physics 1 notes

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Physics 1: University Physics for Scientists & Engineers Please note, this is a work in progress, and as such, will undergo lots of modification until the end of the semester. Most notably, the page breaks, which I want to place at strategic places (so as not to cut off something important into 2 pages), but trying to do it now will only fail as I add and remove lines, so I will do that only at the end. In the meantime, keep this in mind.

Chapter 1: Physics and Measurement o

Chapter 2: Motion in One Dimension o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7

Chapter 3: Vectors o Ex.1 Ex.2

Chapter 4: Motion in Two Dimensions o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6

Chapter 5: The Laws of Motion o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5

Chapter 6: Circular Motion and Other Applications of Newtons Laws o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6

Chapter 7: Energy and Energy Transfer o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6

Chapter 8: Potential Energy o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5

Chapter 9: Linear Momentum and Collisions o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7 Ex.8 Ex.9

Chapter 10: Rotation of a Rigid Object About a Fixed Axis o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7 Ex.8 Ex.9 Ex.10

Chapter 10.9: Rolling Motion o Ex.1 Ex.2 Ex.3

Chapter 11: Angular Momentum o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7 Ex.8 Ex.9 Ex.10

Chapter 12: Static Equilibrium and Elasticity o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7 Ex.8 Ex.9 Ex.10

Chapter 15: Oscillatory Motion o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7 Ex.8 Ex.9 Ex.10

Final Exam Study Guide o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7 Ex.8 Ex.9 Ex.10 o Ex.11 Ex.12 Ex.13 Ex.14 Ex.15 Ex.16 Ex.17 Ex.18 Ex.19 Ex.20

Comment [as1]: Notes for Monday, June 12, 2006 begin here

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I. Chapter 1: Measurements (return to top)II. Chapter 2: Motion in One Dimension (return to top)

A. Purpose: Comment [as2]: Notes for Wednesday, June 14, 2006 begin here.

B. Definitions (and symbols): 1. Distance (d) (scalar) is the total length of space that an object travels. 2. Displacement (D) (vector) is the length of space between your origin and your destination. 3. Speed (s) (scalar) the rate of movement of a particle.

a. Average Speed = the ratio of the distance covered in a certain time

S = dt

i.

b. Instantaneous Speed = how fast you are going at any particular instant. It is also known as the magnitude of the instantaneous velocity. ri.

4. Velocity (v) (vector) the rate of displacement of a particle S(t ) =| v |

a. Average Velocity = the ratio of displacement covered in a certain time

i. vX = x tb. Instantaneous Velocity = the velocity of an object at an instant in time.

vx = limt0xt =

dxdt

i.

5. Acceleration (a) (vector) the rate at which velocity changes a. Average Acceleration = the ratio of velocity covered in a certain time r rvi rvf = vi.

ra

t tb. Instantaneous Acceleration = the acceleration of an object at an instant in timed

ra(t) = d

rvdt

i.

C. Example 1: 1. A particle begins traveling 5 feet to the right and then stops and travels to the left for 2 feet,

traveling the entire distance in 3 seconds. Assuming average velocity, analyze the problem. a. What is the distance (d) and displacement (D)? r

D = 3 ft (or 3ft right, 3 ft east, +3i ft) d = 7 ft i. b. What is the average speed ( s ) and instantaneous speed (s)?

i. s =| rv |= 7

3= 2.33 ft/s s=2.33 ft/s (if constant)

c. What is the average velocity ( v ) and instantaneous velocity (v)? i.

rv = 1i ft/s rv 2.33 ft/s

ad. What is the average acceleration ( ) and instantaneous acceleration (a)?

i. ra = vf vi = (2.33) (2.33) = ft/s21.55

t 3ii. draw the path of the particle

Chapter 2 (Motion in One Dimension)

iii. Regarding average velocity since the displacement is 3 feet in 3 seconds, the average

velocity is 1 ft/s. this would mean that a particle traveling at 1ft/s directly towards the goal (as opposed to the other particle which went forward 5ft and then returned 2 feet), will arrive at the same time as the particle traveling the longer route. As for instantaneous velocity, there is not enough information to figure this out (we dont know if velocity is constant or changing).

D. Example 2: 1. A car goes west for 40 miles in 2/5 hours and then stops for hour. It then goes east for

70 miles in 7/10 hours. Assume the car is going at constant speed and its initial & final speeds are not zero. a. What is the average speed of the car in the first stage, second stage, & the whole trip?

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i. s1 = 402 / 5 = 100mph s2 =70

7 /10= 100mph stot = 1101

2+ 2

5+ 7

10

= 68.75mph

b. What is the average velocity of the car in the first stage, second stage, & whole trip?

i. rv1 = 100mph

rv2 = 100mph

rvtot = 301.6 = 18.75mph

c. What is the average acceleration of the car in the first stage, second stage, & whole trip?

i.

ra1 =

vf (1) vi(1)t1

= 0 (100)2 / 5

= 250mph2 ra2 =vf (2) vi(2)

t= 100 0

7 /10= 142.86mph2

ratot =

vf (2) vi(1)ttot

= 100 (100)1.6

= 125mph2

d. Draw the path of the particle & include all relevant data

e. Graph x vs. t, v vs. t, s vs. t, & a vs. t & show the geometrical meanings of the average values.

E. Example 3:


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x(t) = (t 2 4)(t + 5) 0 t1. A particle is traveling on a path given by the equation from 6

x(t) = t 3 + 5t 2 4t 20x(0) = 20 x(6) = 352x(t) = t=2

x '(t) = v(t) = 3t 2 +10t 4v(0) = 4 v(6) = 164v(t) = 0 0 = 3t 2 +10t 4 t = 0.361x(0.361

seconds. Fully analyze the particles path, its speed, velocity and acceleration. a. Step 1: figure out the significant markers related to the position function (at the beginning, end and

when the function equals zero). To figure out how far the particle will travel and when it will cross the origin (t=0). i. ii. (These will be our initial & final positions) iii.

(this is the time the particle will be at the origin) 0 0=t3 + 5t 2 4t 20

b. Step 2: derive the position function to get the velocity function and its significant markers (again at the beginning, end, and when the function equals zero) to determine its velocity at the beginning and end of its path. Where the function equals zero, the particle has no velocity and could either continue on its path, or change direction. i. ii. (These will be our initial & final velocities) iii. (velocity is zero here) iv. (This means at 0.361 seconds, the position will be at either a local

max or local min, IF the sign changes. ) = 20.745

In this case, the sign changes from negative to positive, meaning x(0.361) is a local minimum. This also tells us our problem will be broken down into two stages, the one with the negative velocity (up to t=0.361) and the one with the positive velocity (from t=0.361 and beyond).

c. Step 3: derive the velocity function to get the acceleration function and its known values. x t v '(t) = a(t) = 6t +10''( ) =i.

ii. (These will be our initial & final accelerations) a(0) = 10 a(6) = 46d. Step 4: find the average speed for stage 1, stage 2, and the average of both

i. S1 = .745.361 = 2.06 m/s S2 =372.745

(6 .361) = 66.1 m/s

Stot = 372.745 (.745)6 = 62.25 m/s e. Step 5: find the average velocity for stage 1, stage 2, and the average of both

i. r

r

v1 = v2 = 66.1 m/s 2.06 m/s3726

= 62 m/srvtot = f. Step 6: find the average acceleration for stage 1, stage 2, and the average of both

i. ra1 = 0 (4)0.361 = 11.08 m/s

2 164 (0)6 0.361 = 29.08 m/s

2 ra2 =

ratot = 164 (4) = 28 m/s26

g. Draw a picture of the particles motion and graph the x, v, s, & a functions.


F. Linear Kinematics Equations (for objects undergoing constant acceleration)

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1. Basic Velocity Function Comment [as3]: Notes for Monday, June 19, 2006 begin here

a. = this is the fundamental velocity equation vf = vi + at

vf2 = vi2 + 2a(x f xi )x f = xi + 12

2. Other Kinematic Equations a. velocity, acceleration, & position

position, velocity & time b. (vi + vf )t

x f = xi + vit + 12 at2 position, velocity, time & acceleration c.

3. Freefall Kinematic Equations since acceleration is -9.8 m/s we can use these: a. velocity & time v f = v

vf2 = vx f = xi + vit 4.9t 2

i 9.8tb. position & velocity i

2 19.6(x f xi )c. position, velocity, & time

G. Example 4: 1. Car A is going at 75mph when it realizes its closing in on the car in front of him (car B). Car

B in front of him is 200ft away and is moving at 65mph when car A hits the brakes, decelerating at 2ft/s. Car B also notices this and 1 second later, accelerates at 3ft/s. a. Is there an accident and if yes, what are their velocities upon impact?

i. Step 1 convert mph to ft/s and find out the velocity & position after 1 second 75mih

1h3600s

5280 ft1mi

= 110 ft / s 65 52803600

= 95.3a) 3 ft / s

car 1 x f = x i + vi t + 12

b) vf = 110 4(1)2 = 106a t 2 = 0 + 11 0 + 1

2c) (4 )(1)2 = 108 ft

car 2 x f = 200 + 95.3d) 3(1)+ 0 = 295.33 ft


Page

car 1 x = x + v t + 1ii. Step 2 determine the position equations for both vehicles from 1 second onward.

a) f i i 2at 2 x = 108 +106t 2t 2

car 2 x f = 295.3f

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b) 3 + 95.33t 12

(3)t 2 x f = 295.33 + 95.33t + 32 t2

x f

108 +106t 2t 2 = 295.3

iii. Step 3 set both vehicles final positions ( ) to equal each other. If they have a solution, then a collision will occur.

a) 3 + 95.33t 32t 2 .667 16.152i

f (t) 108 +106t 2t 2 = 295.3

t = 10 b) A solution cannot have imaginary numbers, so there is no solution = they do not collide.

b. If they do not collide, what is the distance of their closest approach? i. Step 1 find their closest approach by adding both position equations (making them one)

and finding the resultant equations derivative. Then maximize the derivative to find its local max/min.

a)

3 + 95.33t + 32t 2

f (t) 187.33 10.66t + 72 t 2f '(t) 7t 10.66 f '(t) = 0 7t 10.66 = 0

4s (time of max or min distance)t = 1.52car 1 xf = 108 +106(1.524) 2(1.524)2 = 264.879 ftcar 2 x f = 295.3

b)

c) 3 + 95.33(1.524) 32

(1.524)2 x f = 444.104329.137 264.879 = 179.225 ft

d) (this is the closest they will ever get) e) time of closest approach is 2.524 seconds.

