An electromagnet holds a one-inch-diameter steel ball even with the top end of a measuring stick that is slightly longer than two meters. When you press the switch on the small grey box at left on the table in the photograph, this simultaneously cuts the current to the magnet and starts the timer (at right in the photograph). When the ball falls through the photogate (set 1.50 m below the top of the measuring stick in the photograph), it stops the timer. You can set the photogate anywhere along the measuring stick, to a distance of just over two meters from the top. The carpenter’s level in front of the timer allows you to check that the apparatus is vertical. (It will be set vertical before class.) Since the electromagnet sits at about 2.30 meters (7′6.6″) above the floor, the stepladder is provided for those instructors who otherwise would not be able to reach it to set the ball there. Important: When you press the switch, make sure to do so quickly and decisively. Pressing the switch first breaks the connection to the coil and then closes the connection to start the timer. Doing this quickly and firmly minimizes any time lag that might occur between these steps. When you are finished, turn off the power to the coil.
As demonstrations 08.09 -- Cube vs. feather, 08.12 -- Ball vs. feather, and 08.06 -- Tilted air track, show, when released from a particular height in a gravitational field, in this case that of the earth, all objects accelerate downward at the same rate. Close to the earth’s surface (that is, at a distance from the center of the earth that is close to the radius of the earth), this acceleration is 9.81 m/s2 (see, for example, NIST CODATA). Newton’s law of universal gravitation (see demonstration 16.39 -- Cavendish balance) and second law of motion (see demonstrations 12.33 -- Vertical Atwood’s machine, and 12.36 -- Horizontal Atwood’s machine) explain why this is. These are, respectively,
F = GMm/r2
and
F = ma
where in the first equation, F is the (attractive) gravitational force between two bodies, and G is the universal gravitational constant (= 6.72 × 10-11 N·m2/kg2). M is the mass of one of the bodies, in this case the earth (= 5.92 × 1024 kg), and m is the mass of the second body, in this case an object in earth’s gravitational field. r is the distance, from center to center, between the two objects. For objects near the earth’s surface, this is just the radius of the earth, or 6,371 km (the geometric mean radius).
In the second equation, F is the force on an object, m is the mass of the object, and a is the acceleration of the object as a result of the force applied to it. We see that the gravitational force on an object, its weight, is proportional to its mass. Since the force necessary to produce a particular acceleration is also proportional to the mass of the object, the acceleration of objects released from some particular height above the earth is independent of their mass. The second equation becomes F = mg, where g (= GM/r2) equals 9.81 m/s2 as noted above.
The motion of an object in free fall, as that of the steel ball in this demonstration, is linear motion at constant acceleration, in this case, -g. (Gravity points downward.) The acceleration is the time rate of change of the velocity, or g = dv/dt, so that if the initial velocity is v0 at some reference time, then the velocity at any time is
v = v0 - gt
where t is the time elapsed since that reference time. Since the steel ball is initially at rest, v0 = 0, and v = -gt. The velocity, v, is the time rate of change of position, or v = dy/dt, so that if the initial position of the steel ball is y0 at some reference time, then its position at any time is
y = y0 - (1/2)gt2
where t is the time elapsed since that reference time. The vertical scale has its zero at the top, with distance increasing toward the bottom, so y0 = 0, and we have y = -(1/2)gt2. If we wish to determine g from the fall time, we have g = -2y/t2, which for the (-)1.50-meter fall distance and 0.555-second time for the apparatus in the photograph gives g = 3.00 m/0.308 s2 = 9.74 m/s2. This is about 0.71% low compared to the accepted value of 9.81 m/s2. It is, however, only a single measurement. If you wish to obtain a more precise measurment of g, it is best to take measurements for several fall distances, taking multiple measurements at each fall distance, plot the distances versus the squares of the average times, and find the slope of the resulting line, which is g. Making calculations like the one above for single measurements at a variety of fall distances, however, should yield accurate enough values of g to illustrate the motion of bodies in free fall.
There is some variation in g with latitude. Since the earth is oblate, r increases as one goes from either pole to the equator, and since the earth is rotating, objects experience a centripetal acceleration that increases as one goes from either pole to the equator. Both of these cause the apparent weight of an object to decrease as one goes from either pole to the equator. These variations, however, are quite small. For the latitude at which UCSB lies, the deviation introduced by this is likely to be on the order of about one-sixth the deviation of the measurement quoted above.
It bears noting that for g = 9.81 m/s2, we should expect the time for a 1.50-meter fall to be √(3.00 m/9.81 m/s2) = 0.553 s. The value of 0.555 s differs from this by only 2 ms. A timing error of this size could easily arise from insufficiently fast operation of the switch, or hysteresis in the electromagnet. Multiple runs at different fall distances, and consistent operation of the switch, should mostly eliminate the errors introduced by these things.
As for the motion along the tilted air track, we can add some other equations to those above. Since the acceleration is constant, the average velocity for any time interval, t, is
vave = (1/2)(v0 + v)
where v0 is the velocity at the beginning of the interval and v is the velocity at the end of the interval. For the steel ball, v0 = 0, and vave = (1/2)v. As a result, we can add the following to the equations above:
y = y0 + (1/2)(v0 + v)t,
But both y0 and v0 equal zero, so y = (1/2)vt. From the first equation, we find t = -v/g, and if we substitute this for t in the last equation above, we obtain
v2 = -2gy
(The full-blown equation, for y0 and v0 not equal to zero, would be v2 = v02 - 2g(y - y0).)
References:
1) Resnick, Robert and Halliday, David. Physics, Part One, Third Edition (New York: John Wiley and Sons, 1977), pp. 44-6, 344-6.