general physics laboratory manual and ......the quantitative relationship between the physical...
TRANSCRIPT
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ÇANAKKALE ONSEKIZ MART UNIVERSITY
DEPARTMENT OF PHYSICS
GENERAL PHYSICS
LABORATORY MANUAL
AND
WORKBOOK
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TABLE OF CONTENTS
LABORATORY INSTRUCTIONS ……………………………………………………………… 3
REPORT W RITING……………………………………………………………………………… 4
EXPERIMENTAL ERRORS ……………………………………………………………………. 5
SIGNIFICANT FIGURES……………………………………………………………………. 8
GPARHS ...……..………………………………………………………………………………… 10
EXPERIMENT 1: MOTION WITH CONSTANT VELOCITY ………………...…………....... 12
EXPERIMENT 2: STRAIGHT LINE MOTION WITH CONSTANT ACCELERATION AND
MOTION IN A PLANE………………………………………………………………………….. 17
EXPERIMENT 3: NEWTON‟S LAW OF MOTION: AN APPLICATION WITH THE
ATWOOD‟S MACHINE...……………………………………………………….…..….………. 24
EXPERIMENT 4: SIMPLE PENDULUM..………………………………………………….….. 30
EXPERIMENT 5: COLLISIONS AND CONSERVATION OF LINEAR MOMENTUM …... 33
EXPERIMENT 6: ROTATIONAL MOTION……………………………………………….…… 42
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LABORATORY INSTRUCTIONS
1. The most important thing in the laboratory is your safety. The dangers mostly result from a lack of knowledge of the equipment and procedures.
2. Personal safety rules must be obeyed with extreme discipline.
3. When you enter the laboratory never play with the equipment until it has been explained and the instructor has given permission.
4. Keep your experimental equipment and tabletop clean.
5. Report any accident to your instructor immediately.
6. The most of the equipment used in the laboratory is expensive and some of them are delicate. Even after you are familiar with the equipment, always have your experimental set up checked and approved by the instructor before putting it into
operation.
7. If any of the equipment is broken or does not function properly, report it to the instructor.
8. Read and study the experiments before you come to the laboratory.
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REPORT WRITING
The experiment report must consist of the following parts:
TITLE
PURPOSE OF THE EXPERIMENT
PROCEDURE
TABLES
CALCULATIONS
RESULTS AND DISCUSSION
1. TITLE
The title must be as brief as possible.
2. PURPOSE OF THE EXPERIMENT
The purpose of the experiments is outlined in this part.
3. PROCEDURE
The procedure of the experiment and the experimental set up are explained in this part.
4. TABLES
The measured data and the calculated results are presented in the tables.
5. CALCULATIONS
According to the purpose of the experiment, the calculations are done and the graphs are plotted in this part.
First, the formulas must be outlined. If there are multiple formulas of equations each one must be numbered. Second, the units of every parameter and result must be
written.
Appropriately scaled graph papers (logarithmic, linear… ) must be chosen. The axis must be labeled and scaled.
6. RESULTS AND DISCUSSIONS
In this part, the theoretical and experimental results must be compared. The
corresponding errors must be outlined and discussed.
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EXPERIMENTAL ERRORS
Physics is based on the measurements of the physical quantities such as light, time,
mass, speed, velocity, acceleration, force, momentum, temperature…. The measurements must be performed by taking into account the error limits of the measurement set up. As an
example, using a ruler results in an error. The zero mark on the ruler is usually a line, which has a certain thickness. Additionally, the looking direction of the researcher to the zero line also may result in an error. For this reason, a certain measurement is not possible.
Figure 1. Measurement error of a ruler.
In the physics laboratory, many different measurement instruments are used such as ruler, voltmeter, thermometer…. The errors due to these instruments are called as
experimental errors and the quantity of the errors can be estimated. The most accurate result of the measurement can be obtained by considering the error of the measurement.
The experimental error caused by the instrument in measuring can be identified as the smallest unit that can be read from the instrument. For example, for a ruler with a millimeter
division, the error caused by the ruler is 1 mm or 0.1 cm. If the measured length is 2.5 cm, the most accurate result will be cm. Additionally, errors should also be taken into
account when using measurements in calculations.
Let's consider X and Y measurements, that have errors and , respectively. To find R = X + Y with the closest accuracy, ΔR must be calculated and must be used,
such as:
(1)
(2)
Pencil
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(3)
(4)
For, R = X – Y
(5)
When measured values are added or subtracted, the maximum error is the sum
of the errors of both magnitudes.
Let's consider X and Y measurements, that have errors and , respectively. In
order to calculate P=X.Y, ΔP must be calculated to find the result with the closest accuracy. It can be obtained by:
(6)
In the Equation 6. the last term
can be neglected since it is respectively smaller than the
other terms.
(7)
(8)
As a result, Equation 9. is obtained.
(9)
The percentage error can be written as:
(10)
For the division,
, similar to the previous calculations the error „D‟ and the percentage
error can be written as,
(11)
7
(12)
respectively.
For X, Y and Z measurements the percentage error of
the following equation
can be used:
(13)
For the trigonometric calculations such as,
R = sin
the following equation can be used:
R = (14)
where is the error in the angle .
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SIGNIFICANT FIGURES
One of the ways of indicating the error in a measurement is by the number of figures
recorded. For example, for the distance measured with a ruler marked in millimeters we
record 2.550.05 cm, so this distance would be written with three significant figures, 2.55.
In recording measurements, the correct number of significant figures must be used and
when possible, the more definite indication must be added. In order to record the results of the measurement correctly, the digit including the last one estimated are called significant
figures. The number of significant figures has nothing to do with the location of the decimal
point. The number of significant figures in a measurement is determined as follows.
1. The leftmost nonzero digit is the most significant. 2. If there is no decimal point, the rightmost nonzero digit is the least significant. 3. If there is a decimal point, the rightmost digit is the least significant, even if it
is zero. 4. All digits between the least and most significant digits are considered to be
significant. Example:
Some measurements are recorded as; 255 mm; 15.2 cm; 7.09 V; 245000 mA; 0.0523 km, 4.70
g. How many significant figures do we have in these measurements? Answer: We have three significant figures in each of these measurements.
Significant notation: Write the results of a measurement as two factors; the first factor
contains all the significant figures, having one nonzero digit in front of the decimal point, and the second factor, is a power of 10.
Example:
255 mm = 2.55x102 mm 2.04 V = 2.04x100 mm
170000 mA = 1.75x105 mA
Multiplication and division: In the multiplication or division of numerical measurements, retain in the result only as many figures as the number of significant figures in the least
accurate number used in the calculation.
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Addition and subtraction: when carrying out addition and subtraction with numerical
measurements, do not carry out the result beyond the first column (from the left) that contains a doubtful digit.
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GRAPHS
The quantitative relationship between the physical quantities involved in an experiment,
these quantities are presented in tables and/or graphs. A graph must include the following
properties:
1. The title of the graph must be written on the graph paper. 2. The axis must be labeled with the quantity plotted and its proper units must be written. 3. The graph must be scaled in a presentation way that all of the data must be distributed
as wide as possible over the area of the graph paper. 4. The error bars representing the uncertainty of the measurements should be plotted in
each data point.
