### EXPERIMENT: MEASURING THE ACCELERATION OF GRAVITY: ag

Aristotle's idea that falling bodies on earth are seeking out their natural places sounds strange to us today. After all, we know the answer: It's gravity that makes things fall.

But just what is gravity? Newton tried to give operational meaning to the idea of gravity by seeking out the laws according to which it acts. Bodies near the earth fall toward it with a certain acceleration due to the gravitational "attraction" of the earth. But how can the earth make a body at a distance fall toward it? How is the gravitational force transmitted? Has the acceleration due to gravity always remained the same? These and many other questions about gravity have yet to be answered satisfactorily.

Performing this experiment, you will become more familiar with the effects of gravity-you find the acceleration of bodies in free fall yourself and you will learn more about gravity in later chapters.

### ag by Direct Fall*

In this experiment you measure the acceleration of a falling object. Since the distance and hence the speed of fall is too small for air resistance to become important, and since other sources of friction are very small, the acceleration of the falling weight is very nearly ag.

### Doing the Experiment

The falling object is an ordinary laboratory hooked weight of at least 200 g mass. (The drag on the paper strip has too great an effect on the fall of lighter weights. Even here there is significant drag. See experiment results.) The weight is suspended from about 3 meters of paper tape. Reinforce the tape by doubling a strip of masking tape over one end and punch a hole in the reinforcement one centimeter from the end. With careful handling, this can support at least a kilogram weight.

##### *Adapted from R. F. Brinckerhoff and D. S. Taft, Modern Laboratory Experiments in Physics, by permission of Science Electronics, Inc., Nashua, New Hampshire.
A 110-v timer is set up about 2.5m above the laboratory floor. Students must work in pairs or threes since one of the group must climb a ladder in order to place the tape in the timer and do the drop. When the suspended weight is allowed to fall, a 110-v timer will mark equal time intervals on the tape pulled down after the weight. The timer has a frequency of 60 vibrations/sec. Such a small mass affects the timer frequency by much less than 1 vibration/sec. After a few practice runs, you will become expert enough to mark several feet of tape with a series as the tape is accelerated past the stationary vibrating timer. This method can be made to yield a series of dots on the tape without seriously retarding its fall.

Label with an A one of the first dots that is clearly formed near the beginning of the pattern. Count 5 intervals between dots, and mark the end of the fifth space with a B. Continue marking every sixth dot with a letter throughout the length of the record, which ought to be at least 2.5 meters long.

At A, the tape already had a speed of vo. From this point to B, the tape moved in a time t, a distance we shall call d1 . The distance d1, is described by the equation of free fall:
d1 = vot + (ag t2)/2

In covering the distance from A to C, the tape took a time exactly twice as long, 2t, and fell a distance d2 described (on substituting 2t for t and simplifying) by the equation:
d2= 2vot + (4agt2)/2

In the same way the distances AB, AE, etc., are described by the equations:
d3 = 3vot + (9agt2)/2

d4 = 4vot + (16agt2)/2

and so on.

All of these distances are measured from A, the arbitrary starting point. To find the distances fallen in each 6-dot interval, you must subtract each equation from the one before it, getting:

AB = vot + (agt2)/2

BC = vot + (3agt2)/2

CD = vot + (5agt2)/2

and
DE = vot + (7agt2)/2

From these equations you can see that the weight falls farther during each time interval. Moreover, when you subtract each of these distances, AB, BC, CD, . . . from the subsequent distance, you find that the increase in distance fallen is a constant. That is, each difference BC - AB =CD - BC = DE - CD = agt2. This quantity is the increase in the distance fallen in each successive 6-dot interval and hence is an acceleration. Our formula shows that a body falls with a constant acceleration.

From your measurements of AB, AC, AD, etc., make a column of AB, BC, CD, ED, etc., and in the next column record the resulting values of agt2. The values of a agt2 should all be equal (within the accuracy of your measurements). Why? Make all your measurements as precisely as you can with the equipment you are using.

Find the average of all your values of agt2, the acceleration in centimeters/(6- dot interval)2. You want to find the acceleration in cm/sec2. If you call the frequency of the tape timer n per second, then the length of the time interval t is 6/n seconds. Replacing t of 6-dots by 6/n seconds gives you the acceleration, ag in cm/sec2.

The ideal value of ag is close to 9.8 m/sec2, but a small force of friction impeding a falling object is sufficient to reduce the observed value by several percent.

Ql What errors would be introduced by using a timer whose vibrations are slower than about 60 vibrations per second? higher than about 120 vibrations per second?