the definition of angular momentum.

A particle can have angular momentum even if it does not move in a circle. For example, Figure 12.8 shows the location and the direction of the momentum of particle P. The angular momentum of particle P, with respect to the origin, is given by

Figure 12.8. Angular momentum of particle P.

The change in the angular momentum of the particle can be obtained by differentiating the equation for l with respect to time

We conclude that

This equation shows that if the net torque acting on the particle is zero, its angular momentum will be constant.


Figure 12.9 shows object P in free fall. The object starts from rest at the position indicated in Figure 12.9. What is its angular momentum, with respect to the origin, as function of time ?

The velocity of object P, as function of time, is given by

The angular momentum of object P is given by

Therefore

which is equal to the torque of the gravitational force with respect to the origin.

Figure 12.9. Free fall and angular momentum

Figure 12.10. Action - reaction pair.

If we look at a system of particles, the total angular momentum L of the system is the vector sum of the angular momenta of each of the individual particles:

The change in the total angular momentum L is related to the change in the angular momentum of the individual particles

Some of the torques are internal, some are external. The internal torques come in pairs, and the vector sum of these is zero. This is illustrated in Figure 12.10. Figure 12.10 shows the particles A and B which interact via a central force. Newton's third law states that forces come in pairs: if B exerts a force FAB on A, than A will exert a force FBA on B. FAB and FBA are related as follows

The torque exerted by each of these forces, with respect to the origin, can be easily calculated

and

Clearly, these two torques add up to zero

The net torque for each action-reaction pair, with respect to the origin, is equal to zero.

We conclude that

This equation is another way of expressing Newton's second law in angular quantities.

12.5. Angular Momentum of Rotating Rigid Bodies

Suppose we are dealing with a rigid body rotating around the z-axis. The linear momentum of each mass element is parallel to the x-y plane, and perpendicular to the position vector. The magnitude of the angular momentum of this mass element is

The z-component of this angular momentum is given by

The z-component of the total angular momentum L of the rigid body can be obtained by summing over all mass elements in the body

From the definition of the rotational inertia of the rigid body we can conclude that

This is the projection of the total angular momentum onto the rotation axis. The rotational inertia I in this equation must also be calculated with respect to the same rotation axis. Only if the rotation axis is a symmetry axis of the rigid body will the total angular momentum vector coincide with the rotation axis.

12.6. Conservation of Angular Momentum

If no external forces act on a system of particles or if the external torque is equal to zero, the total angular momentum of the system is conserved. The angular momentum remains constant, no matter what changes take place within the system.

Problem 54E

The rotational inertia of a collapsing spinning star changes to one-third of its initial value. What is the ratio of the new rotational kinetic energy to the initial rotational kinetic energy ?

The final rotational inertia If is related to the initial rotational inertia Ii as follows

No external forces act on the system, and the total angular momentum is conserved

The initial rotational kinetic energy is given by

The final rotational kinetic energy is given by


Problem 61P

A cockroach with mass m runs counterclockwise around the rim of a lazy Susan (a circular dish mounted on a vertical axle) of radius R and rotational inertia I with frictionless bearings. The cockroach's speed (with respect to the earth) is v, whereas the lazy Susan turns clockwise with angular speed [omega]0. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops ? (b) Is mechanical energy conserved ?

Assume that the lazy Susan is located in the x-y plane (see Figure 12.11). The linear momentum of the cockroach is m . v. The angular momentum of the cockroach, with respect to the origin, is given by

The direction of the angular momentum can be found using the right-hand rule. The direction of the z-axis is chosen such that the angular momentum of the cockroach coincides with the positive z-axis. The lazy Susan is moving clockwise (see Figure 12.11) and its angular momentum is pointing along the negative z-axis. Its angular momentum is given by

where I is the rotational inertia of the dish. Note that since the rotation is clockwise, [omega]0 is less than zero. The total angular momentum of the system is given by

The rotational inertia of the dish plus cockroach is given by

Since the external torque acting on the system is zero, the total angular momentum is conserved. The rotational velocity of the system after the cockroach stops is given by

The initial kinetic energy of the system is equal to

The final kinetic energy of the system is equal to

The change in kinetic energy of the system is

The change in the kinetic energy of the system is negative, and we conclude that mechanical energy is not conserved. The loss of mechanical energy is due to the work done by the friction force between the surface of the lazy Susan and the legs of the cockroach.