Figure 2. Example of a straight-line graph with error bars.
For equation, where x and y are the variables and m and b are the constants.
When the graph of these two quantities is plotted, the graph is a straight line. In order to plot this graph first, the data must be pointed on the coordinate system, starting from (x1, y1), (x2, y2)… etc. Second, the error bars must be drawn, such as in x-axis and in y-axis. In
practice, the measured points could be scattered because of the error they contain. In such a
case, The best straight line passing through the data points must be drawn. The best straight line can be drawn by using a transparent ruler. The best straight line should pass from as many the data points as possible and equal number of data points should be distributed on
either side of the best line. Additionally, the best straight line must pass inside as many error bars as possible. Third, the worst straight line must be drawn by taking the error bars into
consideration as shown in Figure 1. The slope of the best straight line gives „m‟ and the intersection of the line with the x axis gives „b‟.
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(15)
The error „m‟ is calculated from the absolute difference between the slopes of the best and the worst straight lines.
(16)
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EXPERIMENT 1
MOTION WITH CONSTANT VELOCITY
PURPOSE
The purpose of this experiment is to verify that an object moving under the influence
of no net force moves in a straight line and with constant velocity and to calculate this velocity.
THEORY
Motion is a continuous change of position in time. There are different types of motion. The simplest of these is the motion in a straight line with constant velocity. In this type of motion, the moving object travels equal distances in equal time intervals along a straight line.
The Newton‟s first law states that an object at rest will remain at rest, and an object moving in a straight line with constant velocity will remain so unless a net force acts on the object.
Therefore, an object moves in a straight line with constant velocity experiences no net force. To describe the motion of the object, the object is modeled by point particle. A vector
known as the position vector is introduced to give the position of the moving object with
respect to a given origin of the coordinates. Obviously, this position vector will be a function of time; . This is because a moving particle, that is changing its position with time,
will be at in the instant and in another instant at . This explicit form of the dependence of position on time will be determined by the specific type of motion of the
particle. The average velocity , of the particle is defined as the average change in the
position vector of this particle in a given time interval.
(1.1)
Where means the difference. The instantaneous velocity is defined as:
(1.2)
So the instantaneous velocity is just the derivative of with respect to time and obviously,
it is a vector quantity.
In the straight- line motion, things are a bit simpler. Since this is a one-dimensional motion then by taking the x-axis to be in the direction of the motion, reduces to , which is the displacement of the particle along the x-axis. In this case, the average velocity will be
(1.3)
And the instantaneous velocity will be
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(1.4)
In this motion
must be constant. Therefore, the general form of x(t) will be:
(1.5)
where b and c are constants. Clearly, this constant „b‟ is the instantaneous velocity. The other constant „c‟ is the initial position that is
(1.6)
Finally, the displacement x(t) of a particle moving in a straight line with constant velocity will be given as a function of time in the form
(1.7)
If at t=0, the initial position is the origin ( ), then
(1.8)
From the above relation, It is obvious that for a particle moving in a straight line with constant velocity, if one makes different measurements of position x at different instants of time, and tabulates them as ( etc., and then plots a graph of x versus t using
these data points, then one will get a straight line (see Figure 1-1).
FIGURE 1.1 the x versus t graph of an object moving in a straight line with constant
velocity.
In this experiment, you are going to study, analyze and calculate the velocity of an object moving in a straight line with constant velocity. the puck moving on the surface on the air table will be considered as a particle. A puck released to move freely o n a horizontal
(leveled off) air table can be considered as experiencing no net force since the air table is horizontal and the friction is almost eliminated. Therefore if you push a puck on the surface of
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the air table, it will move in a straight line with constant velocity at the moment you release it.
The x-t record of the position will be provided by the dots on the datasheet (see Figure 1.2). The displacement x of the given dot, first dot, for instance, can be measured directly using a
ruler. The time elapsed to travel this displacement can be determined by counting the intervals from the reference dot and multiplying by the time intervals between two successive dots which is just , where is the frequency of the spark timer.
FIGURE 1.2 The dots produced by the puck on the datasheet.
EQUIPMENT
Air table, ruler, millimetric graph paper.
PROCEDURE
1. Level off the air table.
2. Place the conductive carbon paper and the datasheet on the glass plate of the air
table.
3. Keep one of the pucks stationary by placing it at one corner of the air table and
putting a folded piece of paper under it. In this experiment, only one puck will be used.
4. Set the spark timer frequency to 20 Hz. 5. Activate the footswitch (P) and push the puck diagonally across the surface of the
air table at the moment you released the puck, activate the footswitch (S) to start
the spark timer. Let the puck travel the whole diagonal distance across the air table, and then remove your feet off the (S) and (P) switches.
6. Remove the datasheet from the air table. Number the dots as 0,1,2,… etc. the first dot can be taken as dot 0. Measure the distance of five dots from dot 0 (see figure 1.3) and finding the time corresponding to each dot, fill in Table 1.1. the x and t
measurements in this table should be filled in along with their corresponding errors.
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FIGURE 1.3 Analysis of the data points.
7. Using the data points in Table 1.1, plot the graph of the position x versus the time t on linear graph paper. Plot the position in centimeters along the vertical axis, And the time in seconds along the horizontal axis. Label the axes, and write down the
corresponding unit on each axis. Plot the data points along with the error bars. Note that the distribution of the data points suggests a straight line fitting. Is this an
expected result? drop the best and worst straight lines for your data. 8. Find the slopes of the best line and the worst line . Calculate the error in the
slope . From these slopes find .
9. Using the data in Table 1.2 construct the Tablo 1.2 Find the average velocity for each intervals and fill in the table with the correct number of significant figures.
RESULTS AND DISCUSSION
1. Fill in your x and t measurements with the corresponding errors in Table 1.1.
TABLE1.1
Dot number x (cm) t (sec)
0 0 0
1
2
3
4
5
2. Are the data points on the datasheet evenly spaced? Is this an expected result?
Why?
………………………………………………………………………………………………………………………………………………………………………………
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3. Show how to find the error for only one time measurement below.
………………………………………………………………………………………
………………………………………………………………………………………
4.
5.
6. Using Table 1.1, construct Table 1.2 below.
TABLE 1.2
(cm)
(cm)
(cm)
(s)
(s)
(s)
0 – 1
1 – 2
2 – 3
3 – 4
4 - 5
6. In the space below show how to calculate and the error in for any
intervals (one internal is sufficient). ………………………………………………………………………………………
………………………………………………………………………………………
7. How does the average velocity calculated for each interval in Table 1.2 compared
with the velocity calculated from the graph.
………………………………………………………………………………………………………………………………………………………………………………
8. Write down the comments related to the experiment and/or elaborate on and
discuss any points.
………………………………………………………………………………………
………………………………………………………………………………………
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EXPERIMENT 2
STRAIGHT LINE MOTION WITH CONSTANT
ACCELERATION AND MOTION IN A PLANE
PURPOSE
The purpose of this experiment is to study and analyze the straight line motion
with constant acceleration, and to find this acceleration for a puck moving on an
inclined air table. The motion of a horizontally projected puck on the incline air table
also will be studied and analyzed.