H. Example 5: 1. Train A is going to the left at 30m/s when he sees train B going to the right at 25m/s on the

same tracks 160m away. Train A decelerates at 2m/s and train B decelerates at 3m/s. Is there an accident and if so, what are their velocities upon impact? a. Draw a diagram showing the trains and their relevant values

b. Step 1- calculate the trains x positions and set them equal to each other (same x = collision). This

will return the time at which they collide.

(train A) x f = 160 30t + 12i. (2)t2 = 160 30t + t 2

(train B) x f = 0 + 25t 12

(3)t 2 = 25t 32t 2ii.


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160 2 3iii. 30t + t = 25t 2 t 22 5 t 55t +160 = 02

= 3.45s

tc. Step 2 check the velocities of the trains upon impact. This tells us both the impact velocity and if

validates the equation. If the velocities of the trains keep the same sign, the time of collision is valid

(train A) vf = (30) + (2)(3.45) = 23.1 (sign remains the same)(train B) vf = (25) + (3)(3.45)

i. = 14.65 (sign remains the same)

yf = yi + vit + 12

ii. d. In this case, both trains retain their signs. This means the time of collision is valid. If one of the

trains changed sign, it would mean the train stopped before the aforementioned time which invalidates the equation. In this case, we need to find out when the sign-changing train has zero velocity because this is its final resting point. Once we figure that out, we simply treat the rest of the problem (with the other train) as a simple train hitting the wall equation with the wall being the train that stopped. To find out the time of the REAL collision, set the other trains x position to the same and find the time and velocity of impact!

I. Vertical Motion Problems Including Freefall 1. In the case of freefall, we use the same equations, but now because we always know that g

is 9.8, we use that for all vertical motion problems (except where acceleration is affected by something else).

J. Example 6: 1. A guy on a building 50m high throws a ball up with a velocity of 20m/s. The ball goes up, and

then returns but keeps going to ground level where there is a lake. Upon hitting the lake, its acceleration changes to 2m/s down, taking 12 seconds to reach the floor.

2. Questions: a. What is the depth of the lake? b. What is the final velocity?

3. Solution Strategy: a. Step 1 find out the time the ball hits the water

at 2

0 = 50 + 20t 4.9t 2t = 5.83 (the time it hits the water)v f = 20 9.8(5.83) vf

i.

b. Step 2 find its velocity upon hitting the water = 37.134m / s

Depth

i. c. Step 3 now use that velocity, the time left in 12 seconds,

and the known acceleration in water to figure out the depth of the lake and velocity upon impact.

= Vit + 12 at2

= (37.134)(12 5.83)+ 12

(2)(12 5.83)2i. = (37.134)(6.17) (6.17)= 267.161m

2

Lake vf = 37.13+ (2)(6.17) = 49.47m / s

ii.

Chapter 2 (Motion in One Dimension) K. Example 7:

1. A rocket goes up for a constant acceleration for 8 seconds, then its engines fail and it goes into a freefall, hitting the ground with a velocity of 200ft/s.

2. Questions: a. What is the total time for this trip?

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vf = vi + at = 0 + a(8) = 8a

yf = yi

b. What is the acceleration of the rocket? c. What is the maximum height of the rocket?

3. Solution Strategy: since we have multiple unknowns, we have to figure out a way to relate them to each other. a. Step 1 figure out how the velocity relates to the height

i. (This is the final velocity of the rocket during its firing stage. This will become the initial velocity of our rocket in its freefall stage)

+ vit + 12 at2 = 0 + 0 + 1

2a(8)2 = 32aii.

(This will be the final height of the rocket during its firing stage. This will become the initial height of our rocket in its freefall stage)

b. Step 2 we now have enough to calculate the descent since we have all variables but time. We can use the velocity squared function (the one that excludes time) to put it all together.

i.

Keep in mind, the a in this final equation is

vf2 = vi2 + 2a(yf yi )

2002 = (8a)2 + 2(32)(0 32a)a = 13.68

NOT the same as the a in our previous two equations. Since this equation is based on the freefall, the a is -32ft/s, whereas the previous a was based on the rockets propelled ascent.

c. Step 3 with our acceleration, we can now figure out the time at which the rocket has zero velocity (the top) and from there, get the position at the top ( ) y f

v f = vi0 = 8(1

yf = yi + vit + 12i.

+ at3.68) + (32)t

t = 3.42s

at 2 yf = 32a + 8at 16yf = 32(13.682) + 8(13.682)(3.42) + (16)(3.42)2yf = 625.02 ft

III.

t 2

Chapter 3: Vectors (return to top) A. Coordinate Systems

1. Background - There are various systems for denoting vectors, such as Polar Coordinates, bearing (N, S, E, W) & magnitude or component vectors. a. Polar coordinates example:

i. 10m @ 260 b. Bearing & magnitude example:

i. 10m @ 10W of N c. Component vectors example:

i. which is equivalent to: 10Cos(260)i10Sin(260) j 10Cos(20) j

10Sin(20)i

Chapter 3 (Vectors) B. Example 1:

ax

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1. A box has the following forces applied to it ( = 4N , ay = +2N

(4)2 + (2)2

). Draw a picture of the box with its vectors and find out what is the resultant vector in polar format.

2. Solution Strategy: a. Step 1: add up the vectors

= 4.472

Tan1( 24

i. b. Step 2: find the angle

i. ) = 26 = 180 26 = 154 C. Example 2:

1. A car goes 40 miles at 35 N of W in hr and then turns around and goes 30 miles at 60 E of N in 3/5hr.

2. Questions: since this is a two stage problem, we will find all relevant data for all stages) a. Find s1, s2 , stot b. Find

rv1,

rv2 ,

rvtot

c. Find ra1,

ra2 ,

ratot

A = 40 40Sin(145) j = 32.766i + 22.943 j B

d. Draw the path of the car

3. Solution Strategy: a. Step 1 determine the unit vectors for both lines and add them up ur

i. Cos(145)i +ur = 30 30Sin(22) j = 27.816i +11.238 j32.766i + 22.943 j + 27.816i +11.238 j= 4.95i + 34.183 j

C

Cos(22)i +ii. iii.

b. Step 2 find the resultant vectors magnitude and direction ur4.952 + 34.1832 = 34.540=i.

Tan1(34.1834.95

) = 81.76ii. c. Step 3 - find the velocity and speed for all three stages:

v1 =x f 1 xi1

t= 40 0

2 / 3= 60mph @35 N of W i.

v2 = 30 03 / 5 = 50mph @ 68 E of Nii.

vtot = C

ur

ttot= 34.540

2 / 3+ 3 / 5iii. = 27.268 @ 81.76 N of W

s1 = v1

iv. Since the speed is the absolute value of the velocity, your speed for stage 1 and stage 2 are the absolute value of the velocity, but the third speed is NOT. Unlike the total velocity that is the total displacement over the total time, the total speed is the total distance over the total time.

= 60mph s2 = v2 = 50mphv.

Chapter 3 (Vectors)

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stot = dtott30 +

vi. tot

402 / 3+ 3 / 5 = 55.26mph

d. Step 4 find out the acceleration of the car i. Because neither the velocity nor the direction changed a1 = 0

a2 = 0

ii. Because neither the velocity nor the direction changed [50Cos(22)i + 50Sin(22) j] [60Cos(145)i + 60Sin(145)]

19 /15iii.

ratot =

vf vittot

=

a3 =vf vittot

= (46.359i +18.730 j) (49.149i + 34.415 j)1.26

95.508i 15.684 j1.26

= 75.401i 12.382 j (Total acceleration in vector components) iv.

ratot = (75.401)2 + (12.382)2 = 76.411 = Tan1 12.382

75.401

(Total acceleration)

v. = 9.326 = 350.674 (theta of Total acceleration)

IV. Chapter 4: Motion in Two Directions (return to top) A. Background:

Comment [as4]: Notes for Wednesday, June 21, 2006 begin here

1. This chapter is really an extension of Chapter 2 (motion in one dimension) and Chapter 3(vectors). From chapter 2, we know that the motion of a particle is a function of time, from which we can infer its velocity, and acceleration. From Chapter 3,

B. Distance (d)(in 2 dimensions) 1. When measuring the distance of an object moving in two dimensions, we must include all

distances traveled by the object. a. if the path is linear or has linear segments, calculating its distance is a matter of adding up its

segments. b. If the path is curved and we know the function that represents the curve, we can use the arc

length integral to determine its distance from point a to point b of the function.

dcurve = 1+ f '(x)2i. a

b

dx

Displacement (

C. Displacement (D)(in 2 dimensions) 1. Displacement is the distance between two points. In vector form, this works for any vector,

whether it be position, velocity, or acceleration. r r rrinitial = (x f xi )i + (yf yi ) j = rD) = rfinal ra. v t

rb. Average Velocity

i + dydt

c. Instantaneous Velocity vr = drdt

= dxdt

j = vxi + vy j

d. Average Acceleration r vf vi

t f ti= vta

a dtdv

e. Instantaneous Acceleration

Chapter 4 (Motion in Two Dimensions) D. Example 1:

1. Problem: An object is traversing a path given by the equation y(x)=x from (-1, 1) to (2, 8) feet. It is traveling at such a rate that dx/dt = 2 ft/s and it makes the journey in 1.5 seconds.

2. Questions: a. Calculate the total distance traveled and the average speed. b. Calculate the total displacement and the average velocity. c. What is the initial & final velocity, initial & final speed, initial & final acceleration, and

average acceleration? 3. Solution Strategy:

a. Nota bene: a common mistake most people make is to confuse the position and the velocity vectors and thus get the wrong result and cant figure out whyif your results are not what you expected (but sometimes by coincidence they might work until you get to the acceleration vector), be aware of which are needed for what!

b. Step 1 calculate the distance for this curve using the arc length integral, which you can use to calculate the average speed.

d = 1+ (3x2 )21

2

dx = 10.178i.