12.7. The Precessing Top

Figure 12.12. The precessing top.

A top, set spinning, will rotate slowly about the vertical axis. This motion is called precession. For any point on the rotation axis of the top, the position vector is parallel to the angular momentum vector.

The weight of the top exerts an external torque about the origin (the coordinate system is defined such that the origin coincides with the contact point of the top on the floor, see Figure 12.12). The magnitude of this torque is

The direction of the torque is perpendicular to the position vector and to the force. This also implies that the torque is perpendicular to the angular momentum of the spinning top. The external torque causes a change in the angular momentum of the system

This equation shows that the change in the angular momentum dL that occurs in a time dt must point in the same direction as the torque vector. Since the torque is at right angle to L, it can not change the magnitude of L, but it can change its direction. The result is a rotation of the angular momentum vector around the z-axis. The precession angle d[phi] is related to the change in the angular momentum of the system:

This shows that the precession velocity is equal to

This equation shows that the faster the top spins the slower it precesses. In addition, the precession is zero if g = 0 m/s2 and the precession is independent of the angle [theta].

13. EQUILIBRIUM AND ELASTICITY13.1. Equilibrium

An object is in equilibrium if the linear momentum of its center of mass is constant and if its angular momentum about its center of mass is constant:

P = constant

L = constant

An object is in static equilibrium if its linear momentum and angular momentum is equal to zero:

P = 0 kg m/s

L = 0 kg m2/s

13.2. Requirements for Equilibrium

If a body is in translational equilibrium then dP/dt = 0, or

If a body is in rotational equilibrium then dL/dt = 0, or

In summary, the following equations must be satisfied for an object in static equilibrium

If we restrict ourselves to two dimensions (the x-y plane) the following equations must be satisfied:

13.3. Equilibrium and the Force of Gravity

Figure 13.1. Weight of an object balanced by a single force.

Figure 13.1 shows a body of arbitrary shape balanced by a single force. The origin of the coordinate system is defined such that it coincides with the center of gravity of the object, which is the point upon which the balancing force acts. An object that is supported at its center of gravity will be in static equilibrium, independent of the orientation of the object. If the body is in equilibrium, the net force acting on it must be zero. Figure 13.1 shows that

Since the body is in equilibrium

and therefore

In obtaining this result we have assumed that the gravitational acceleration is the same for every point of the body. The net torque acting on the body is given by

Since the body is in static equilibrium

and therefore

This shows that rcm = 0 or rcm is parallel to g. We conclude that for a body to be in equilibrium, its center of mass must coincide with its center of gravity.

Sample Problem 13-1

A uniform beam of length L whose mass is m, rest with its ends on two digital scales (see Figure 13.2). A block whose mass is M rests on the beam, its center one-fourth away from the beam's left end. What do the scales read ?

Figure 13.2. Sample problem 13-1.

For the system to be in equilibrium, the net force and net torque must be zero. Figure 13.2 shows that

Here we have replaced the force acting on the beam with a single force acting on its center of gravity. The net torque of the system, with respect to the left scale, is

This shows immediately that

From the equation of the net force we obtain

Sample Problem 13-3

A ladder with length L and mass m rests against a wall. Its upper end is a distance h above the ground (see Figure 13.3). The center of gravity of the ladder is one-third of the way up the ladder. A firefighter with mass M climbs halfway up the ladder. Assume that the wall, but not the ground, is frictionless. What is the force exerted on the ladder by the wall and by the ground ?

The wall exerts a horizontal force FW on the ladder (the normal force); it exerts no vertical force. The ground exerts a force Fg on the ladder with a horizontal component Fgx and a vertical component Fgy. If these two components were not present, the system would not be in equilibrium. The net force in the x and y directions is given by

and

The net torque, with respect to O (which is the contact point between the ladder and the ground), is given by

Figure 13.3. Sample Problem 13-3

This immediately shows that

We can now calculate the force Fg:

and

We observe that Fgx depends on the position of the firefighter. Suppose that the firefighter is a distance f L up the ladder. In this case Fgx is given by

If the coefficient of static friction between the ladder and the ground is us, than the maximum distance the firefighter can climb is reached when

or

This shows that

13.4. Stacking Blocks

Two bricks of length L and mass m are stacked. Using conditions of static equilibrium we can determine the maximum overhang of the top brick (see Figure 13.4).