THEORY
In this experiment, you were concerned with the motion in which a puck on an air
table was moving along the x-axis with a constant velocity and you found a linear relationship between displacement and the time for this motion.
In this experiment, we will consider the motion of a puck moving in straight line in
such a way that its velocity changes uniformly. Consider an air table whose backside is raised so as to form a smooth inclined plane as shown in Figure 2.1a. If we put a puck at the top of
the incline and allow it to move down, we observed that the puck still moves in a straight path, but the dots produced on the datasheet are no longer evenly spaced, as shown in Figure 2.1b. This means that the puck‟s velocity increases as it goes down the incline. If the velocity
of the puck changes with time, we say that it has acceleration. Just as the velocity is the rate of change of the position, acceleration is the rate of change of the velocity.
(a) (b)
Figure 2-1. (a) The set up for the puck moving down on an inclined air table b) The dots produced by the park on the datasheet.
Note that, the positive x-axis is taken to be in the direction of the puck‟s motion. The type of motion that you have observed is straight- line motion with constant acceleration.
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Suppose that at time t1 the puck is at A and has velocity v1 and that at a later time t2 it
is at point B and has velocity v2. The average acceleration of the puck in the interval t is defined as:
(2.1)
Similar to the definition of the instantaneous velocity the instantaneous acceleration of
the puck in the x-direction is:
(2.2)
The acceleration is also a vector quantity and is always in the direction of v. It may
not be in the direction of the motion. Suppose that at the time t1=0, the pack is at position x0
and his velocity v0, and that at a later time t2=t. It is at position x and has velocity v. If the acceleration of the puck is constant, the average exclamation and instantaneous acceleration
are equal to each other, and therefore we find,
(2.3)
or,
v = v0 + a.t (2.4)
The expression for the position x of the puck is a function of time can be written as
(2.5)
where xo=x(t = 0 ) is the position of the puck at t = 0.
This equation may is easily be checked by taking to derivative dx/dt and comparing it with the equation for the velocity v. If the puck starts from rest (vo = 0), then it's position at any instant of time is given as
(2.6)
Therefore if the graph of x versus t2 is plotted, we've obtained a straight line that has a slope 1/2a and intercept x0. If, in addition, x0 =0, then this straight line will test through the origin.
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(a) (b)
FIGURE 2.2. Horizontally projected puck on an inclined air table. (a) Schematic diagram of the trajectory. (b) The data points produced by the puck on the datasheet.
The other type of motion that we are going to investigate in this experiment is the
motion of a horizontally projected object. Here, the puck is projected horizontally with an initial velocity v0 = v0x as shown in
Figure 2.2a. The dots produced on the datasheet will look like those shown in Figure 2.2b. In
order to analyze the motion, we will investigate the motion of the puck along the horizontal and the vertical axis independently. For this purpose, we drove the x and y-axes as in Figure
2.2b taking as the origin the first dot. The positive direction of the y-axis is taken downwards. If we project the x and y-components of each dot to the axes, then this will look as in Figure 2.3.
FIGURE 2.3. The projection of the 2 dots on the datasheet along to the x and y-axes.
Note that the intervals between the x-projections of the dots are equal, meaning that
the motion in the horizontal direction is just a straight line motion with constant velocity. In other words, the x-component of the velocity of the puck is constant. As for the motion along
the y-axis, note that the distance between the y-projections of the puck increases with time. this situation is identical to the one we encountered earlier, in this case of motion with
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acceleration. Actually, the acceleration of the puck the along the y-axis is nothing but the
acceleration a of a puck released to fall from rest down the same inclined air table. Therefore, quantitatively, the motion along the x-axis will obviously be given by the equations
(2.7)
and
x = (2.8)
As for the motion along the y-axis, we have
(2.9)
and
(2.10)
If we eliminate the time t in Equation 2.10 using Equation 2.8, then we get y as a function of x and vox :
(2.11)
This is the equation of a parabola that passes through the origin in the x-y plane, which is indeed the shape of the trajectory of the puck.
EQUIPMENT
An air table, a puck shooter, wooden blocks, a ruler, millimetric graph paper. PROCEDURE
This experiment has two parts, both of which will be carried out using the inclined air
table. One side of the datasheet will be used at the time. First, level of the air table, and then tilt it to an inclined position by using the wooden block. Note that the sine value of the
inclination angle is written on the wooden block. Keep one of the pucks at rest over the
carbon paper at the lower right corner of the air table. In both experiments, only one puck will be used.
PART A: STRAIGHT LINE MOTION WITH CONSTANT ACCELERATION
1. Put the puck at the top of the inclined plane off the air table. Activate only the (P) switch. check that the puck false freely down the plane. Set the spark timer frequency
to 20 Hz. Put the puck at the top of the inclined plane. Put the (P) and (S) switches on top of each other and activate them simultaneously. Take your foot off switches then
the puck reaches the bottom of the inclined plane.
2. Remove the datasheet from the air table and examine the dots produced on it. Have
your data approved by your instructor. Take the trajectory of the puck as the positive x-axis. Number the data points starting from the first dot as 0, 1, 2, …, 5. Take x = 0
and t = 0. At the first dot, and measure the distance of the remaining four dots from the
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dot 0. Also, determine the time t for each of these dots. Fill in your data in Table 2.1,
along with their corresponding measurement errors.
3. Using the data in Table 2-1, plot x versus t2 graph on a linear graph paper. Draw best and worst lines, and find the acceleration ( a Δa) of the puck.
FIGURE 2.4 Adjusting the shooter on the edge of the inclined air table.
PART B: HORIZONTAL PROJECTION
1. Use the other side of the data sheet in this part. 2. Keep one of the pucks stationary by placing it on a folded piece of paper at the lower
right corner of the inclined air table. 3. Attach the shooter to the left side of the inclined plane of the air table about 10 cm
from the upper edge, and adjust the shooting angle to 0 degrees.
4. Activate only the (P) switch and place the puck into the shooter, and make a few test shots to adjust the tension in the rubber belt of the shooter to give a convenient
trajectory 5. Now activate the (P) switch and place the park into the adjusted shooter, then release it
while simultaneously activating the S switch to trigger the spark timer. Keep the
switches on until the puck comes to the bottom of the datasheet and then resing move your feet off them.
6. Before removing the datasheet, place the puck opposite to, but outside, the shooter, and activating both the P and S switches simultaneously, make it slide freely down the inclined plane.
7. Now take the datasheet off the air table, and examine the trajectories you have got. you should get something like the trajectory's illustrated in Figure 2.5 below. Denote
the two trajectories by A and B as in the figure. If your data points are inconvenient for analysis, make another run and get new data.
8. Circle the data of both trajectories, and starting from the first dot, number them as 0,
1, 2, 3, …, 5, etc.
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FIGURE 2.5 The dots on the datasheet.