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ii. s = 10.1781.5

= 6.79 ft s c. Step 2 calculate the total displacement and average velocity r ri = (2i + 8 j) (1i 1 j) = 3i + 9 ji. D = rf

ii. rv =

rDt= 9

2 + 321.5

= 9.4871.5

= 6.325 ft s iii. = Tan1(9 / 3) = 71.57

if

dxdt

d. Step 3 calculate the component-position function in order to derive the velocity and acceleration functions. i. Since y(t)=x, you need to figure out x(t) because your velocity function depends on both the

y position and the x position:

= 2 x = dxdt

dt = 2dt = 2t + ca) This equation will give us the x-position of the particle as a function of time but we still need to figure out the constant c. we do this by remembering that at time = 0, x = -1 (see position graph)

1 x(t) whe 2(0)+c=-1 cn t=0, x=-1, so x(0)=-1 x(0)= = = 2t 1

y(x) = x3

y(t) =

b) Now we know the x function and only need to recall the y function.

c) This equation was given. It gives us the y-position as a function of x-position. We can make it a function of time by sticking the x function inside it

once weve done this, were ready to put both functions together.

y(x(t)) = (2t 1)3

d) rr(t) = x(t)i + y(t) j = (2t 1)i + (2t 1)3 j This equation creates a component-position function that tells us where the x and y components of the particle are at any point in time. For example, at 1.5 seconds, it tells us the particle is at (2,8) which matches our position function. This function is used to arrive at the velocity since the velocity is a function of the absolute position over time.

ii. Once we have the position function r(t), we can derive it with respect to time to get the velocity function:

Chapter 4 (Motion in Two Dimensions)

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a) rv = d

rrdt

= 2i + 3(2t 1) (2) j = 2i + (24t 24t + 6) j 2 2

iii. From here, we continue the process to get to the acceleration function: ra) a = (48t 24) j

e. Step 4 with all necessary functions, calculate their values at the beginning and end of the trip and any other significant landmarks (for example, if and when the velocity and/or acceleration went to zero, we might want to know that in some cases) i. We can see that the initial and final positions were given (which is

where we got our equations from, but if you want to check, plug them back into the component-position function:

a) ri = r(0) = (0 1)i + (0 1)3 j = i jrf = r(1.5) = (2(1.5) 1)i + (2(1.5) 1)3 j = 2i + 8 j

vi = v(0vf = v(1

si = vi

b)

ii. Now we calculate initial and final velocity: a) ) = 2i + (0 + 0 + 6) j = 2i + 6 jb)

This tells us the velocity at the beginning and the end of the particles trajectory.

.5) = 2i + (24(1.5)2 + 24(1.5)+ 6) j = 2i + 96 j

iii. Since the speed is the magnitude of the velocity vectors we get: a) = 2i + 6 j b) s f = vf = 2i + 96 j

iv. Now we do the same for the acceleration: a) ai = a(0) = (48(0) 24) j = 24 j

a f = a(1 48 j

aavg =vf it

b) This tells us the acceleration at the beginning and end of the particles trajectory.

.5) = (48(1.5) 24) j =

f. Step 5 this allows us to figure out the average acceleration:

i. vtot

= (2i + 96 j) (2i + 6 j)1.5

= 901.5

j = 60 j

E. Example 2: 1. Problem: a particle is going at a constant speed in a counterclockwise circular motion with a

radius of 22ft. it starts at an angle of 25 and ends up at 287 in 10 seconds.

2. Questions: a. What is the average speed and average velocity? b. What is the instantaneous acceleration? c. What is the average acceleration and show that it points to the

center when placed at the midpoint of the path. 3. Solution Strategy:

a. Step 1 find the average speed by calculating the distance i. d = r = 22[(287 25) 180 = 100.6 ft ii. s = d t = 100.610 = 10.06 ft / s

b. Step 2 find the velocity by completing the triangle to get the displacement (keep in mind that since both sides of the triangle are the same length, you have an

isosceles triangle, which tells you two of its legs are the same angle (the

Comment [as5]: The velocity picture has an absolute placement

you , so

r ere

which might get messed up ifmove the whole thing aroundkeep this in mind

Comment [as6]: Notes foMonday, June 26,2006 begin h

Chapter 4 (Motion in Two Dimensions) two legs that make angles with the leg of different length). We can use this to complete the triangle.

Page 13 of 59

c2 = a2 + b2 + 2abCos( )c = 22

i. Using the law of cosines a)

2 + 222 + 2(22)(22)Cos(98) = 33.21Sin(a)a

b)

ii. Using the law of sines

= Sin(b)b

Sin(41)22

= Sin(98)x

x =a) 33.21

iii. rv = D t 33.2110 = 3.321 ft/s=

c. Step 3 the instantaneous acceleration is centripetal given its circular motion:

ac = v2

r=

223.3212 = 0.501 ft/s2i.

d. Step 4 the average acceleration is (like in the previous example), is the result of the final minus initial acceleration over time. The trick here though is finding the final and initial acceleration! One thing to remember is that the velocity doesnt change throughout this curve. i. The initial & final velocities require that we figure out their

vectors. Since the velocity vector in a circular motion is the tangent (90 ) to the particles angular position, our calculations only require simple math (as long as you understand the pstatement)

revious

va) initial = xinitial + 90 = 25 + 90 = 115 v

b) final = x final + 90 = 287 + 90 = 17

vf = s f v f = 10.06 17 = 10.06[Cos(17)i + Sin(17) j] = 9.620i + 2.941 jvi = si = 4.251i + 9.117 j

ii. Once we know the direction, we calculate the initial and final velocity by multiplying the speed by the direction to get a component (vector) of the velocity.

a)

b) vi = 10.06 115 = 10.06[Cos(115)i + Sin(115) j]iii. from here, calculating the average acceleration vector is the same as other examples:

a) = (9.620i + 2.941 j) (4.251i + 9.117 j)10

= 13.872i 6.176 j10

ra = vf vi

ttot

ra = 1.3

87i 0.618 j b)

ra = 1.3872 0.6182 = 1.518Tan1

.6181.387

(The magnitude in case we need it)

c)

= 24 or 336

F. Projectile Motion 1. Background: Projectile motion is really an extension of our previous two-dimensional motion

problems, except simplified. The initial velocity vector is broken down into horizontal and vertical components and these components are solved in their own kinematic equations. In most cases, the airtime of the particle will determine the trajectory of the particle. This tells us that we want to find out how long the ball is in the air (y-kinematics) and from that, we can figure out how far the ball will go (x-kinematics) since both are related by time (the same time, duh).


2. In this example, we can see a number of notable properties of a ball in projectile motion:

a. The horizontal velocity does not change (if we ignore wind resistance) b. The vertical velocity is the same at the same y-position. So the catcher of the ball in this example

feels the same velocity with which the thrower threw it. c. The point at which vertical velocity is zero is the apex of the curve, many problems will require

that you figure this out. d. The only acceleration is vertical (g) which affects the velocity of the ball throughout the trip.

G. Example 3: 1. Problem: a basketball is thrown from a height of 610 towards

a basket 25 away and 10 high. The initial velocity of the ball is 28ft/s and the angle is 42.

Page 14 of 59

2. Questions: a. Does the person make the basket? What is the final velocity? b. If not, what angle should they aim at?

3. Solution Strategy: a. First we figure out the components of the initial velocity

i. 28Sin42 = 18.73628

ii. Cos42 = 20.808

x f = xi + vit 25

b. Now we calculate how long it takes the ball to travel 25 feet in the

x direction. Since this time will be our limiting time, we will then use it to figure out our y altitude at that time. If it is exactly 10 ft, we have a basket. Any other value will mean the ball is over or under the basket, thus missing.

20.81t t= = 1.20s yf = yi + v t + 12

i.

at 2 y = 6.83fii. i 3 +18.736(1.2)16(1.2)2 = 6.276

10 = 6.83

iii. at x=25ft, y is only 6.3 feet, falling short of the basket. c. Since the ball fell short, we try to determine an angle that (with the same velocity) will give us the

required (x=25, y=10) values. We set up both x and y equations and set them equal to each other 3 + 28Sin()t 16t 2

25 = 28Cos()t t = 2528Cos()

i. (The y-equation)

ii. (The x-equation)


10 = 6.833 + 28Sin( )28Cos( )

25 25 16 28Cos( )

2

3.16

iii.

6 = 25Tan() 16(625)2(784)Cos ( ) 3.166 = 25Tan() 12.755Sec2 ( ) no solution

3.16

d. Since there is NO angle at which we can launch it, we now keep the angle the same and change

the velocity to find out a velocity at which both would be the same(x=25, y=10).

Page 15 of 59

i. 6vi

2Cos2 (42)= 25Tan(42) 16(625) vi = 30.596

H. Example 4: 1. Problem during a soccer game, a penalty corner kick is awarded to the attacking team.

The kicker, trying to get past the goalie, wants to kick the ball to his teammate who can then head it into the goal. This header is at an angle of 10 from the kicker and right down the middle of the field (32.5 meters).

2. Questions: 3. Solution strategy:

I. Kinematics of Circular Motion 1. Background: Centripetal (towards the center) acceleration occurs whenever a particle turns

a corner or otherwise travels in a circular arc or a whole circle. Calculation of centripetal acceleration is the result of the direction changing (because while the object moves, its acceleration keeps pointing to the center, constantly changing in angle), while the speed usually remains constant; this is uniform circular motion. If both the speed and direction change, then the particle undergoes non-uniform circular motion.

Comment [as7]: Notes for Wednesday, June 28, 2006 begin here

a. Centripetal acceleration:

r r r = v

2

rv vr r r =

vvCos(0)r rac =

2r = this is used in uniform circular motion

b. Tangential acceleration: at = dvdt this is used in non-uniform circular motion along with the centripetal acceleration equation, as both are components of an objects trajectory.

2. In uniform circular motion, since there is no tangential acceleration, centripetal acceleration is also the total acceleration. In non-uniform circular motion, the tangential component is simply tangential (at a 90 angle) to the radial component. The resultant vector is the total acceleration. Another way to denote total acceleration (for both types of circular motion) is to use the notation, but please note, that the , r r unit is directed outwards, so if using it, you must reverse the sign of the centripetal vector.

