\twelvebf Physics~133~---~Summer~2002

# Physics 133 - Summer 2002

 Experiment #2: Newton's Second Law

## Objectives

In this experiment we will verify the validity of Newton's Second Law by studying the motion of the system depicted in Fig. 1. By measuring the kinematical value of the acceleration and comparing it to the value predicted by the Second Law we will obtain another value for the acceleration of gravity, g.

## Theory

Dynamics is the science that studies the causes of motion; Newton's Second Law is its fundamental principle. Newton's Law is a cause-effect relationship: any dynamical system subjected to external forces will obey

 SF = ma.
(1)
On the left-hand side we have the cause, the total or net force acting on the system; on the right we have the effect: the system accelerates in the direction of the net force, and the amount of acceleration is directly proportional to the amount of force; the constant of proportionality defines the system's mass.

 Figure 1. Experimental setup

If you did the homework, you should already be acquainted with the system whose motion we will study. We have two masses tied together with a long piece of tape, which we can take to be massless and rigid. One of the masses rests on a frictionless horizontal surface; the ribbon is draped over a frictionless ``pulley'' in such a way that the other mass hangs freely.

Newton's Second Law predicts that the system will move with uniform acceleration, the value of which we can find using the following argument: On one hand, the only unbalanced external force that acts on-and thus drives-the system is the weight mg of the hanging mass; on the other hand, the inertial mass of the entire system is m+M, and so the acceleration must be

 a = FexternalMtotal = mgm+M .
(2)
Both masses will move with the same acceleration because the tape doesn't stretch. We wish to compare the acceleration thus predicted to one that we can measure directly. One of the equations that describe the motion of a body with uniform acceleration tells us its speed after it has covered a distance x:

 v2 - v02 = 2 a x.
(3)
If the object is initially at rest, the square of the speed is then simply v2=2ax, whence a=v2/2x. If we compare this last expression to Eq. (2), we find the equality

 v22x = mm+M g,
(4)
which tells us that if we plot v2/2x vs. m/(m+M) we will obtain a straight line with slope g and no intercept.

## Experimental Procedure

A linear air track is a hollow beam pierced by many small holes that emit gentle streams of air. The glider of mass M rests on the beam and, when the air is turned on, floats on a thin cushion of air and moves with negligible friction. Affixed to the top of the glider is a black card of length L. On the same end of the track as the ``pulley'' is a light source shining on a photogate, shown in Fig. 1 by a circle.

An electronic clock has been set so that it will start whenever the light beam is interrupted by the passing black card and will stop when the beam is restored. The time of interruption, t, is displayed in milliseconds on an LED device. (Don't forget to reset the timer before each data run.) During this time the glider advances a distance equal to L, the length of the card. The quantity L/t is a good approximation to v, the glider's velocity at the point where the light beam is interrupted by the very center of the car; at this same point, the acceleration is approximately given by v2/2x.

1.
Start by finding the mass of the glider using a scale. Write down the value (and don't forget the units ):

M0 =

2.
Measure L, the length of the card, using a ruler.

L =

3.
Move the glider on the track towards the photogate until you have it right at the point where the timer is about to start running. Record the position of the glider (use the pin on the glider's side as your pointer). Refer to Fig. 1: you have found point b.

b =

4.
Keep moving the glider towards the pulley until the timer is about to stop running. Record the position; now you have point a.

a =

5.
Subtract b from a. The number you obtain, denoted in Fig. 1 by c, should be equal to L as found in Step 2 (why?). Does it? If the values are not identical, which one do you trust the most? Why? Would you rather use an average of the two? Why?

c =

6.
By averaging b and a you can find the position of point d. We want x = 50 cm throughout the experiment; assuming that the photogate won't move, we can release the glider from the same initial position every time. What is this position?

x0 =

7.
Now you are ready to take data. Hang a 5-gram mass from the tape. Make sure there are no obstacles on its way. Slide the glider to x0 and hold it still for a few seconds. Let it go. Record the readout from the timer; don't forget to convert it to seconds. Vary M and m and fill the data table: the first column is Ma, the mass added to the glider (zero in this first case); the third is the hanging mass; the second is M=M0+Ma. The mass ratio will have the same value regardless of whether you use grams or kilograms, as long as you are consistent.

## Data Analysis

1.
Make a graph showing v2/2x on the y-axis and m/(m + M) on the x-axis. Use all of the points. (Don't forget to label your axes; whenever something has units, display them. Give your graph a title. ) Perform a linear fit. Write down the values you obtained for the slope m and the intercept b.

m=     b=

2.
Repeat Step 1 but just use the points where m is smallest. Write down the slope m and intercept b that you obtain.

m=       b =

3.
Repeat Step 1 using the points where m is largest.

m=   b =

4.
Repeat Step 1 using the points where M is smallest.

m=

5.
Repeat Step 1 using the points where M is largest.

m=    b =

## Discussion

1.
The hanging mass m has two forces acting on it: its weight mg and the tension T of the string. Newton's Second Law then says that the equation of motion for this mass is mg-T=ma. In the space below, draw the mass and the forces that act on it.

2.
Three forces act on mass M: its weight, Mg, the tension T of the string, and the normal force N exerted by the air track that keeps the mass from falling through. We now have two equations of motion: Mg=N and T=Ma. In the space below, draw the mass and the forces acting on it.

3.
Why have we taken the tension to be the same in both cases? Why have we assumed that the acceleration is the same for both masses?

4.
Write down the value of g that you found in the preceding section. Express it in m/s2.

g =

5.
Look at your graphs from the preceding section and select the one whose slope is lowest. Which one is it? What is the slope? Call this slope (s1) and write it down. Find g-s1, using g from Step 4.

s1=                                                         g-s1 =

6.
Now select the graph whose slope is highest. Which one is it? What is the slope? Call this slope (s2) and write it down. Find s2-g, using g from Step 4.

s2=                                                         s2-g =

## Conclusions

1.
The values calculated in the last two steps of the preceding section give us an estimate of the uncertainty in our value of g. We can say that g=[`g]±Dg, where [`g] is the value that includes all points; as for the uncertainty, given g-s1 and s2-g we can take Dg to be the larger of the two. Write down yet again the value you obtained for g, but now add the uncertainty (which has units).

g=                                                     ±

2.
The uncertainty in the measurement can also be expressed as a fraction of the final value: g=[`g]±p%, where p=Dg /[`g]×100. The uncertainty is now dimensionless.

g=                                                     ±

3.
Compare your final g to the accepted value, g » 9.8 m/s2. Is your measured value too low or too high? What would be the effect of friction and air drag? Could these explain why your result was too low or too high?

4.
Which of the plots from the section on Data Analysis gives the value of g that is most ``wrong''? Which one gives the ``best'' value? Could this information help us know how to get a more accurate value for g?

5.
We have not discussed the intercept of any of the plots. We know that they should be very close to zero. Is this the case? What does this intercept mean?

 Data Sheet for Experiment #2: Newton's Second Law

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On 21 Jun 2001, 19:12.