9. Draw the x and y-axes for trajectory A. This can be done by first drawing a line parallel to the trajectory B that passes through the first dot of trajectory A. This line will give the y-axis. Then draw a horizontal line perpendicular to it from dot 0 to have
the x-axis. Take the positive y-direction to be downwards. 10. Now, draw from each that a trajectory A normal to the x and y-axes to get the x and y-
projections of these dots. You will get something like Figure 2.5. What type of motion does the puck have along the horizontal and vertical axis?
11. Measure and record the time of flight tf (The total time that elapsed during the motion
of the puck), and the range R of the projectile (the horizontal distance traveled during the motion). Using R and tf find the projection velocity v0x.
12. Starting from dot 0 of trajectory A, measure the distance from the y-projections of 5 data points from this point. Determine also the time corresponding to each of these points. Record your results in the left margin of Table 2.2. Also, measure and record
the distances of the first dots in trajectory B from dot 0 along with the corresponding times. put these in the right margin of Table 2.2.
13. Taking y to be the distance of dot 5 from dot 0, use Equation 2.10 to find the accelerations aA and aB for both trajectories A and B. Compare these with acceleration you found in Part A of this experiment.
PART A: STRAIGHT LINE MOTION WITH CONSTANT ACCELERATION
1. How do the dots you've got in this experiment differ from those you got in experiment 1?
………………………………………………………………………………………………………………………………………………………………………………
2. What type of motion did you have in each of these experiments?
………………………………………………………………………………………………………………………………………………………………………………
3. Fill in your data in the table below. report your measurements with the correct number
of significant figures and include the errors.
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TABLE 2.1
Dot Number x (cm) t (sec) (sec 2)
0 0 0 0
1
2
3
4
5
4. In the space provided below show how to find the error in t2 for only one data
point.
………………………………………………………………………………………………………………………………………………………………………………
5. in the space provided below report the slopes of the best and worst lines of the x-t2 .
m = ……………………..cm/ sec2
m = ……………………..cm/ sec 2
6. Find the acceleration of the puck along with its error from the slopes a bow and record it with the correct number of significant figures.
a …………………..cm/sec2
PART B: HORIZONTAL PROJECTILE
7. What kind of motion does the horizontally projected puck have along the x and y-axes? explain your answer.
………………………………………………………………………………………………………………………………………………………………………………
8. Report tf and R.
tf = …………..sec.
R=……………cm. 9. Calculate the projection velocity v0 = vox.
V0 = ……………………..cm/sec. 10. Record your measurements for the trajectories A and B in the table below. report them
with the correct number of significant figures and errors.
TABLE 2.2
Trajectory A (y-projection ) Trajectory B
Dot Number 0 0 0 0 0
1
2
3
4
5
8
10. Using Equation 10, and taking y is the distance of the fifth point, find the accelerations
aA and aB, and report them with the correct number of significant figures. compare these two accelerations and compare them with the acceleration you found in Part A too.
aA = ………………………cm/sec2
aB =………………………..cm/sec2
11. Write down any comments related to the experiment, and/or elaborate on and discuss any points.
………………………………………………………………………………………
………………………………………………………………………………………
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EXPERIMENTS 3
NEWTON’S LAW’S OF MOTION:
APPLICATION WITH THE ATWOOD MACHINE
PURPOSE
This experiment investigates the relationship between motion and its causes. In this experiment, we use Atwood‟s machine arranged on an inclined air table to study Newton‟s
second law of motion.
THEORY
In order to move an object that is initially at rest, we must apply a force on it. Force is
a vector quantity and its SI unit is the Newton (N).
The vector some of several forces acting on an object is called the ir resultant. A resultant force is required to accelerate an object. remember that acceleration is the rate of change of velocity. We know from experience that a resultant force must act on an object
initially at rest to set it into motion and eventually the object speeds up. Similarly, a resultant force is required to slow down or stop an object that is already in motion. Certainly, we need
to apply the resulting force on a moving object to change the direction of its motion. In all of these cases, the object accelerates (changes its velocity) under the action of the resultant force.
The acceleration of an object is directly proportional to the magnitude of the resultant
force exerted on it. When we doubled the force, the acceleration is also doubled. This
means that the ratio of the magnitude of the force to the magnitude of acceleration is a constant. This ratio is called the mass m of the object. Therefore, we may write
F = ma (3.1)
This last relationship is called Newton‟s second law of motion. Note that both ve are vectors and they are in the same direction. When several forces act on an object moving in
the xy-plane, the method of components yields Σ Fx= max , Σ Fy= may (3.2)
Elementary Atwood‟s Machine consists of two different masses m1 and m2 (m1 < m2)
connected by a court passing through a pulley as shown in Figure 3.1a below. When the system of two masses is released from rest, the heavier mass m2 moves downward with constant acceleration and m1 moves upward with the same acceleration. The forces acting on
each of the messages are shown in Figure 3.1b. T is the tension in the cord. Since m2
accelerates downwards, then this means that it experiences a resultant force in this direction,
so m2g>T. As for the mass m1 we obviously have m1g<T.
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(a) (b)
FIGURE 3.1 Atwood‟s machine. (a) The setup. (b) The forces acting on each puck.
Obviously, since the acceleration a of the system is constant, and since both pucks
start for motion for rest, then the relation
holds. To make an elementary Atwood‟s
machine on an inclined air table, the angle of inclination being , we set up the system shown
in Figure 3.2a. The two pucks act, as the two masses is required for the machine. Putting extra mass on it increases the mass of one of the pucks.
(a) (b)
FIGURE 3.2 The experimental set up of an Atwood‟s machine on an inclined air table.
The greater mass m2 Is shown in Figure 3.2b; 2 forces along the inclined plane act it
on: the tension T in the cord pulling upward and the component of its weight m2gsin. Since
the mass accelerates downward, the tension T is smaller than mgsin, and hence the resultant
force acting on mass m2 is
m2gsin – T = m2a (3.3)
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T – m1gsin = m1a (3.4)
Adding the two equations side by side and eliminating T, we find the acceleration
a = [(m2 – m1)gsin] / (m1 + m2) (3.5)
Solving for the tension in the cord, we obtain
T = [2m2m1gsin ] / (m1 + m2) (3.6)
Where g is the acceleration due to gravity (=9.8m/s2) and Is the angle of inclination of the
air table.
EQUIPMENT
An air table, pulleys, extra masses, wooden blocks, a cord, a ruler, millimetric graph paper.
PROCEDURE
This experiment is carried out on the inclined air table. First, level the air table of, then
tilt it to an inclined position by raising its backside and placing a wooden block underneath it. Note that the sine value of the angle of inclination is written on the wooden block.
1. Attach the pulley said to the middle of the upper side of the air table and hang the cord with two masses at the ends of the pulleys as shown in Figure 3.2a. Note that the puck
on the right has extra mass. 2. Put the puck on the left (with smaller mass) at the lowest position at the other puck at
the highest position. Activating only the pump (P), release the system from rest and
observe the motion of the two pucks. Repeat this several times to get familiar with the motion.