(atot ) = atacatot = ac2 + at 2a. (for non-uniform circular motion) (r)+ at ()atot = acb.

J. Example 5: 1. Problem A car starts out at an initial velocity of 6ft/s around a circle of 60ft radius. It

accelerates at a rate 3t+1 ft/s.

Chapter 4 (Motion in Two Dimensions) 2. Questions:

a. What are the initial centripetal acceleration, initial tangential acceleration and initial total acceleration (including its angle)?

b. What are the centripetal acceleration, tangential acceleration and total acceleration (including its angle) after 12 seconds?

3. Solution Strategy: a. Figure out the centripetal and tangential components to calculate the resultant vector:

ac = 62

60

Page 16 of 59

i. = .6 ft/s at = 3(0) +1 = 1 ft/s

atot =

.62 +12 = 1.1662 ft/s ArcTan 1.6

= 59.04ii.

b. For question b, the challenge is figuring out the centripetal acceleration at 12 seconds. Since the object is speeding up as it moves, its centripetal component will grow dramatically larger over time. Remembering that velocity is the integral of acceleration, we simply integrate the acceleration function with a time of 12 seconds and add 6 seconds (because the initial velocity was 6 ft/s). From there, it's the same as the previous equations:

i. v(12) = (3t +10

12

)dt + 6 = 234 ft/s ac = v2r = (234)2

60= 912.6 ft/s2

at = 3(12) +1 = 37 ft/s2

atot =

912.62 + 372 = 913.35 ArcTan912.6

37 = 2.32ii.

K. Relative Velocity and Relative Acceleration 1. Background: this topic comes from the idea that observations of movement are often

related to the reference frame of the observer. For instance, a person walking down the moving walkway between Caesars Palace and the Mirage would appear to be going much faster to a stationary observer while an observer standing on the moving walkway would only see him going at a normal walking pace. This relativity of speed forms the basis of calculation of speeds where we would want to figure out the movement of an object (an airplane) in a moving medium (air), relative to a static position (the ground). Just like the speed of the walkway walker relative to the ground is simply the addition of the speed of the belt and the speed of the walker, we could calculate an airplanes speed relative to the ground by adding its velocity relative to the air with the velocity of the air relative to the ground.

L. Example 6: 1. Problem An airplane is headed from Seattle WA to Miami FL just in time for the hurricane

season. The distance between the 2 cities is 2400 miles and the angle is 40 S of E. A wind rises up during the flight with a speed of 70mph @ 80 N of W

2. Questions: a. If the plane wants to make the trip in 4 hours, what should their average airspeed and direction

be? b. If the pilot forgets to account for the wind, where will the plane end up in 4 hours?

3. Solution Strategy: a. Make an image of the exercise to get an idea of the solution:


b. First we figure out the vectors for all three:

2400mi4h

Page 17 of 59

i. rs

= 600 mi/hr

vag = 60 vagy = 600Sin(320) = 385.673

(40) = 459.627

vwg = 70 vwgy = 70Sin(100)

ii. 0@40vagx = 600Cos

=@100

68.937

vwg = ?

vag vvagy vwgy = vawyvagx =

= 600Sin(320) 70Sin(100) = vawy

600Cos(320) 70Cos(100) = vawx

=454.61471.78

(454.612 ) + (471.782 )

vwgx = 70Cos(100) = 12.156 iii.

iv. c. Next we use the pictures we made to figure out the unknowns for question a:

i. wg = vawii. vwgx vawx

= 655.17 ArcTan 454.61471.78

iii. = 43.9 = 316.1

vag + v awvagy + vwgy = vawyvagx + v = v

= 600Sin(320)+ 70Sin(100) = vawy

600Cos(320)+ 70Cos(100) = v

=316.74447.47

(316.742 ) + (447.472 )

d. After that, we solve question b in the same way (but with its own equation of course) i. = vwgii.

wgx awx awx

= 548.22 ArcTan 316.74447.47

iii. = 35.3 = 324.7

dtotal = 548.22

iv. 4 = 2192.91 milesV.

Chapter 5: The Laws of Motion (return to top) A. Background:

1. How is Kinematics different from Dynamics: In Kinematics, youre asked to analyze a motion (i.e. circular, left/right, up/down, projectile). In Dynamics, you are asked to examine the forces which produce the motions you analyze in kinematics (gravity, normal, centripetal).

2. Fundamental Dynamic Equation: F = maB. Example 1:

1. Problem - You have a block with mass 2kg, which is being moved with an initial velocity of 5m/s for 15m. The force moving it is 15N.

2. Questions:

Chapter 5 (The Laws of Motion) a. What is the final velocity? b. Assume the same block but now it weighs 2lb, its initial velocity is 5ft/s, the force is 6lbs, and the

distance is 15ft. what is the final velocity now? 3. Solution Strategy:

a. Draw a diagram of the system

b. Set up the dynamic equation to analyze the forces:

i. F = ma 6 = (2)a a = 3m / s2vf

2 = vi2 + 2a(x x ) v 2 = 52 + 2(3)(15) v = 10.73m/sc. Use kinematic equations along with the acceleration figure to calculate the final velocity:

i. f i fd. For question b, we simply switch values, but keep in mind the weight is not the mass, we need to

convert it to mass to get an accurate value! 2lb 1slug32 = 116

Page 18 of 59

i. lb slugF = ma 6lb = (116

ii. slug)(a) a = 96ft/s2

F = ma (6 19.6) = (2)a a = 6.8vf

2 = 52 + 2(6.8)(15) vf = 13.3i

F = ma (6lb 2lb) = 116

C. Example 2: 1. Problem You have a block with mass 2kg, which is being moved vertically with an initial

velocity of 5m/s for 15m. the force moving it is 15N 2. Questions:

a. What is the final velocity? b. Assume the same block but now it weighs 2lb, its initial velocity is 5ft/s, the

force is 6lbs, and the distance is 15ft. what is the final velocity now? 3. Solution Strategy:

a. In this instance, keep in mind that gravity is playing a part, so include it in our initial dynamic calculations. i.

ii. iii. since we get an imaginary number, this means with the given forces, the

block will never make it to 15m. b. for part b, we again remember to convert the lbs to slugs before figuring out the

acceleration. i. a a = 64ft/s2

vf2 = 5

ii. 2 + 2(64)(15) vf = 44ft/sD. Weight, Mass, and the Normal force

1. Although the normal force is sometimes hard to figure out, an intuitive way of thinking about it is think of it as your apparent weight because it is the normal force that gives us our feeling of weight. For example, a person in a rapidly accelerating and rising elevator will feel lighter because the normal force is higher (their weight will be the same). A person in the same elevator, which is rapidly accelerating and going down, will fell very little apparent weight, and as a matter of fact, if the acceleration of the elevator is the same as gravity, the elevator passenger will feel weightless.

E. Example 3 (friction): Comment [as8]: Notes for Monday, July 3, 2006 begin here

Chapter 5 (The Laws of Motion) 1. Problem You have a 3kg block being pulled with a force F. the coefficients of friction are

s= 0.6 and k= 0.4 2. Questions:

a. If the force applied is 5N, will the block move? If not, what is the magnitude of the force of static friction?

b. What is the minimum force needed to move the block? c. Once the block is in motion, what is the minimum force needed to keep it in motion without

accelerating it? d. If the applied force is 1N greater than the answer in b, what will the blocks acceleration be? e. If the applied force is 1N greater than the answer in c, what will the blocks acceleration be?

3. Solution Strategy:

Page 19 of 59

fs = Fapplied = 5N

(Fmin = fsmax )

a. See question b first, it will answer both a and b together.

i. (Based on b) No. ii.

b. We first need to use the equation of the force of static friction to figure out the maximum static friction force before the block slips. i. sn fsmax (.6)(29.4) = 17.64N

(Fmin =

c. Now we need to use the equation in part b but with k instead: fk ) kn fk (.4)(29.4) = 11.76N

F = =18.64 (17.64) = 3a a = 2.29

i. d. We figure out the acceleration (if any) by figuring out its net force:

i. ma F sn ma

3m/s2

F =12.76

ii. e. We do the same here except with the kinetic force:

i. ma F kn = ma2(11.76) = 3a a = .3m/sii.

F. Example 4: 1. Problem the problem is the same as in example 3, but with the external force pulling at an

angle of 50 above the horizontal 2. Questions:

a. Same as in example 3. b. Same as in example 3. c. Same as in example 3. d. Same as in example 3. e. Same as in example 3. f. What is the minimum force, which will lift the block? What will its acceleration be then?


a. See question b first, it will answer both of these questions i. (Based on b) No. ii. fs = 5Cos50 = 3.21

Fy = mg n + FSin50 = 29.4 n = 29.4 FSin50

b. Like before, we need to figure out the minimum force which can move the block, but this time, we need to keep in mind it is made up of two components, both of which affect the equation, so we will need to solve for two equations at once: i. Since the external force is reducing the normal force, we need to take it into account when

calculating the vertical forces: a)

Chapter 5 (The Laws of Motion)

Page 20 of 59

Fx = ma FCos50 sn = 3aFCos50

ii. When calculating the horizontal forces, we again need to calculate the horizontal component of the external force, this will combine both equations and give us the answer to b:

a)

(.6)(29.4 FSin50) = 0 .6428F 17.64 + .4596F = 01.103

b) F F= 17.64 =c) 16.002

FCos50

c. This is the same as the previous equation but the coefficient of static friction is replaced by the

coefficient of kinetic friction: (.4)(29.4 FSin50) = 0 .6428F 11.76 + .3064F = 0

.9492F = 11.76 Fi.

= 12.389F = 17

ii. d. For this question, we simply increase the force by 1N and solve it as a dynamic equation:

(17)Cos50 (.6)[29.4 (17)Sin50] = 3ai. ii. 10.927

F = 134 17.64 + 7.8137 = 3a a = .367m/s2

e. We do the same, except we use the kinetic values instead: .389 (13.389)Cos50 (.4)[29.4 (13.389)Sin50] = 3ai.

ii. 8.6063 2

N + FS

11.76 + 4.1026 = 3a a = .3163m/sf. In order to solve this, we set the normal force to zero and calculate the vertical acceleration. After

this, just plug it into the horizontal equation: in50 = 29.4 N = 0i.

ii. Fy =Fx = ma 38.378Cos50 (0) = 3a a = 8.223m/s2

ma FSin50 = 29.4 F = 38.379Niii.

