Experiment 7: Rotation
You will check the work-energy theorem in a rotating system, and also
check Newton's second law of rotation, τ = Iα. A metal disk, of known
mass, is suspended from strings tied to its axle. You wind it up, then
study its motion as it unwinds down from rest. Observing the disk’s
radius, change in elevation, the number of revolutions it turns, and the
time for the trip lets you calculate the potential energy at the top and
kinetic energy at the bottom. To find the work done in between, you continue to watch the disk
after it bounces off the bottom; the height it returns to will be less than where you released it, and
the difference tells you the energy lost to friction. Iα. can be calculated from this same data. This is
compared to the torque, rF, exerted by the strings. You measure the radius around which the strings
are wound; F approximately equals the system’s weight.
Procedure.
Distances from the lowest position make
things simpler than using distances from the
floor: Subtract off z before recording h1 or h2.
Wind the disk up about 15 or 20 cm,
measuring this distance, h1. To measure θ,
the angle turns as it unrolls, count revolutions
as you wind it up. This will be the same
number as on the way down, but easier to
count. Wind it so that the string is always
wrapping around the axle itself, not on top of
another layer of string. Keep the disk centered between the strings as you wind to keep it from
rubbing them on the way down.
Release the disk, and measure the time, t, for it to reach the bottom. Do not let it rub on the strings.
It will then wind back up; measure the height it rises, h2. Repeat, for a total of three trials, and
average. Estimate uncertainties from the amount of variation between trials. (This isn’t a
statistically significant number of trials, so you can’t calculate a standard deviation.) For example,
if you got 16.2, 16.9 and 16.4 for one of the measurements, the average is 16.5 and a good estimate
for the uncertainty would be + .5. All the trials fit within 16.5 + .4, but with only three of them you
can’t be sure that you might not be a little farther off, so be
generous.
Record the mass of the disk, M, from the tag attached near its
edge, and measure its radius, R. The uncertainty in M is
practically zero; estimate how far off you might be with R. Also
measure the moment arm, r, which the string unwinds from. As
shown, you can't be sure whether the string tension force is acting at the left side of the string or at
the right, or somewhere in between. So, the moment arm could be anywhere from the axle's radius,
to the axle's radius plus the thickness of the string. Measure the axle’s diameter, d, with a
micrometer. Find its radius for the minimum moment arm. Add the string’s thickness to get the
maximum.
Calculations:
For the trip the disk made downward: Compute its average angular velocity. What is its original
angular velocity? From the average and original, find the final angular velocity. Compute its
angular acceleration. Compute the moment of inertia of the disk alone. (The axle has a much
smaller radius, so its I is practically zero.) With your answer, show the formula used to find it.
Remember that when multiplying or dividing, you add percentages of uncertainty. Just leave the
results as percents here.
Energy: Calculate Et, the disk/axle system’s energy at the top, and Eb, the system’s energy at the
bottom, with their uncertainties in joules. Note that:
1. For the potential energy, include the total mass of both the disk and the axle. (The axle is 48
grams.) You left out the axle’s mass in finding the moment of inertia because of its much
smaller radius, but the radius is not involved in the potential energy.
2. The kinetic energy in this situation is primarily in the form of rotational kinetic energy.
From the amount of height lost between h1 and h2, determine how much potential energy the system
lost between those points. Remember, when subtracting you add the uncertainties themselves.
Then, convert to a percent when you multiply.
“Energy lost to friction” or “work done by friction” are pretty much two ways of saying the same
thing. The only difference is a sign: Energy lost would be negative work. Let’s assume half of this
work was done on the way down and half going back up. That would mean that you get the work
done on the way down by halving the energy loss you just calculated and slapping a minus sign on
it.
Calculate Et + W for comparison to Eb.
Torque: Compute the quantity Iα. Separately find τ, the torque causing
the disk to rotate, and see if they come out the same, within uncertainty:
Compute the total weight of the disk and axle. Because the disk's
acceleration is small, this weight is nearly equal to the tension in the
string, F. Its uncertainty is small enough to call it zero. Find the torque of
the string tension, which is the torque causing the disk to rotate.
Within the uncertainties, does the initial energy plus the work done agree
with the final energy, as the work-energy theorem states? Does τ agree with Iα ?
PHY 131
Experiment 7: Rotation
Data:
θ = _______________ ± 0
trial 1
z = ______________
trial 2
trial 3
average
h1 =
±
t=
±
h2 =
±
M = _______________ + 0
R = _______________ + _______________
d = _______________
rmin= _____________, rmax= _____________, so, r = ______________ + _____________
Calculations:
ωav =
ωi =
ωf =
α=
I=
Et =
Eb =
E lost between h1 and h2 =
Work =
Et + W =
Iα =
Weight = F =
τ=