3. Set the spark timer frequency to 20 Hz. You may, however, change this to 10 Hz if your data was not convenient.
4. Repeat now by putting the two foot switches on top of each other and the placing them
simultaneously to produce the spark record for both pucks. Observe that the puck on the right goes down while the other goes up the incline.
5. Remove the datasheet and observe the dots produced on it. What kind of a trajectory do the pucks have? Do they have the same type of motion?
6. Starting from the first dot, number the dots produced by each puck on the datasheet as
0, 1, 2, 3, … etc., the first dot of each trajectory is to be taken as dot 0. Use this point as the reference point as 0 position and 0 time.
7. Taking the positive y-axis to be the direction of motion, measure the position and the time of five data points in each trajectory relative to dot 0, and fill in Table 3.1 below.
8. Use the data tabulated in Table 3.1 to plot the y versus t2 graph for either of the masses
m1 and m2. Draw the best and the worst lines, and find the acceleration a from the slope of the graph.
9. Measure the masses m1 and m2. Using the acceleration you obtained sin, m1 and m2 calculate the acceleration due to gravity.
12
RESULTS AND DISCUSSION
1. Record the position and time measurements of five data points of the pucks m1 and m2
relative to dot 0 in Table 3.1 below. TABLE 3.1.
Dot Number
(cm)
(for m1)
(cm)
(for m2) (sec) (sec2)
0
1
2
3
4
5
2. In the space below, show the details of the calculation of the errors Δt veΔt2 in t ve t2 for any data point of the above table.
………………………………………………………………………………………………………………………………………………………………………………
3. Examine the dots on the datasheet and in the above table, what kind of motion does
each puck have? Why?
………………………………………………………………………………………………………………………………………………………………………………
4. Can you say that both pucks have the same motion? Explain your answer. …………………………………………………………………………………………
……………………………………………………………………………………
5. Write down the slopes of the best and worst lines you found from the y-t2 graph with
the correct number of significant figures. Mb = ………………
Mw = ………………
………………………..
13
6. Report below with the correct number of significant figures, the acceleration of the
puck you found from the y – t2 graph. ………………. cm/sec2
7. Calculate the gravitational acceleration g, and we called it with the correct number of
significant figures and units. ………………………………………………………………………………………………………………………………………………………………………………
8. Calculate the tension T in the cord. report the result with the correct number of
significant figures and in the SI units.
………………………………………………………………………………………………………………………………………………………………………………
9. Write down any comments related to the experiment, and/or elaborate on and discuss any points.
………………………………………………………………………………………………………………………………………………………………………………
14
EXPERIMENT 4
SIMPLE PENDULUM
PURPOSE
The main purpose of this experiment is to study kinetic and potential energy changes
and conservation of the energy of the simple pendulum on a tilted frictionless air table.
THEORY
A simple pendulum is a weight (body) suspended from a string which is negligible
mass. As seen on Fig. 4.1, while the motion is happening on vertical plane due to gravity, the
string end is fixed. And all figures (a to e) of Fig. 4.1 define complete motion or one period of the pendulum. The system try to keep equilibrium position. So if the body move from
equilibrium position and release, the gravity will work for accelerating the body through the equilibrium position. This makes an oscillation around the equilibrium position.
FIGURE 4.1. A period of the simple pendulum
Total energy is conserved in isolated system.
EKU (4.1)
From Fig. 4.1, total energy is conserved; however potential and kinetic energy would
change over time (from a to e). Because the body will have variable velocity and height. While the body is at the bottom; consider the potential energy is zero. From Figure 4.1, the
velocity is at maximum at (c) position and the kinetic enery is at maximum too. At (a) and (e) positions, the body is at maximum height; so the potential energy is at
maximum. At these positions, the velocity of the body is zero; so the kinetic enery is zero.
15
While simple pendulum is placed on the air table tilted at an angle, φ; the force vector
will shift (Figure 4.2).
FIGURE 4.2. The forces on a body on the tilted plane.
Let say, h for the height of the body on a tilted plane and calculate potential energy:
)()( 0yUyUU (4.2)
yy
y
dymgdyyFyUyU
0
0 )sin(0)()()(
0
(4.3)
sin)( mgyyU (4.4)
Equation (4.4) defines the potential energy of a body with a mass, m on the tilted plane. Pay attention to the potential energy here is calculated, does not give the potential energy of the tilted plane (airtable) itself. We won‟t count for the height of the tilted plane.
The kinetic energy of the body is
2)(sin2
1mK (4.5)
Let‟s write the equations (4.4) and (4.5) to its place of the equation (4.1). Here, we
define conservation of energy for the simple pendulum:
2
2
1sin mmghE (4.6)
EQUIPMENT
Airtable, wedge, puck, string.
PROCEDURE
1. Take the wedge and give elevation to the table at a tilted angle, . Place first
the carbon paper on the airtable, then place data paper. 2. We will use only one puck for this experiment; so place one of the puck to the
corner of the table. Tie a puck which is its mass known, with a proper lenght of a string and fix its string end to the upper side of the airtable.
3. Move the puck to a new position besides the equilibrium position by keeping string strained. Release the puck and watch its movement without pressing the air pedal.
16
4. Set the frequency of the sparkle to f = 20Hz (or T = 0,04 seconds).
5. Press the air and sparkle pedal simultaneously until the puck stops. Turn back your data paper vertically. Draw motion axis for spark points.
6. Mark the first point of the motion as zero point, “0”. Write numbers down in order for all points by starting from the first point of the motion. Input time and distances to the motion axis of each points to the Table 4.1.
7. Calculate slope of the airtable and get angle. Gravity constant can be
assumed as g = 10 m/s2
RESULTS AND DISCUSSION
1. Calculate velocities ( ), heights (y), potential (U), kinetic (K) and total (E) energies
of the puck at the each points. Input the results to the Table 4.1. Explain all your calculations at your report.
i
i
xit
x
and
i
i
yit
y
Here,
22
yx
TABLE 4.1
t (s) x (m) y (m) h (m) x (m/s) y (m/s) (m/s) U (joule) K (joule) E (joule)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
2. Plot potential and kinetic enery variations over time on same graph paper and explain the variations at your report. What hills and valleys of the plot point out?
Explain. ……………………………………………………………………………………………
…………………………………………………………………………………
3. Plot total energy over time to a new graph paper. Is total energy conserved? Explain
by using the graph.
17
……………………………………………………………………………………………
…………………………………………………………………………………
4. Explain expectations and findings. Comment your results. ………………………………………………………………………………………………………………………………………………………………………………
1
EXPERIMENT 5
COLLISIONS AND CONSERVATION OF LINEAR
MOMENTUM
PURPOSE
The purpose of this experiment is to verify the linear conservation of momentum in different types of collisions in an isolated system, to study the motion of the center of mass
the collusion of a two-puck system and to investigate the conservation of kinetic energy in elastic and completely inelastic collisions.
THEORY
The linear momentum of an object is defined as the product of its mass and velocity,
( 5.1)
Therefore, an object at rest will have a zero linear momentum. Also, from the above definition, it is clear that an object that has a constant mass will have a constant momentum, unless its velocity changes. Linear momentum will be referred to shortly as momentum from
here on. We know, however, that the velocity of an object changes only when a net external
force is applied. This means that the momentum of an object will change only then that
object experiences a net external force. This fact can actually be seen from Newton‟s second law. For an object with constant mass, we have Newton‟s second law.