G. Example 5: 1. Problem the problem is the same as in example 3, but with the external force pulling at an

angle of 50 above the horizontal and the block is on a 20 incline. 2. Questions:

a. Same as in example 3. b. Same as in example 3. c. Same as in example 3. d. Same as in example 3. e. Same as in example 3.

3. Solution Strategy: (here one important tip is to accurately record the component vectors for all relevant forces. A good idea is to rotate the inclined plane so that the incline is the x-axis. All angles would need to be readjusted, but this often makes it easier)

a. Like the previous two, we need to go to b to figure out a:

fs (5) =i. sn fs (5) = (.6)(27.6270 (5)Sin30) 15.07N =a) Since the force of static friction at 5N is less than the minimum force needed to move the block up

the incline (fsmax), the block will not move up. ii. Since the block is on an incline and we know it will not move up, we now need to know if the

block will move down. If not, then we need to find the force that keeps it from going down.

Chapter 5 (The Laws of Motion)

Page 21 of 59

fs(up) + F(5) = FDownRampfs(up) + (5fs(up) = 10fs(up) = 5.7253

Fy = ma n + FSin30 = mgCos20 n = 27.6270 0.5FFx = ma FCos30 (mgSin20 + sn) = 0

0.866F +

a) We need to figure out the force that stops the block from moving down. Since the pull is down and friction acts against this force too, the friction force (which pulls up) plus the external force pulling the block up should equal the force pulling the block down. 1)

2) )Cos30 = 29.4Sin203) .0554 4.33014)

b) So the force of static friction up the block (against gravitys pull) fs(up) is the force keeping the block from sliding down the ramp.

b. As in example 4, we must remember the force is at an angle, and will affect both the x and y components of the equation, so we need to keep this in mind. i.

ii. 0.5F)][10.055 (.6)(27.6270 = 0iii.

iv. 0.866F 26.631+ 0.3F = 0 1.16F = 26.631 F = fsmax = 22.8388

Fx =0.866F

c. As in example 4, but we change our equation for the kinetic force:

i. ma FCos30 (mgSin20 + sn) = 0 [10.055 + (.4)(27.6270 0.5F)] = 0ii.

0.866F 21.106 + 0.2F = 0 1.066F = 21.106 F = 19.799iii. d. Here we increase the minimum static force by 1 and calculate any acceleration. Keep in mind that

since we are exceeding the static force, we use the kinetic coefficient F =

Fy = 06Fx =

4.1098 = 3a a = 1.3700m/s2

23.8388i. ii. ma n = 27.620 0.5(23.8388) n = 16.20iii. ma (23.8388)Cos30 [10.055 + (.4)(16.2006)] = 3aiv.

e. Here again, we increase the minimum kinetic force by 1 and calculate any acceleration Fi. =

Fy =Fx = ma (20.799)Cos30 [10.055 + (.4) (17.2205)] = 3a

1.0693

VI.

20.799 ii. ma n = 27.6270 0.5(20.799) n = 17.2205iii.

iv. = 3a a = 0.3564m/s2Chapter 6: Circular Motion and Other Applications of Newtons Laws (return to top)

Comment [as9]: Notes for Wednesday, July 5, 2006 begin here

A. Background: 1. Objects traveling in a circular path can often be analyzed from a Newtonian perspective if we

keep in mind that Newtons 2nd law still applies. In this chapter, we will look at forces in horizontal and vertical uniform circular motion. We will also analyze through a viscous medium, which could be considered an extension of the principles of friction presented in the previous chapter.

B. Horizontal Circular Motion 1. When calculating Horizontal Circular Motion, we will usually be dealing with a y-component

that is in equilibrium (thus giving us a reference point from where to solve our problems). It is usually best to solve the y-component first for this reason.

Chapter 6 (Circular Motion and Other Applications of Newtons Laws) C. Example 1:

1. Problem A 6lb ball is being rotated about a rod. The ball is being held by two lines, the higher one at 40 to the horizontal and the lower line at 30 to the horizontal. The lower lines length is 4ft. The highest tension in either string is 40 lbs before snapping.

2. Questions: a. What is the shortest period that this ball can have?

3. Solution Strategy: a. First we keep in mind this is an extension Newtons 2nd law in that

both the horizontal and vertical components should be set against each other to determine the forces acting on them. We use the fact the y-components are in equilibrium, and the fact one of the two strings will reach 40lbs first (since angles and lengths are unequal between T1 and T2). Our guess as to which one reaches first should be a matter of instinct (in this case, T1 because T2 and the ball are pulling down on it), but failing that, we can still figure it out.

b. First we set the y-components equal to each other and set one to 40lbs. i. Fy = 0 (T1)(Sin40) = (T2 )(Sin30) + 6 (set T1 = 40)

T2 = (40)(Sin40) 6Sin30

Page 22 of 59

ii. T2 = 39.4230

Fx = r

iii. This proves our guess was right, when T1 hits 40, T2 will be close, at 39.42lbs c. Now that we know T1 and T2, we can figure out the shortest period they can have by doing a few

algebraic manipulations on our centripetal acceleration equation keeping in mind that its velocity is the same as its angular frequency () times its radius, and the angular frequency is a function of its period (T):

ma = mv2 Tand v = r and = 2i.

ii.

Fx = m(r )2r = mr 2 = mr 2T 2

= 42mr

T2

(T1)(Cos40) + (T )(Cos30) = 4

2 (6 / 32)(4Cos30) T = 0.629siii. 2 T2D. Example 2:

1. Problem You have a car driving around a circle of radius 120 ft. the coefficient of static friction is (.8)

2. Questions: a. What is the fastest the car can go without slipping?

3. Solution Strategy: a. Like all horizontal motion problems, we should start

out by figuring the vertical forces to get friction. n = mgi.

b. For horizontal forces, we recall friction is acting as a centripetal force.

ac = fs = sn ac = mv2

rsn = mv

2

ri.

Chapter 6 (Circular Motion and Other Applications of Newtons Laws)

(.8)(mg mv2

) =120

v = (.8)(120)g v = 55.42ft/s ii.

55.42ft/s 3600(sec/hr)5280(ft/mi)

Page 23 of 59

iii. = 37.8mi/hr

Fy = 0 N = mgCos( )Fx = ma mgSin( ) sn = 0

mgSin(

E. Example 3: (Review of Chapter 5) 1. Problem: (this review example will help us better understand how to solve example 4). You

have a block of mass m on an inclined plane. The coefficient of static friction is 0.6 and the coefficient of kinetic friction is 0.4.

2. Questions: a. What is the maximum inclination angle the plane can have before the block slips down? b. If is 10 higher than in question b, what will the acceleration be?

3. Solution Strategy: a. First we solve the y-component to figure out the normal force:

a)

ii. Now we figure out the x-component to get : a)

iii. Combine both equations: s[mgCos( )] ) =a)

mgSin()mgCos( )b) = Tan( ) = .6 = 30.96

n = mgCos(

s

b. For this question, we simply plug in 40.96 or 41 (for simplicitys sake) and get the answer. Dont forget to replace the coefficients of friction!

mgSin( ) kni. ) + = magSin(

ii. ) k[gCos( )] = a 9.8Sin(41) (.4)[(9.8)Cos(41)] = a iii. a = 3.47

4. Lessons: a. The next problem will be a car navigating a banked turn. The problem will be a combination of

exercise 2 (horizontal circular motion) and exercise 3 (a banked road). The main thing to figure out is the normal force.

F. Example 4: 1. Problem You have the same car from ex.2 but now the road is banked at an angle of 20. 2. Questions:

a. What is the maximum velocity the car can achieve without slipping? b. What is the minimum velocity the car can achieve without slipping?


Comment [as10]: This images on this page were done at 300dpi. The ones before it were only done at 72dpi which caused them to print poorly. The only thing is the file size is 4 times greater I decded to keep the image at 200dpi because it was the best tradeoff between size and graphics. The only way to improve the image would be to up it to 600dpi, but the file will become 5-10 times bigger and even worse, it will take forever to scroll through the document.

a. In order to figure out the maximum, we need to consider the fast car scenario: i. Time to figure out the vertical components to get the normal force:


Page 24 of 59

(n)Cos( mga) ) (s )(n)Sin( ) mg = 0 n = Cos( ) (s )Sin( )

ii. Now we set up the horizontal components to figure out the centripetal acceleration and plug in the y-based normal force equation to solve it. We remember that the force of static friction pulls the car back into a circular path, so one of its velocity vector components is centripetal. We also notice the normal force has a centripetal vector component too, so we put them together and form our centripetal acceleration equation:

mac = (n)Sin( ) + (s )(n)Cos( ) mv2

ra) = n[Sin( )+ (s )Cos( )]

mv2

r

iii. Now plug in the equation for the normal force and solve: mg

Cos() (s )Sin()

[Sin()+ (s )Cos()]=a)

mg[Sin() + (s )Cos()]Cos() (s )Sin()

v =

(g)(r)[Sin()+ (s )CoCos() (s )Sin(v

2 = rm

s()])

b)

v = (32)(120)[Sin(20) + (.8)Cos(20)]Cos(20) (.8)Sin(20) v 79.409ft/s=

79.409ft/s 3600(sec/hr)5280(ft/mi)

c)

d) = 54.14 mi/hr

n =

Cos( (s )Sin( )

b. The slow car scenario is actually quite simple; notice the only difference is the force of static friction is acting in the opposite direction, if you do the math, you will notice it means all you do is switch the signs in the final equation we have:

mgi. )+ mac = (n)Sin( ) (s )(n)Cos( )

mv2

r

-

= mgCos() s )Sin()ii. + (

[Sin() (s )Cos()]

v = (g)(r)[Sin() (s )Cos()]s )Sin()

-

-+ (

iii. Cos()

v = (32)(120)[Sin(20) (.8)Cos(20)]

8)Sin(20)-

+ (.iv.

