Lab 10 : Rotational Energy and Angular Momentum

10 Rotational Energy and Angular Momentum
Introduction
In the process of adding rotational motion to our models of kinematics and dynamics, we
have introduced the concepts of rotational kinetic energy and angular momentum.
We must include rotational kinetic energy in order to apply the principle of conservation of
energy to systems involving rotational motion. Angular momentum is a new (to us) quantity
thatis conserved inasimilar way tolinear momentum. That is, thetotal angularmomentum
of an isolated system is conserved even when energy is not.
Energy of a Mass-pulley System
I
R
m B
ω
v
d
m A
v
Figure 1: A hanging mass-pulley system.
Figure 1 shows a diagram of a system of two hanging masses m A and m B connected by a
string running over a pulley with rotational inertia I. If the hanging masses are not equal,
then the system will accelerate when released from rest. The pulley has rotational inertia, so
the kinetic energy of the system is a combination of linear and rotational terms. The kinetic
and potential energies of the system are expressed as
K = 1 2 m Av 2 + 1 2 m Bv 2 + 1 2 Iω2
U = m A gy A +m B gy B (1)
The hanging masses are connected by a string that does not stretch. If mass 1 moves upward
a distance ∆y 1 = d as shown in Fig. 1, then mass 2 moves the same distance downward

∆y B = −d. A convenient coordinate system for describing changes in the vertical positions
of the masses (for calculating potential energy, for example) is therefore
y A = −y B (2)
Furthermore, the string does not slip over the pulley, so the distance d is related to the
angular position of the pulley in radians via
and the angular and linear velocities are related via
With Eq.’s 2-4, the energy expressions of Eq.’s 1 become
d = Rθ (3)
v = Rω (4)
K = 1 (
mA R 2 +m B R 2 +I ) ω 2
2
(5)
U = (m A −m B )gRθ (6)
If no work is done by forces other than gravity, and energy is conserved, then the total
mechanical energy of the system E = K+U is constant. All of the quantities in Eq.’s 5 and
6 can be readily measured. You will have an opportunity to test conservation of energy for
this system using a rotational motion detector and Logger Pro.
A Rotational Collision
Before
After
ω = 0 Β1
ω Α1
I B
I A
ω 2
I A
+
I B
Figure 2: A perfectly inelastic rotational collision.
Fig. 2 shows a perfectly inelastic (“sticky”) rotational collision in which an initially stationary
object with rotational inertia I B is dropped onto another object with rotational
inertia I A initially spinning with angular velocity ω A1 . After the collision, both objects spin
together with angular velocity ω 2 .

The total angular momentum of an isolated system is conserved. The angular momentum
of an object with rotational inertia I and angular velocity ω is given by
L = Iω (7)
Conservation of angular momentum for a two-body rotational collision is expressed
L A1 +L B1 = L A2 +L B2 (8)
For the perfectly inelastic rotational collision shown in Fig. 2, Eq. 8 becomes
I A ω A1 = (I A +I B )ω 2 (9)
You will test conservation of angular momentum for perfectly inelastic rotational collisions
between the ring and disk you considered in Lab 9.
Experiments
Note : In order to analyze these measurements, you will need the rotational inertia of the
aluminum disk and the heavy ring with uncertainties, which you measured as part of Lab 9.
Energy of a Mass-pulley System
1. Use the Vernier caliper to measure the diameter of the largest spool on the rotational
motion sensor (RMS).
2. Mount the RMS on the horizontal bar, away from the table so that the masses can
hang freely from the pulley. The aluminum disk should be mounted on the RMS.
3. Hang the two 50 g mass hangers over the largest spool of the RMS using the thread
provided.
4. Run Logger Pro and load the file RotationalEnergy.cmbl.
5. Add a 10 g mass to one of the mass hangers, and measure the motion of the system as
it accelerates. The system will slow down and stop on its own once the heavier mass
hits the floor.
6. Collectmeasurements offivecleandrops. (Rejectdropsduringwhichthemassescollide
as they pass one another.)
A Rotational Collision
1. Mount the RMS on the vertical bar so that its rotational axis is aligned vertically.
2. Measure the mass of the ring.
3. Run Logger Pro and load the file RotationalMotion.cmbl.

4. Start acquiring RMS data with Logger Pro, and start the aluminum disk spinning at
ω 1 ≈ 30 rad/s.
5. Carefully hold the ring, smooth-side-down, centered over the disk. Gently drop the
ring onto the disk from a height of no more than a few millimeters.
CM of the Ring
a
h
Axis of Rotation
b
Figure 3: A typical collision in which the ring lands a distance h off-center.
6. It is very unlikely that the ring will be centered on the disk after the collision. Instead,
it will usually end up with its center of mass some distance h from the axis of rotation
of the system as shown in Fig. 3. You will need to correct the rotational inertia of
the ring for this offset using the parallel axis theorem. The distance h is difficult to
measure directly, but it isrelated tothe shortest andlongest distances, aandbbetween
the outside edges of the ring and disk by
h = b−a
2
After you have your angular velocity data, and before you remove the ring from the
disk, measure a and b as precisely as you can by using the metal rod that emerges from
the end of the Vernier caliper.
7. Collect data for at least five rotational collisions. Make sure that you have angular
velocity data both before and after each collision.
Analysis
Energy of a Mass-pulley System
Note : The aluminum disk provides rotational inertia, but does not act as the actual pulley,
so in the analysis of your measurements, the radius R in Eq.’s 3-6 is the radius of the largest
spool of the rotational motion detector, not the radius of the aluminum disk.
1. Put your measured m A , m B , I, and R values into the calculation of the kinetic energy
in Logger Pro. Click on Data -> Column Options -> Kinetic energy to bring up
a Calculated Column Options widow. Choose the Column Definition tab. In the
‘‘Equation’’ box, you should see
(10)

1/2*((0.05+0.06)*0.02^2 + 0.0001)*"Angular Velocity"^2
which follows the form of Eq. 5 with m A = 50 g, m B = 60 g, I = 0.0001 kgm 2 and
R = 2 cm. Replace the values of I and R with your measured values, and click Ok.
2. Putyourmeasured m A , m B , I, andRvaluesintothecalculationofthepotentialenergy
in Logger Pro. Click on Data -> Column Options -> Potential energy to bring
up a Calculated Column Options widow. Choose the Column Definition tab. In
the ‘‘Equation’’ box, you should see
(0.05-0.06)*9.8*0.02*abs("Angle")
which follows the form of Eq. 6 with m A = 50 g, m B = 60 g, I = 0.0001 kgm 2 and
R = 2 cm. Replace the value of R with your measured value, and click Ok.
3. Devise and carry out a strategy to determine whether or not your observations are
compatible with the principle of conservation of mechanical energy within uncertainty.
A Rotational Collision
1. For each collision, devise and carry out a strategy for determining ω A1 and ω 2 from
your measurements using Logger Pro.
2. Use Excel to calculate L 1 and L 2 for all of your collision measurements. In calculating
L 2 , we must use the parallel axis theorem to correct for the fact that the ring was
off-center, so
L 2 = (I A +I B +m ring h 2 )ω 2 (11)
3. Don’tworryaboutevaluatingexperimentaluncertaintiesforthesemeasurements. These
collisions are not repeatable, so it’s difficult to use a statistical approach here.
Questions
1. Discuss the compatiblility of your observations of the mass-pulley system with the
principle of conservation of energy.
2. Are your measurements of perfectly inelastic rotational collisions compatible with with
the principle of conservation of angular momentum within 10% In collisions where L
was not conserved according to your analysis, what do you think may have happened