( 5.2)
where m is constant, this can obviously be written as,
( 5.3)
This above equation means that if no net force acts on an object then its momentum will be
conserved., i.e. constant in time. That is if =0, then
( 5.4)
or
( 5.5)
Here, constant means that the momentum does not change with time., i.e. the object
will have the same momentum at all times. the above result can we generalized to a system of
2
N-particles with masses . the total momentum of the system of particles at
any instant of time is by definition:
( 5.6)
Where … etc. The sum in Equation 5.6 is obviously a vector, not an algebraic sum. In this case, Equation 4.3 generalized to,
(5.7)
where means a net force external to the system of particles, i.e. any force other than the force that the particles of the system exert on each other. This force might be the friction
force, the gravitational force, … etc. Therefore, is no such net external force acts on the system of particles the total momentum of the system will be conserved ., i.e.
( 5.8)
or
( 5.9)
Recall that the above sum a vectorial sum. Therefore, a system of particles experiencing no net external force, or an isolated system, will her it's total momentum the same at any instant in time, regardless of any collisions among the particles.
In this experiment, we are going to investigate the conservation of momentum of the system
of two pucks moving on a leveled air table. The air table being horizontal, and with the friction almost eliminated, obviously produces no net external forces on the pucks on it. Therefore, we expect the total momentum of the pucks to be conserved. The pucks will be
allowed to collide, and they have told her momentum therefore and after the collision will be measured and compared. the forms of the points that we will have on the datasheet are shown
in Figure 5.1.
FIGURE 5.1. The data points in the elastic collision of the two pucks o a leveled air table.
The velocities of the two pucks Before the collusion will be and , And after the
collusion and
. Since the system is an isolated system, the total momentum will be
conserved, and we will have at any time;
3
( 5.10)
so,
( 5.11)
where … etc. Since the masses of the pucks are identical, the above relation reduces to
( 5.12)
The sum in equation 5.12 is also vectorial, And a method of finding this sum is geometrically
is explained below. In the completely inelastic collision, momentum will obviously be conserved too, since the system is still an isolated one. The two pucks in this collusion will
stick together and form a single object with mass 2m that moves with velocity . The dots on the datasheet will look like those in Figure 5.2 below.
FIGURE 5.2. The data points in the completely inelastic collision of two pucks on a leveled
air table.
The conservation of momentum in the course of the collusion will be given as
( 5.13)
or
( 5.14)
Another concept that we are going to introduce and study in this experiment is that of the center of mass (CM). Intuitively, you can guess that the CM of a homogeneous cube (Figure 5.3a) or a sphere (Figure 5.3b.) Will be at the geometrical center of this symmetrical
objects. Also, you can guess that CM of the dumbbells shown in Figure 5.3c. will be at the meet point of its rod. Thus, the CM of two identical homogeneous spheres will be at exactly
the midpoint of the line joining their centers (Figure 5.3d.). If one of the spheres, however, were heavier, the CM would be shifted towards the heavier sphere as shown in Figure 4.3e. The amount of shifting would be determined by how much the mass M is larger than m. From
the above examples, it is clear that it is possible to guess the position of the CM or some symmetrical mass distributions. For example, it is not difficult to guess that the CM of two-
puck system of this experiment with lie at the midpoint of the line joining their centers.
4
m m
(a) (b) (c)
FIGURE 5.3. The center of mass of some symmetrical homogeneous objects.
For more arbitrary mass distributions, however, one has to define the CM more formally. The
center of mass‟ position vector of a system of N particles with masses
whose position vectors are, respectively is defined as,
( 5.15)
FIGURE 5.4. The center of mass of a distribution of masses.
As the particle change their positions in time, the position of the Cm will also change., and this rate of change of the position vector of the CM can be thought of as the velocity of the
CM.
( 5.16)
For particles with constant masses, taking the time derivative of both sides of Equation 5.15 we get:
(5.17)
or
5
(5.18)
Where the dots in Equation 5.17 mean the time derivatives which are just the velocities. The
able formalism, when applied to our two-puck system, gives
( 5.19)
( 5.20)
where we have canceled out the masses in moving to (Equation 5.20) since the pucks have identical masses. The velocity of CM will then be
( 5.21)
The above equation has important implications. First, note that the numerator of the right-hand side is a constant in the case of the two-puck system of the leveled air table since momentum is conserved. This means that the velocity of the CM, in this case, is constant. In
other words, the CM moves with constant velocity. So, for an isolated system for which the total momentum his conserved, the CM of the system always moves in a straight line and with
constant speed. Moreover, our case, Equation 5.21 implies that the velocity of the CM is half the vectorial sum of the velocities of the pucks at any instant in time. Therefore, we have for our two-puck system before and after collision,
( 5.22)
or
( 5.23)
In this experiment, we are also going to investigate the conservation of the kinetic energy of the pucks in the course of the collision. remember the definition of the kinetic energy K often object of mass m and linear velocity .
( 5.24)
Therefore, the total kinetic energy of there two-puck system before an elastic collision will be
( 5.25)
and after the collision
( 5.26)
in the completely inelastic collision, however, where the two pucks Stick together to form a
single object with mass 2m and velocity , The total kinetic energy after the collision will be
( 5.27)
6
Since the kinetic energy is a scalar quantity, thus the sums in Equations 5.25 and 5.26 are obviously algebraic sums. on the other hand, while the kinetic energy is almost conserved in
an elastic collision, it is not concert by definition in the completely inelastic collision. The fractional loss of the kinetic energy is defined as
and using this, we can define the percentage fractional loss in kinetic energies
.100%
EQUIPMENT
An air table, Velcro bands (To stick the pucks together in the completely inelastic collision), ruler.
PROCEDURE
This experiment consists of two parts. Part A is the elastic collision and Part B is the completely inelastic collision. The experiment will be carried out on a horizontal leveled air
table. Therefore, before you start the experiment, level off the air table as described in the first part of this manual.
PART A: ELASTIC COLLISION
1. Activate only the pump switch (P) and project the two pucks diagonally towards each other across the air table so that they collide somewhere at the middle of the table. Repeat this procedure several times until you get satisfactory collision. Don't project
the two pucks too slowly, or too fast, But push them so that they move with a moderate speed. Now, choose a proper spark timer frequency (20 Hz, for example), and then project the pucks across the air table while activating the switch (P), and once
you release them, activate the spark timer switch (S). keep both switches depressed until the two pucks complete their motion.
2. Remove the datasheet and examine the dots produced. they should look like those and Figure 4.1a. Number the dots for each puck as 0, 1, 2, … etc. Have your data approved by your instructor before you go on.