Cos(20) v = 36.01i a) The imaginary result simply means there is NO minimum speed. You can stop on the ramp because

the friction force alone is enough to keep the car in place.

G. Vertical Circular Motion: 1. Vertical Circular Motion, unlike horizontal circular motion has a constantly changing Normal

force, which makes calculating much more complicated. Since the Normal force is gravitationally influenced, it is different at every point in the curve from its maximum at the bottom (where n and g vectors are at 180) to its minimum at the top (where n and g are pointing in the same direction). When solving equations with vertical circular motion, it is important to remember the direction of the normal force in relation to gravity (which is always pointing down and always the same magnitude).

2. Roller Coaster analogy: a. Revolution Type: in this type, the roller coaster does a loop-the-loop on the inside of the track (if

it does it on the outside, the normal force would point in the opposite direction) and the normal force points to the center of the circle. At the bottom, the normal force is greatest because it needs

Chapter 6 (Circular Motion and Other Applications of Newtons Laws) to counteract the force of gravity plus provide enough force result in centripetal acceleration. At the top, the normal force is least because both the normal force and gravity are pointing towards the center of the circle, thus, the normal force doesnt need much more to provide centripetal acceleration.

b. Colossus Type: in this type, the train goes up and down hills. The normal force is also greater at

the bottom and least at the top, but for slightly different reasons. At the bottom of the turn, the normal force acts just like it did in the revolution-type example, with both forces opposing each other and the normal force pointing to the center of the radius. At the top though, unlike the loop, the normal force points straight up (180 away from the center of the circle) so only gravity is providing centripetal acceleration if at all. If the velocity is high enough up the hill and the rider lacked a restraint, gravity would be unable to provide centripetal force (pushing the rider back into the curve), the normal force would be zero, and the rider would fly out and become a projectile. i. Nota Bene: dont try to memorize the equations above, if you understand the forces acting

on the car, you can easily figure them out anyways, and if you cant, then memorizing isnt going to help much at all either trust me!

H. Example 5: 1. Problem A designer is making a roller coaster similar to the one above with a revolution-

type loop and a few colossus-type hills and valleys. 2. Questions:

a. How fast should the ride be so the rider feels half their weight at the top of the hill? b. How fast should the ride be so the rider feels weightless at the top of the loop? c. How fast should the ride be so the rider feels twice as heavy at the bottom of the loop? d. How do we calculate the normal force at any given position in the circle?

3. Solution Strategy: a. Since the weight a person feels is actually the normal force, we just set the normal force to and

solve for the equation: normally n=mg so now n =

Page 25 of 59

i. 12mg

mg ntop = mv2

r

= mv2

rg = v

2

r mg mg

2 g 1

2ii.

g) v = rg2

v2 = r(12

= 4.9riii. b. Here again, we just set the normal force to zero and solve the top-of-loop equation:

ntop + mg = mv2

0 + mg = mv2

v2 = rmgi. r r m

v = rg = 9.8rii. Guess what, this is also the minimum speed to do the loop! c. Do you notice a pattern?

nbot mg = mv2

2mg mg = mv2

v2 = rmgm

i. r r

Chapter 6 (Circular Motion and Other Applications of Newtons Laws) v2 = rg = 9.8r Hey, thats the same value as at the top of the loop! ii.

d. This is just a matter of remembering that the normal force points to the center and is equal to the force of gravity and the cosine of the angle it makes with the vertical (n=mgCos) because the sine would actually be the tangent of the radius. Since at the bottom, we already know the equation, we simply add the cosine to allow for the change in angle actually; we could add it to the top or bottom, as long as the angles origin begins at the vertical down (e.g. the +x axis would be 90 and the x axis would be -90) and it will work fine.

Page 26 of 59

i. n mgCos = mv2

rn = m gCos v

2

r

I. Motion in the Presence of Resistive Forces (air, liquids) 1. Similarly to friction, resistive forces (air or water drag) oppose the objects direction, but

unlike friction, resistive forces in liquids tend to grow dramatically faster (the coefficient of kinetic friction can also vary with speed, but the change isnt notable until the object goes very fast).

Comment [as11]: Notes for Monday, July 10, 2006 begin here

a. In the case of slow moving particles or particles in a viscous fluid, the equation is: i. R = bv

R = 12

b. And in the case of fast moving objects (in air):

i. DAv2

when mg = bv vTerminal = mgb

a) D= drag coefficient (dimensionless) b) (rho)= density of air

2. Terminal Velocity: when an object is initially dropped, its acceleration is equal to gravity. Over time, the resistive force increases until it equals the downward force: this is terminal velocity. Terminal velocitys dependence on mass is the reason why heavier objects have a higher terminal velocity than lighter objects and why they hit the ground faster:

3. Developing the Terminal Velocity Equation: though its not essential to know how the equation is made (its not going to be on any tests), understanding how we got the equation provides a nice chance to review basic calc skills and to see how we tie together both the terminal velocity and force equations a. First we set up the conditions

if Fy = mg bv then ma = mg bv so m dvdt = mg bvdvdt

b. Now a few simple algebraic manipulations to prepare our integrate for time

= g bmv becomes dv

g (bm)(v) = dt

g

Next, we integrate this combined terminal velocity/force equation dv(bm)(v)

c. 0 0

Now, we set up the substitution necessary to integrate this

v

= dtt

let u=g (b m)(v) so du = bm v and dv =

mb

dud.

Now we put it all back together and manipulate the logs


Page 27 of 59

mbe.

udu = t m b

0

v

( )lnu g m(g) (b ) v( )m

b

f. ( ) ln[g (b m)(v)] ln(g)[ ]= t ln g bm( ) v( )g

=

bt m

Get rid of the natural log by raising it to the power of Eulers number g bmg. ( )(v)

g= e bt m( ) 1 bv

mg= e bt m( ) 1 e bt m( ) = bv

mg

h. And voila! The velocity of an object under drag as a function of time!

mgb 1 e

bt m( )

= v(t)

if t =

i. We should also note that: 0 then v = 0 if t v then mg b

4. Air-resistance modified kinematic equations: a. a(t) = v '(t) = ge bt m(

y(t) =

) The acceleration of an object under air drag mgb

[1 e bt m( )] dt =

mgb t +

mb e

bt m( ) mb

b. The position of an object under drag

J. Example 6: 1. Problem The instructor dropped a paper with a mass of 4.7g from a height of 2m. The time

it took for the paper to fall is 1.37s. 2. Questions:

a. Calculate b (the coefficient of resistance) from this data b. What is the terminal velocity for this paper? c. What is the final velocity of the paper and did it reach terminal velocity?

3. Solution Strategy: a. Since we have the position (y), time (t), and mass (m), we can plug this into the position terminal

velocity equation.

t + mbi. y(t) =mgb e

bt m( ) mb

2 = (.0047)(9.8)b

ii. [(1.37)+ .0047( )b

e(1.37 )(b )(.0047 )

(.0047)b

] b = 0.02764

mg = bv (.0047)(9.8) = (.02764)v vt

b. Just plug in all your known values into the terminal velocity equation = 1.6688

v(t) =

i. c. Use a kinematic equation to find the final velocity (velocity at time t) and compare

(.0047)(9.8)(.02764)

1 e (.02764 )(1.37 )(.0047 )( ) = 1.6659m/si.

vfvt

= 1.66591.6688

= 99.82% of terminal velocityii.

VII. Chapter 7: Energy and Energy Transfer (return to top) A. Background

1. Work: when a force is applied to an object and moves it, the resultant displacement of the object can be quantified as work. In this way, work is the product of a force and the distance

Chapter 7 (Energy and Energy Transfer) it moves the object. There are many ways to calculate it, but all are based on this basic concept. This should make it obvious but its worth stating: a force that does not result in displacement does no work. r r

W = F r = F r

Page 28 of 59

a. Cos

W =rF drr = ( rF)drr Cos 0( ) = (m

2. Relationship between Work and Kinetic Energy: Kinetic Energy has the same units (Joules) as work and is can actually be derived from the basic work equation to prove its equivalency if we assume the angle between the force and the system is zero, our equation works out like this

r r a)dr (1)

W = r r d

a. r r

b. ma( )dr = m vdt

drdt

vf 1 dt = mv( )dv

vi = 2 mv f2 - 21 mvi2

KE =

12mv2

K f = K + W 12

c. 3. Work-Energy Theorem: if work is done on a system by external forces and the only change

in the system is the speed, then the work is just the change in energy of the system. a. i mvf

2 = 12mvi2 + W b. This means that if a 1kg object is moving at 2m/s and its final velocity is 4m/s, its Work is the

same as its change of energy, which is 6 Joules. 4. Power: is the amount of energy (work) transferred during an interval or instant in time.

= Wt P instantaneous = limt0Wt =

dWdt

Paveragea.

B. Example 1: (comparison of dynamic versus work approaches) 1. Problem - A 2kg block with an initial velocity of

3m/s is moved 15 meters with a force of 10N. 2. Questions:

a. What is the final velocity of the box? b. What is the average power?

3. Dynamic (chapter 5) Solution Strategy: a. Figure out the acceleration (dynamics) and then

figure out the velocity (kinematics): F = ma 10 = 2a a = 5vf

2 = vi2 + 2ad vf2 = 32 + 2(5)(15) vf = 159 i.

=

W = F rr = (10) (15) = 10( ) 15

ii. 12.61m/sb. This question cannot be answered with Dynamic equations

4. Work (Chapter 7) Solution Strategy: a. To figure this out, we find the total work and then plug it into the work-energy theorem.

i. Set up the basic work equation and fill in the variables: r ( )Cos 0( )= 150J1

2

a)

ii. Now that you have work figured out, use the work-energy equation to get final velocity:

f2 = 12mvi2 + W 12 (2)vf2 = 12 (2)(3)2 +150 v f = 12.6mva) 1

vf = vi

b. To find the power, find the time the work was done and then solve + at 12.61 = 3+ 5t t = 1.925si.

Paverage = W =150J

1.925s78.13 wattii.

tC. Example 2: (how is work affected by a change in direction?)

Chapter 7 (Energy and Energy Transfer) 1. Problem The same block as the one above is pushed with three different forces

(everything else remains the same). The first force pushing at 40 to the horizontal, the second is pulling at 40 from the other side (140) and the last one is pulling up (90).