The two forces acting on the top brick are the gravitational force Fg and the normal force N, exerted by the bottom brick on the top brick. Both forces are directed along the y-axis. Since the system is in equilibrium, the net force acting along the y-axis must be zero. We conclude that

Figure 13.4. Two stacked bricks.

If the top block is on the verge of falling down, it will rotate around O. The torque exerted by the two external forces with respect to O can be easily calculated (see Figure 13.5). The gravitational force Fg acting on the whole block is replaced by a single force with magnitude m g acting on the center of mass of the top block. The normal force N acting on the whole contact area between the top and the bottom block is replaced by a single force N acting on a point a distance d away from the rotation axis O. The torque of the normal force and the gravitational force with respect to O is given by

The net torque acting on the top brick is given by

If the system is in equilibrium, then the net torque acting on the top brick with respect to O must be zero. This implies that

Figure 13.5. Forces acting on top brick.

or

This equation shows that the system can never be in equilibrium if a > L/2 (since d < 0 in that case). The system will be on the verge of losing equilibrium if a = L/2. In this case, d = 0. We conclude that the system can not be in equilibrium if the center of mass of the top brick is located to the right of the edge of the bottom brick. the system will be on the verge of losing equilibrium if the center of mass of the top brick is located right over the edge of the bottom brick. Finally, if the center of mass of the top brick is located to the left of the edge of the bottom brick, the system will be in equilibrium.

14. GRAVITY14.1. The Gravitational Force

Gravity is the weakest force we know, but it is the force of gravity that controls the evolution of the universe. Every body in the universe attracts every other body. Newton proposed that the magnitude of this force is given by

where m1 and m2 are the masses of the particles, r is the distance between them and G is a universal constant whose value is

G = 6.67 x 10-11 N m2/kg2

The gravitational forces between two particles act along the line joining them, and form an action-reaction pair (see Figure 14.1).

Figure 14.1. The gravitational force.

In real life we are not dealing with point particles; instead we are dealing with extended objects. To evaluate the gravitational force between extended objects, theshell theorem can be used:

"A uniform shell of matter attracts an external particle as if all the shell's mass were concentrated at its center"

Proof:

Figure 14.2 shows a shell located a distance r from a particle with mass m. The radius of the shell is R and its mass is M. The mass density of the shell is given by

All points on the small hoop indicated in Figure 14.2 have the same distance to the particle m. The magnitude of the gravitational attraction between any of these points and the mass m is therefore the same. The net force between the

hoop and mass m acts along the axis connecting the center of the shell and mass m. The area of the hoop is given by

Figure 14.2. Shell theorem.

and its mass m is equal to

The net force is equal to

The angles [theta] and a can be eliminated by using the following relations:

and

Differentiating the first of these two equations with respect to [theta] we obtain

or

Further more we see that

The total force acting on mass m can now be obtained easily

The shell theorem immediately shows that a sphere of uniform density (and mass M) attracts an external particle as if all the mass of the sphere is concentrated in its center.

In a similar fashion we can proof that a uniform shell of matter exerts no gravitational force on a particle located inside it .

14.2. The gravitational constant G

The strength of the gravitational force depends on the value of G. The value of the gravitational constant can be determined using the Cavendish apparatus. Two small lead spheres of mass m are connected to the end of a rod of length L which is suspended from it midpoint by a fine fiber, forming a torsion balance. Two large lead spheres, each of mass M, are placed in the location indicated in Figure 14.3. The lead spheres will attract each other, exerting a torque on the rod. In the equilibrium position the gravitational torque is just balanced by the torque exerted by the twisted fiber. The torque exerted by the twisted wire is given by

Figure 14.3. The Cavendish Apparatus.

The torque exerted by the gravitational force is given by

where R is the equilibrium distance between the center of the large and the small spheres. If the system is in equilibrium, the net torque acting on the rod is zero. Thus

All of a sudden the large spheres are rotated to a new position (position B in Figure 14.3). The net torque acting on the twisted fiber is now not equal to zero, and the system will start to oscillate. The period of oscillation is related to the rotational inertia and the torsion constant [kappa]

The angle between the two equilibrium positions is measured to be 2[theta]. This, combined with the measured torsion constant, is sufficient to determine the torque [tau] acting on the torsion balance due to the gravitational force. Measurements show that G = 6.67 x 10-11 Nm2/kg2.