3. Find the velocity of each puck before and after the collision by measuring the length of two or three intervals of each trajectory and dividing by time. Label the two
trajectories of the pucks by A and B before collision and and after collision. 4. Find the vectorial sums and
. To find , for example, extend
the trajectories A and B until they intersect. Then, so starting from the intersection point, draw vectors along the directions of and , and with lengths proportional to
the magnitude of these velocities. You can, for instance, draw a 1 cm vector to represent a velocity of 10 cm.s-1, (see Figure 4.5). Then, completing the
parallelograms find the sum of these velocities. Repeat the same procedure to find
.
7
FIGURE 5.5. Vector sum
5. Identify the dots produced at the same instant of time before and after the collision,
and joining these determine the position of the CM along the line which joins every pair of dots. Doing this, you will get a record of the position of the CM during the
course of the collision. 6. Using the record you obtained for the CM in number 5 above find its velocity before
and after the collision.
7. Find the total kinetic energy of the two pucks before and after the collision, and compare them.
PART B: COMPLETELY INELASTIC COLLISION
1. Wrap the Velcro band firmly around the two pucks (see Figure 5.6); Make sure that the rims of the bands are not in contact with the surface of the datasheet. Activate only
the pump switch (P), and project the two pucks towards each other across the air table so that they collide and stick together somewhere at the middle of the table. Make sure that the pucks do not rotate after the collision. Repeat several times until you get a
convenient collision. 2. Now, activating the pump (P), project the pucks toward each other, and the moment
you release them activate the spark timer switch (S). Keep both switches depressed until the pucks complete their motion. The dots on the datasheet will look like those in Figure 5.2. Find the velocities and of the pucks before the collision and the
common velocity of the two pucks stuck together after the collision.
3. Employing the method described in number 6 of part A above, find the vector sum and verify the conservation of momentum.
4. Find the total kinetic energy of the pucks before and after the collision, and calculate the fractional loss, and thus the percentage fractional loss.
8
RESULTS AND DISCUSSION
PART A: ELASTIC COLLISION
1. Write down to velocities of the two pucks before and after the collision. = …………………. =………………….
=………………….
=………………….
2. Find the sums and
, and discussed the conservation of momentum.
=………………………….
=………………………….
3. Find the velocity of the CM before and after the collision using the time record you
produced on the datasheet.
=…………………. =………………….
4. What kind of motion does the Cm have during the course of the collision? Compare
the results you reported above for and with
and
, respectively.
…………………………………………………………………………………………
…………………………………………………………………………………………
5. Find the total kinetic energy of the two pucks before and after the collision. Is the kinetic energy then considered with the correct number of significant figures
conserved?
K =………………….. =………………………
PART B: COMPLETELY INELASTIC COLLISION
6. Write down the velocities of the two pucks before the collision, and the common velocity of the pucks stuck together after the collision.
= …………………. =………………….
=………………….. 7. Find the sum And verify the conservation of linear momentum.
=………………
8. Find the total kinetic energy of the pucks before and after the collision. Is the kinetic energy conserved? If not, find the percentage fractional loss of energy.
……………………………………………………………………………………………………………………………………………………………………………………
9
9. Write down the comments related to the experiment, and/or elaborate on and discuss any points.
……………………………………………………………………………………………………………………………………………………………………………………
10
EXPERIMENT 6
ROTATIONAL MOTION
PURPOSE
The purpose of the experiment is to investigate the dynamics of a disk that rotates
about an axis passing through its center and to calculate the angular acceleration, angular velocity and the moment of inertia of a disk and to investigate the conservation of mechanical energy.
THEORY
So far, we have studied the kinematics and the dynamics of an object having linear
motion. In this experiment we are going to study the rotational motion of a rigid body. A
rigid body is an idealized model for an object that has definite and unchanged shape and size. To study the rotational motion of a rigid body, we first need to introduce some new physical
quantities and concepts that are essential to the description and understanding of this type of motion.
(a) (b)
FIGURE 6.1. The same net force F is acting on two objects with masses m and M (M>m). The object in (a) will suffer a larger change in its motion than that in (b).
The first quantity that we are going to introduce is the moment of inertia of a rigid body. To understand this, we go back to linear motion to make an analogy. We know that in
linear motion, if the same force were applied to two objects with masses m and M (M>m), (see Fig. 6-1), the one with the smaller mass would have greater acceleration. In other words, the force would cause a greater change in the motion of the object with the smaller mass
compared to that with the larger mass. So, the mass of an object is actually a measure of the inertia (resistance) that an object exhibits against changes in its motion, and the larger this
mass is, the larger is the inertia. This is why the mass appearing in Newton‟s second law F=ma is usually referred to as the inertial mass.
(a) (b)
FIGURE 6-2. The same person (i.e. the same force) trying to rotate a pencil (a) and a heavy metallic rod (b). It is easier to put the pencil into rotation than the rod.
11
Consider now the situation described in below of the Figure 6.2, where a pencil and a heavy
metallic rod are put into rotation by the same person, i.e. by the same force. Obviously, it will be easier to rotate the pencil than it is to rotate the rod. In physical terms,
we say that the moment of inertia of the rod is larger than that of the pencil. You might, in view of the above examples, rush to the conclusion that the moment of inertia of a rigid body is determined solely by its mass. To see that this is not true, consider the situation described in
Figure 6.3. The same person about two different axes puts the two rods with the same mass M into rotation.
(a) (b)
FIGURE 6.3. Two identical metallic rods put into rotation around two different axes by the same person (force). It is easier to rotate the rod in (a) than it is in (b).
It is clear that it is harder to rotate the rod in Figure 6.3b. That is, it has a larger
moment of inertia compared to the other one, although they both have the same mass M.
Therefore, the moment of inertia depends on the mass distribution with respect to the a xis of rotation in addition to its mass. Figure 6.4 below gives the moment of inertia of some
symmetrical homogeneous rigid bodies.
FIGURE 6.4. The moment of inertia of some symmetrical homogeneous objects about different axes.
12
We know that according to Newton‟s second law of motion when a net force acts on
an objects that is at rest, it puts in into motion. So, can any force put a rigid body into rotation? To answer this question, see Figure 6.5 and its description.
(a) (b)
FIGURE 6.5. A force F acting tangentially on a disk that can rotate freely about an axis passing through its center as in (a) can rotate it. The same force acting as in (b) can not cause
rotation.
The net force tangentially on the disk in Figure 6.5a causes it to rotate about the axis
passing through its center. On the other hand, when the same force acts on the same disk so that its line of action passes through its center as in Figure 6.5b, obviously it can not rotate the
disk. From the above observation we see that a net force is not sufficient to cause a rigid body to rotate; the point where the force acts is also important. In physical terms, we say that a rigid body will experience a change in its rotational motion if a net torque of that force about the
rotation axis (or point) exists. The torque of a force about a point or an axis is defined as:
=r x F (6.1) Where r is the vector from the axis of rotation to the point where the force is applied (see Fig
6.6). Clearly the torque is a vector quantity.
FIGURE 6.6. The angle θ in the equation = rFsin.
From the equation 6.1, and the definition of the vector (cross) product, we have for magnitude of
= rFsinθ (6.2)
13
Where θ is the angle between r and F (see Fig. 6.6). The above equations explain why the net
force in Figure 6.5a causes rotation, while that in Fig. 6.5b does not. The angle θ in the latter case in zero.