2. Questions: a. What is the work done with the force in the 40 direction? b. What is the work done with the force in the 90 direction? c. What is the work done with the force in the 140 direction?

Page 29 of 59

3. Solution Strategy: a. Figure out the work done at 40: r rW = F r = (10)(15)Cos(40) = 114.906

(10)(15)Cos(90)i.

b. Figure out the work done at 90: = 0

(10)(15)Cos(140)

i. a) The force doesnt move it, and so does no work

c. Figure out the work done at 140: = 114.9

W = 12

(The force is losing) i. ii. Is the result reliable?

mvf2

21

a) mvi2

114.9 = 12

)vf2 1

2(2)(3)2 = 10.29i(2b)

iii. We got an imaginary number, which means there is no solution. That means the block never makes it to +15m and probably moves backwards instead.

iv. If it never makes it to 15m, then where does it stop? W = K f Ki (x) 10( )Cos 40( )= 12 )(0)2 12 (2)(3)2(2a) x = 1.17 this is where the block stops moving forward (and then goes backwards). b)

D. Zero and Negative Work: 1. When a force is perpendicular to the displacement, no work is done.

a. The Normal force is always perpendicular to displacement, so it never does work. b. Any force acting as a centripetal force does no work.

2. When a force is greater than 90 to the displacement, it does negative work. a. Friction always does negative work.

E. Example 3: (work done when force varies) 1. Problem same block as in the previous

example with the force at 40 to the horizontal and the force is 10x

2. Questions: a. What is the work done on this system?

3. Solution Strategy: a. Since the force varies as a function of displacement, we integrate it to get the work:

i. W = (F)dr Cos ( )= 10x( )0

15

dx Cos(40) = 861.7994. Note:

a. In this example, the force (10x) will actually end up lifting the block at the 3.04 m mark, thus causing work in the vertical direction too. Since this is an introductory example, we wont cover the y-component, but in tests, if asked for total work, you must include the sum of both the horizontal and vertical work.

Chapter 7 (Energy and Energy Transfer) F. Example 4: (work with a non-conservative force)

Page 30 of 59

1. Problem the same block as before (2kg, vi=3m/s, xf=15m) but the force is now

rF = 10xi + 3x2 j

Fxi + Fy j( ) dxi + dyj( )0

15

= 10x( )dx0

15

Cos(0)i + 3x2( )?

?

(a non-conservative force) 2. Questions:

a. What is the total work done in the x and y directions?

3. Solution Strategy: a. Solve by integratingbut wait!

Comment [as12]: a.I need to get back to this example to finish the vertical component if possible.

i. dyCos(90) j = 1125i + 0 j4. Notes:

a. The work in the y direction is zero because displacement is zero. b. Although in theory, the 3x would result in a lifting of the block at 2.55 feet, the problem does not

state

G. Example 5: (work with a variable force across a limited path) 1. Problem This time the block from the previous examples is going in a straight line from the

points (1, 4) to (10, 2). The force is varying in both the x and y directions, but the path is limited to the straight line between the previously mentioned points.

2. Questions: rF = 10xi + 3y2 j ) a. Find the work done by a conservative force ( r

F = 10xi + 3 jx2b. Find the work done by a non-conservative force ( ) 3. Solution Strategy:

a. Here we integrate each based on their paths:

i. W = f (x)dxxi

xf

+ f (y)dyyi

yf

= 10x( )dx1

10

+ 3y2( )dy4

2

= 439Jb. The trick here is to solve the y parameter. Since it depends on the x parameter, the easiest thing

would be to integrate it on the x-scale (you can integrate it to the y-scale, but it would be unnecessarily difficult and time-consuming, especially given the fact you already have enough to solve it on the x-scale).

i. (Since we need to integrate it to x, we need to find and equivalent equation) 3x2( )dy?

?

y y1 = y2 y1x2 x1

(x x1)ii. (This is about the most no nonsense way to get the slope, since its based on the points were already given no need to find functions based on the forces because the path is limited by the straight line)

Chapter 7 (Energy and Energy Transfer)

Page 31 of 59

y 4 = 2 4iii. 10 1 x 1 y = 9

2 x 92 + 4 y =

92 x +

934

Now the trick We differentiate implicitly to get our slope: dy = 2 9iv. dxW =

Now we can replace the dy with the dx in our integral and solve

3x2( ) 2 9dx( )= 1

10

222

Wtotal =

v.

The work is negative since the positive force is working against negative displacement. Now we add the x work and the y work and were done

vi. Wx +Wy = 495 222 = 273Jc. Extra: we want to change the path from a line to a parabola and compare the work done by

conservative and non-conservative forces on this path. i. The conservative force would do the same work as before (439J) ii. The non-conservative force (3x) though, would require a whole recalculating of the path

times the force as we did in the first half of this example. a) First we use the quadratic formula for our model of a parabola and meld it with our current x-

equation. Since our points are given, we can create two equations that we can add together to give us one variable, from which we will then get our second variable.

b) y = ax2 + bx4 = a(1)2 + b(1)

= a(10)2 + b(10) 2 = 100a +10b10(4 = 0

c) At y=4, the equation is: d) At y=2 (the end), the equation is:

2e) Set up one equation to cancel a variable in the second equation and you get:

a + b) 4 10a= +10b(2 40) = (100a

f) Subtract the second equation from the first one to get:

10a) (10b 10b) 38 = 90a a + = 19 45 ) + b bg) From which we get: 4 = ( 19 = 199 4545

x2 + 199 45y = 19 45h) Which gives us the equation of the parabola: x

y = 19 45

iii. Now we differentiate (like last time) to get the equivalent x-equation that we can integrate: x2 + 199 45 x dy = (2)(19 45)(x)dx + 199 45dx

dy = 38 45 xdx + 199 45dx dy = ( 38 45 x + 199 45)dx

Wtot =Wx +Wy = 495J + 3x2

iv. Now we put it back together and integrate on the x

38 45 x + 199 45( )dx = 495 1914.9 = 141

10

19.9 v. As is obvious, the conservative path is still the same because it only depends on the

beginning and end points. The non-conservative path, though, is totally dependent on the path, so we need to integrate the force on the parabolic path. So we decide on a quadratic equation (we could chose any equation for the parabola, but a quadratic is easier because of its familiarity) and fit the y parameter to conform to the x parameter. Once we get the path, we differentiate it implicitly to get the equivalency between y and x slopes (remember, derivatives are slopes?). This, times our non-conservative force can now be integrated over the x to get the work based on the path and the force on every point of that path.

H. What makes a force conservative? 1. Conservative Forces are forces that only depend on the initial and final position to

determine work. The path is irrelevant. Conservative forces always have a potential energy (chapter 8) associated with them.

Comment [Wednesday,here

as13]: Notes for July 12, 2006 begin

Chapter 7 (Energy and Energy Transfer) 2. Non-conservative Forces are dependent on the path taken and so any irregular path

must be taken into account when calculating work done. They do NOT have a potential energy associated with them.

I. Work Done by Gravity Forces (conservative) 1. Gravity is a conservative force which we can prove by comparing

three scenarios: 2. Work done in a freefall: r

Wg = F drr = F d Cos = (mg)(Hi H f )Cos(0) = mgH

Wg = mgH = 12

a. b. If it hits the ground(to find the velocity upon impact)

mv2 v f = 2ghi. 3. Work done (against Gravity) going back up:

Wg = F d Cos

Page 32 of 59

a. = (mg)(H H )Cos(180)f mg = i H

Wg = F

4. Work done along a frictionless incline: a. d Cos = (mg)(d)Cos( )

dCos(

b. ) = HWg = (mg)H = mg

Ws =

rF drr = (kx)dx

0

xf

= 12

So c. H

5. Conclusion we can see that work done by or against gravity is conservative because it doesnt depend on the path (straight drop or a drop on an incline). The fact that raising the block performs negative work also is part of why gravity is able to store potential energy.

J. Work done by Spring Forces (conservative) 1. A Spring is another example of a conservative force, which we can prove by calculating the

work done based on Hookes Law (F=-kx): 2. Work Done in compression/extension:

a. kx f2

Ws = kx( )dxxf

0

= 12

3. Work done by the spring returning to its origin:

a. kx f2

4. Conclusion: work done on a spring is the same as work done against gravity; it results in negative work (which is where a conservative forces potential energy is stored. When the spring returns to its natural position, the work it does then is positive.

K. Work done by Friction Forces (non-conservative) 1. This work is non-conservative because it always

depends on the path taken. Since the energy used up by friction is only stored as heat, it cannot return to the system and is essentially wasted; so no conservation of forces occurs.

2. Calculation of friction work (flat surface):


Wfk =

r rF dr = fk dxCos(180)0 = kn dx

0 = knd

d d

a.

b. Conclusion: as we can see, frictions work is negative because it always in a direction opposite to the displacement. We also see that the path of friction directly affects the work output.

L. Calculation of velocity on rough slopes of different lengths (showing friction depends on path)

1. Work done on the block by the time it gets to the bottom of the ramp

mvf2 12mvi2 = 12mvf2 0 = 1 2Wblock = 12

Page 33 of 59

mv f2

Wblock =Wg +Wf 12

2. Calculating the work done by gravity minus friction to get our final velocity:

a. mvf2 = mgH + (knd)

1

mv mgH mgCosf2 = k d 12b. 2 vf2 = gH kgCosd

12

vf2 = g(H kCosd) v f = 2g(H - kdCos )c.

3. Conclusion: we can see in our final equation that the work done by gravity is dependent on the path and so the greater path (the longer slope) will result in greater friction work, and as a result, end with a lower final velocity.

M. Example 6: (work with a variable force, Pt II) 1. Problem A block is being pulled with a variable force of 2x at an angle of 40 for a distance

of 20m. The coefficient of kinetic friction is 0.1. 2. Questions:

a. What are the work done by the force (F), by friction (fk) and the total work? b. Find the final velocity. c. How long can this block be pulled before it is lifted?