14.3. Free-fall Acceleration

If the mass density of the earth depends only on the distance from the center of the earth (homogeneous shells), we can easily calculate the net gravitational force acting on a particle of mass m, located at an external point, a distance r from the center of the earth:

where M is the mass of the earth. For a particle on the earth surface, r = Re, the gravitational force is given by

We conclude that the free-fall acceleration depends on the mass of the earth and its radius:

The measured value of g = 9.8 m/s2 and Re = 6.37 x 106 m gives

which is in good agreement with the accepted value of 5.98 x 1024 kg. In reality, the situation is more complicated:

The earth's crust is not uniform. Precise measurements of the variations of the free-fall acceleration give information about non-uniformaties in the density of the earth. This can suggest the presence of salt domes (which often indicated the presence of oil).

The earth is not a sphere. The earth is an ellipsoid. It is flattened on the poles and bulging at the equator (difference in radius is 21 km). The free-fall acceleration is larger at the poles than it is at the equator.

The earth is rotating. The centripetal acceleration will change the free-fall acceleration.

To illustrate the effect of the rotation of the earth on the gravitational acceleration, consider a mass m located on a scale at the equator (see

Figure 14.4). The mass m will carry out a uniform circular motion with a period T equal to 24 hours. The radius of the circle is equal to the radius of the earth Re. The corresponding centripetal acceleration is given by

The following forces act on the mass:

1. The gravitational force m g0 (downwards)

2. The force W exerted by the scale on the mass (upwards)

The net force acting on the mass must be equal to the centripetal force required for the circular motion:

Figure 14.4. Mass located at equator.

The effective free-fall acceleration, obtained from the measured weight W, is given by

For the earth,

g0 - g = 0.034 m/s2

14.4. Gravitational Potential Energy

In chapter 8 we have discussed the relation between the force and the potential energy. Consider two particles of masses m1 and m2, separated by a distance r. In the gravitational field it is convenient to define the zero potential energy configuration to be one in which the two particles are separated by a large distance (infinity). Suppose the two masses are brought together (distance r) from infinity, along the path connecting the centers of the two masses. The work done by the gravitational force can be calculated as follows

(note that the force F and the position vector r are pointed in an opposite direction, and the angle between them is 180deg.). The potential energy U(r) is now given by

The potential energy is always negative and is a property of the two masses together rather than of either mass alone. We can verify our calculation by using U(r) to calculate the gravitational force

which is of course equal to Newton's law of gravity.

The work done by the gravitational force depends only on its initial and its final position, and not on the actual path followed. For example, a baseball travels from point A to point B (see Figure 14.5). The work done by the gravitational force on the baseball along the arcs is zero since the force and displacement are perpendicular. The only segments that contribute to the work done are those segments along the radial direction. The work done is negative if the force and the displacement are pointing in the opposite direction; if the force and the displacement are pointing in the same direction the work is positive Therefore the net work done if we travel along the radial direction back-and-forth (initial and final points coincide) is zero. We can now easily show that the net work done by the gravitational force on the baseball is just determined by its initial radial position and its final radial position.

Figure 14.5. Work done by the gravitational force.

Figure 14.6. A system of three particles.

If the system contains more than two particles, the principle of superposition applies. In this case we consider each pair and the total potential energy is equal to the sum of the potential energies of each pair. This is illustrated in Figure 14.6 for a system consisting of 3 particles. In calculating the total potential energy of a system of particles one should take great care not to double count the interactions. The total potential energy of the system shown in Figure 14.6 can be easily calculated:

The total potential energy of a system of particles is sometimes called the binding energy of the system. The total potential energy is the amount of work that needs to be done to separate the individual parts of the system and bring them to infinity.

Example

The gravitational potential can be used to calculate the minimum initial speed that a projectile must have to escape from the earth. Suppose a projectile of mass m has a speed v. Its initial kinetic energy if given by

The initial potential energy of the projectile is given by

and its initial mechanical energy is equal to

In deep space the potential energy of the projectile will be zero, and its minimum kinetic energy will also be equal to zero. We conclude that the minimum mechanical energy of the projectile must be zero. Therefore

and

This initial speed is called the escape speed. For the earth we obtain

vcrit = 1.1 x 104 m/s

14.5. Motion of planets

Suppose a planet with mass m is in a circular orbit around the sun, whose mass is M. The radius of the orbit is r. The gravitational force between the sun and the planet is given by

This is the force that keeps the planet in its circular orbit and its magnitude should therefore be equal to the centripetal force FC:

This implies that

or

This shows that for circular orbits, the square of the period of any planet is proportional to the cube of the radius of the orbit (law of periods). The constant depends only on the mass of the sun (M) and the gravitational constant (G).