Net force acting on an object gives it a net linear acceleration; net torque acting on a rigid body gives it an angular acceleration α defined by
= I α (6.3)
Where I is the moment of inertia of the rigid body about the axis of rotation. Now, let us define angular velocity and angular acceleration. Consider a rigid disk that is rotating about an axis passing through its center as shown in Fig. 6.7 below.
FIGURE 6.7. The change in the distance ds and the angle dφ in an infinitesimal time interval
dt for a rigid disk rotating about an axis passing through its center, O
Consider a point on the rim of the disk, and take the origin of the coordinates to be at the center of the disk. In a time interval dt, such a point will travel a distance ds along the rim of the disk, and the radius R will sweep an angle dφ. The magnitude of the linear velocity of
the point on the rim of the disk will therefore be
(6.4)
Consequently, the linear acceleration, which is the rate of change of the velocity, will be
(6.5)
The angular velocity of the rotating disk is defined as the rate of change of the angle
φ (the angle measured in radians). So
(6.6)
And the angular acceleration in thus
(6.7)
14
In the above geometry, the lenght of the arc element ds is obviously ds=Rd φ (φ in radians),
therefore with R constant;
, (6.8)
or
v=Rω (6.8)
Consequently,
a=Rα (6.9)
The above two relationships define the dimensions (i.e. units of ω ve α). Since in the
SI system of units v is given in m/s and R in m, then from Eq. (6.8) ω has the unit 1/s.
Similarly, α has the unit of 1/s2.
From the above developments, it is clear that there is a close analogy between the kinematical and dynamical quantities in linear and rotational motions. Table 6.1 summarizes this analogy.
TABLE 6.1. Analogy between kinematical and dynamical quantities in linear and rotational motions.
Linear Motion Rotational Motion
v Ω
a Α
m I
F=ma =I α
v(t)=vo+at ω (t)= ωo+ t
FIGURE 6.8. The experimental set up. Puck m performs linear motion downwards causing the puck M to rotate.
15
In this experiment we are going to have the set up shown in Fig. 6.8, above. A solid disk of mass M that can rotate freely about an axis passing through its center is mounted to the
upper edge of the inclined air table. A cord is wrapped around this disk and the free end of the cord is attached to a puck of mass m as shown in the figure (In this experiment only one puck will be used). As the system is released from rest, the suspending puck m will accelerate down
the plane of the inclined air table putting the disk M into rotation. The tension T in the cord will act as the force whose torque causes the disk to rotate.
The forces acting on the suspended puck are shown in Figure 6.8 above. Applying
Newton‟s second law, we have
mgsinΦ - T=ma (6.10)
Where Φ is the angle of inclination of the air table, and a is the linear acceleration of the puck. The angular acceleration of the rotating disk is related to the above linear acceleration through
Eq. (6.9);
a=Rα Where R is the radius of the rotating disk. The torque caused by the tension T is given by
= RT = Iα (6.11)
Where I is the inertia of the disk. If the system is released from rest, then at a later time t, the
linear velocity of the puck will be given by v = at
Similarly, the angular velocity of the rotating puck will be
= α t (6.13)
The puck has kinetic energy due to its translational (linear) motion and the disk has
kinetic energy due to its rotational motion. The total kinetic energy of the system at any instant of time will thus be
(6.14)
As the puck falls down the plane, its potential energy changes into translational and rotational
kinetic energies of the puck and the disk, respectively. Conservation of energy- with friction neglected- requires that ΔK= -ΔU, where Δ means the change in energy. Therefore, at any
instant of time after the system is put into motion, we have
mg(Δd)sinΦ =
(6.15)
Here, Δd is the change in the distance of the puck travelling down the inclined plane.
16
EQUIPMENT
An air table, a wooden block, a rotating disk, a cord, and a ruler, milimetric graph paper.
PROCEDURE
This experiment is carried out on an inclined air table. Therefore, first level off the air table, then bring it into the inclined position using the wooden block. The sine of the inclination
angle is written on top of the block.
1. Attach one end of the cord to the rotating disk M. Then wrap the cord several times
around the rim of the disk. Attach the other free end of the cord to the suspended puck.
As only one puck is going to be used in this experiment, keep the other puck still at the
lower corner of the air table by folding a piece of paper and putting it under the puck.
2. Adjust the suspended puck so that the rope is stretched, then pressing only the foot
switch (P), allow it to fall down the inclined plane. Observe the rotational motion of
the disk. Repeat several times until you get a convenient motion.
3. Now, adjust the suspended puck again as in step 2, and choose a suitable sparktimer
frequency (20 or 10 Hz). Put the foot switches S and P on top of each other and press
them simultaneously. Keep the switches depressed until the suspended puck reaches
the bottom of the inclined plane.
4. Remove the data sheet, and observe the data produced on it. Is the trajectory of the
puck a straight line? Are the dots evenly spaced or does the distance between
successive dots increase with time? What kind of trajectory and inter-dot spacing was
expected? Does the data you obtained agree with what you have expected? In the view
of the above questions and your answers to them, assess the data that you have got. If
you feel that your data is inconvenient, repeat the experiment again. Consult your
instructor if you have any further problems.
5. Starting from first dot, number the first five dots as 0,1,2,...etc. Measure and record
the time and the distance of each dot from dot 0 along with the corresponding errors.
Record your data in Table 6-2.
6. Using the data in Table 6-2, plot x vs t2 on a graph. Take the positive x-direction in the
direction of motion. Draw both best and worst lines and find the slope of the graph and
from this find the linear acceleration a of the suspended puck.
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7. Measure the radius of the rotating puck, and using Eq. (6.9) find the angular
acceleration a of the rotating disk.
8. Measure the mass m of the suspended puck, and using Eq. (6.10) find the tension T in
the cord. Using Eq. (6.11) then find the torque of this tension, also calculate the
moment of inertia I of the disk using = RT = Iα.
RESULTS AND DISCUSSION
1. Tabulate your data in Table 6.2.
TABLE 6.2
Dot number (cm) (s) (s2)
0
1
2
3
4
5
2. From the x vs t2 graph, calculate the lineae acceleration of the suspending puck.
a ± Δa=.............................
3. Measure and record the radius of the rorating disc.
R ± ΔR=.............................
4. Calculate the angular acceleration α of the rotating disk. Show your calculations
and report α.
α =...............................................
5. Find the tension T in the cord, the torque and the mon-ment of inertia I of the
rotating disc. Show your calculations.
T=...............................
=..............................
I=.................................
6. Find linear velocity v, and the angular velocity ω. Show your calculations.
v =.....................................................
ω =......................................................
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7. Find both the translational and rotational kinetic energies of the suspending and rotating pucks. Show your calculations.
………………………………………………………………………………………………………………………………………………………………………………
8. Verify the conservation of energy.
………………………………………………………………………………………………………………………………………………………………………………
9. Write down the comments related to the experiment, and/or elaborate on and
discuss any points. ………………………………………………………………………………………………………………………………………………………………………………