3. Solution Strategy: a. Finding the first part is a simple work integration,

friction isnt much more complicated:

i. WF = 2x( )dx0

20

Cos40 = 306.418Jn + 2xSin

ii. Calculate the normal force and use it to figure out friction (keeping in mind k is constant): = mg n = (3)(9.8) 2xSin40 n 29.4 = 2xSin40

Wfk = fk d = kn d = k n( )dxCos(0) =0

20

Wfk = xSin ( )dx

20

= J

WTotal =WF +Wfk = (306.418) + (

(.1) 29.4 2 400 33.0885

iii. Put them all together: 33.0885) = 273.3295J

Wtotal

b. Remember the work-energy theorem = K 273.330 = 12 (3)vf2 0 v f = 13.50 m si.


Page 34 of 59

2c. Just set the normal force equal to gravity to find the position at liftoff.

xSin40 = 29.4 x =i. 22.869VIII.

Chapter 8: Potential Energy (return to top) A. Background:

1. Conservative Forces: as stated in the earlier section on conservative forces, a force that is path independent (gravity, spring) is conservative and by its nature, stores energy done against it as potential energy. a. Path Independent: This is the first condition of a conservative force. The work done does not

depend on the path, only the displacement between initial and final position. b. Work on a closed path: This is the second condition of conservative forces. When an object goes

around a complete path under the influence of a conservative force, then the total work is zero ri. A d

rr = 0c. Component notation: for conservative forces, the component and its magnitude will be from the

same direction. rF = 2x i j

i j2i (3y + 5) j Is conservative. Notice the is dependent on the x and the is

dependent only on the -3y-5. If instead, the were dependent on the y and/or the were dependent on the x in any way, the equation would become non-conservative.

i.

d. Relationship to Potential Energy: there is always a function associated with conservative forces such that:

i.

rFc = U = dUdr =

dUdx

i dUdy

j dUdz

k = e. Units of Energy: the typical units are Newton Meters (Nm) aka Joules (J)

2. Non-Conservative Forces: if a force is NOT path independent and/or if the work done by a particle traveling a closed path is NOT zero, then the force is Non-Conservative (in other words, if any of the two conditions for conservative forces is not met). And non-conservative forces have no potential energy function associated with them. a. Component Notation: non-conservative forces will have at least one component dependent on a

magnitude from a different direction. ri. F = 2x i 3 j j

rFNonConservative drr 0

2 x The is dependent on the x instead of the y. b. Work on a closed path: unlike conservative forces, work always adds up as a path is traversed, so

work on a closed path will NEVER equal zero.

i.

B. Example 1: r1. Problem - you have the following functions: F = 2x2i (3y + 5) j

, 2. Questions:

a. Find the associated Potential Energy function of this force. 3. Solution Strategy:

a. We go back to the relationship between a force and potential energy:

i. rFc = U = dUdx i

dUdy

j

dUdx

= 2x2 and dUdy

ii. = (3y 5)

then Ux = 2x3

3

and Uy = 3y2

2 5yiii.

Chapter 8 (Potential Energy)

Page 35 of 59

so 2x3

iv. U(x, y) = 3

3y2 2 5y + C = 3

2x3 +2

3y2 + 5y + C Joules C. Example 2: (Comparison of Methods of Calculation; Ch5, Ch7, & Ch8)

1. Problem You have a block which is dropped from an initial height of yi, 2. Questions:

a. Calculate its final velocity using Newtons Laws (Chapter 5) b. Calculate its final velocity using the Work-Kinetic Energy Theorem (Chapter 7) c. Calculate its final velocity using the Potential Energy Method (Chapter 8)

3. Solution Strategy: a. Actually, our solution harkens back to Chapter 2 (Kinematic Equations)

vf2 = vi2 + 2a(x f xi ) = 0 2g(0 yi ) = 2gy v f = 2gy i.

b. Here we use the fact that work is a change in kinetic energy i. Wg = K mgh = 12mv2 12f mv2 = 12i mv2 0

gh = 12f

vf2 v = 2ghii.

c. Even though we saw this in Chapter 7, including the U makes it more Chapter 8-like i. WNonCon

0 = (K+WExtForces = K + U

ii. since no non-conservative or external forces are acting on the block f Ki ) + (U f Ui ) = (K f 0) + (0 Ui ) = K f Ui

0 = 12

vf2 = gh v f = 2ghmv mghf2 12

D. Example 3: putting it all together 1. Problem you have a block resting on top of an incline of 80ft at an angle of 25 with a

coefficient of kinetic friction of 0.2. The block will slide down the ramp and travel through a loop with a 10m radius and a coefficient of static friction of 0.4. After passing the loop, it will compress a spring with a stiffness constant of 250 lb/ft. In this example, ignore the constant change of normal force in the loop and assume it is 10.

2. Questions:

a. What is the final velocity at the top of the circle? Will it complete the circle? b. How much would the spring be compressed by if the height were 100ft?

3. Solution Strategy: a. We calculate total work with the Work-Kinetic-Potential energy equation:

i. WNonCon +WExtForces = K + Uii. Time to figure out the variables involved in the inclines friction calculation:

k1 = (0.2) n1 = 6Cos25 d1 = 100Sin25

Chapter 8 (Potential Energy) iii. Now we figure out the variables in the circles friction calculation:

(keep in mind the distance traveled is really the circumference at this point)

Page 36 of 59

k2 = (0.4) n = (10) d = (2)1 1 (10)2 = 10

K = K f Ki = 12

iv. There is only final kinetic energy, which is where we will get our velocity from 6

32( )vf2 0 = 3 32vf2U =U

v. Total potential energy is the sum of the initial height minus the loops height. f Ui = mghf mghi = mg2r mghi = 6( ) 2( ) 10( ) 6( ) 80( )= 360

0 +

vi. Now we set up the equation with the known variables: k1n1d1( ) k2n2d2( )= 12mvf2 12mvi2( )+ U f Ui( )

0.2( ) 6Cos25( ) 80 Sin25( ) 0.4( ) 10( ) 10( )= 3 32( )+ 360( )vf2 vii. And the solution is: , but does it make the loop? vf = 17.424 m/sviii. Time to use the equation we developed back in chapter 6, example 5:

Vmin = rg = (10)(32) = 17.88 the blocks speed is a tad smaller than the minimum required to travel the loop. This means that at some point before it hits the top of the circle, the normal force will be less than the weight of the block and for a few moments, the block will behave like a projectile until it encounters the other side of the loop and returns on its path. Given the tiny difference, this is likely to be a tiny amount, but the math required to figure this out would be far greater than the math required to figure out the example, so we will ignore that for now and leave this problem at that.

b. Now we reuse the equation we previously had, but keep in mind the height is now 100 feet and now the block is following the entire circle (not just half of it). Also, since the initial and final speeds of the block will be zero (its velocity is zero at maximum spring compression), we need only calculate the potential energy, and for final potential energy, we replace it with the spring potential energy equation we got from Chapter 7, section J. i. k1n1d1( ) k2n2d2( )= U f Ui( )

0.2( ) 6Cos25( ) 100 Sin25ii. ( ) 0.4( ) 10( ) 210( )= 1 2 250x2( ) 6( ) 100( ) iii. And we get x=0.8547ft.

E. Equilibrium: 1. Equilibrium and Potential Energy: all potential energy functions have points of equilibrium,

but the equilibrium can have three different types: a. Stable Equilibrium when moved away from this equilibrium, the particle will return to

it. Graphically, this looks like a U-shaped graph and the point of equilibrium is at the bottom.

b. Neutral equilibrium when moved away from this position, the particle will stay where it is so long as where it is moved is also a point of equilibrium. This is like a U shaped graph where the bottom of the U is extra long, looking like a floor.

c. Unstable equilibrium when moved away from this point of equilibrium, the particle will not return to it and instead will move away from it. Graphically, this looks like an upside

down U-shaped graph with the particle at the top.

F. Example 4: 1. Problem: you have a the potential energy function for a spring which is - kx 2. Questions:

a. Find the associated force function b. Find the points of stable & unstable equilibrium

Comment [as14]: Notes for Monday, July 17, 2006 begin here

Chapter 8 (Potential Energy) 3. Solution Strategy:

a. When trading between energy and force, we can use this simple relationship (if you notice example 1, it follows this form):

Page 37 of 59

Uderive

integrate F

U(x)i.

F(x) 12 kx2 F(x)dUdx

ii.

iii. 12 kx2 = (k) dx

dU1

2 x2 = kx

b. The points of stable & unstable equilibrium are the local max and min of U(x) or F(x): i. There is only one point of equilibrium and its at zero and stable. U(x) =

(x) = x3 + 6x2 7x +12J

erivative r

F(x) = 0 3x2 +12x 7 = 0

0 @ x=0

G. Example 5: 1. Problem: You have the Potential Energy function u 2. Questions:

a. Calculate the points of stable & unstable equilibrium. b. Calculate the associated force function. c. What is the U(x) for the points of equilibrium. d. What will happen if this particle is placed at x=0, x=-2, x=2

3. Solution Strategy: a. First, refer to step b which gives us the force, from which we solve to

get the zeros of the force. Once we find those, we run the 2nd dtest on the zero forces to see if they are local maxima (unstable) ominima (stable). i.

F xii. ( ) = 0 at x = 0.709, 3.291U ''(

x F x x) = '( ) = 6 +12 use this for local max/min

calculations iii.

F '(0.709) = +7.7 positive 2nd derivative = concave up = local min = stable

iv.

F '(3.29

dUdx

1) = 7.7v. negative 2nd derivative = concave down = local max = unstable

b. Here we just remember the relationship: u(x) derive F(x)

i. x 7 = F(x)

U(.709 1)

3 + 6x2 7x +12( )= 3x2 +12x c. Simple plug-n-chug

) = 9.697 U(3.29 =i. 18.303

IX.

d. A view of the graph is the easiest thing to use to figure this out. Imagine the potential energy graph

is a path and a marble was placed at any of the points in question. i. If its placed at x=0, it would oscillate from x=0 and x=1.59 (the two places where its y-value

is 12) and if friction was a factor, it would eventually stop at 0.709. ii. If a marble was placed at x=2, it would oscillate between x=2 and x=-0.236 (the two places

where its y-value is 14). iii. If the marble was placed at x=-2, it would roll towards the right, but since the

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