In reality none of the planets carry out a circular orbit; their orbits are elliptical. The general equation of an ellipse is given by (see Figure 14.7)

The parameter a is called the semi-major axis of the ellipse (if a > b). It corresponds to the longest distance between the center of the ellipse (x=0,y=0) and the trajectory. The parameter b is called the semi-minor axis of the ellipse (if a > b). It corresponds to the shortest distance between the center of the ellipse (x=0,y=0) and the trajectory. An ellipse has two focuses (see Figure 14.7): each focus is located on the x-axis, a distance (e a) away from the center of the ellipse. The parameter e is called the eccentricity of the ellipse and is equal to

We see that for a circle the eccentricity is equal to zero, and the semi-major axis is equal to the radius of the circle. The shortest distance between the focus and the ellipse is called the perihelion distance Rp. It is easy to see that this distance is given by

Rp = a (1 - e)

Figure 14.7. The ellipse.

The largest distance between the focus and the ellipse is called the aphelion distance Ra which is given by

Ra = a (1 + e)

Figure 14.8. Trajectory of planet around sun.

The planets move about the sun in an elliptic path with the focus at the position of the sun (see Figure 14.8). The elliptical shape of the trajectory of the planet is a result of the 1/r2 nature of the gravitational force and the initial conditions. Under certain conditions the trajectory will be hyperbolic and the planet will approach the sun only once in its lifetime. Examples of hyperbolic trajectories are the trajectories of satellites that use the gravitational fields of the planets to change direction. The law of periods, previously derived for the special case of circular orbits, also holds for elliptical orbits, provided we replace r by a, the semi-major axis of the ellipse.

Sample Problem 14-8

Comet Halley has a period of 76 years and, in 1986, has a distance of closest approach to the sun of 8.9 x 1010 m. (a) What is the aphelion distance ? (b) What is the eccentricity of the orbit of Comet Halley ?

The semi-major axis of the orbit of Comet Halley can be found using the law of periods:

where

M is the mass of the sun (= 1.99 x 1030 kg) G is the gravitational constant (= 6.67 x 10-11 Nm2/kg2) T is the period (= 2.4 x 109 s).

Substituting these numbers we obtain a = 2.7 x 1012 m. The perihelion distance Rp is related to the semi-major axis a and the eccentricity e:

Rp = a (1 - e)

This equation shows that the eccentricity of the orbit can be calculated easily:

The aphelion distance can now be calculated

14.6. The Law of Areas

The trajectory of a planet about the sun is described by an ellipse with the sun in one of its focuses. Figure 14.9 shows the position of the planet at two instances (t and t + [Delta]t). The shaded wedge shows the area swept out in the time [Delta]t. The area, [Delta]A, is approximately one-half of its base, [Delta]w, times its height r. The width of the wedge is related to r and [Delta][theta]:

Figure 14.9. Area swept out by planet during a time [Delta]t.

[Delta]w = r [Delta][theta]

We conclude that the area [Delta]A is given by

If the time interval [Delta]t approaches zero, the expression for [Delta]A becomes more exact. The instantaneous rate at which the area is being swept out is

The rate at which the area is being swept out depends on the velocity of the planet and is also related to its angular momentum L. Figure 14.10 shows how to calculate the angular momentum of the planet. The angular momentum of the planet can be calculated as follows

Figure 14.10. Angular momentum of planet.

Substituting this in the expression obtained for dA/dt we conclude that

Since no external torques are acting on the sun-planet system, the angular momentum of the system is constant. This immediately indicates that dA/dt also remains constant. We conclude that

" A line joining the planet to the sun sweeps out equal areas in equal time "

This shows that the velocity of the planet will be highest when the distance between the sun and planet is smallest. The slowest velocity of the planet will occur when the distance between the sun and the planet is largest.

14.7. Orbits and Energy

Suppose a satellite of mass m is in orbit around the earth (mass M). The radius of the orbit is given to be r. The kinetic and potential energy of the satellite can be easily expressed in terms of r. The potential energy of the satellite is given by

The velocity of the satellite can be found by requiring that the magnitude of the gravitational force is equal to the centripetal force:

The kinetic energy can therefore given by

The total mechanical energy can now be calculated

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