Energy and momentum conservation Lab Writeup
(E.J. Zita for Physics of Astronomy, Feb 2006)
1. Purpose: to investigate conservation of energy, momentum, and angular momentum
2. Questions: How can we measure changing energies and momenta in a dynamic
experiment? To what extent are energies and momenta conserved in our experiment?
3. Method: We used the Rotational Dynamics Apparatus: A disk (M,R) floating on air is
spun up by a falling mass (m) hung over a “frictionless” pulley (mp, rp). We dropped the
mass from rest and measured the spin frequency f(t) of the disk (using an optical sensor) and
the changing position (x) and speed (v) of the mass (using a sonic pulse emitter and sensor).
The resultant data are analyzed in Sec.5 below to answer the questions in Sec.2.
m
Predictions:
(a) Total energy should be conserved, insofar as the system is frictionless. Rotational kinetic
energies of the disk and pulley + translational Kinetic energy of the falling mass should
increase at the expense of Potential energy (and vice versa as the mass rises)
(b) Angular momentum should not be conserved, since there is an external torque on the
system: gravity pulling the hanging mass down, then string tension pulling it up. Similarly,
linear momentum is not conserved, but increases and decreases with the kinetic energy.
4. Data and observations:
Observations: As the hanging mass falls, sure enough, the disk spins up. When the hanging
mass hits bottom, the disk winds the string back up, pulling the mass up. If we let the
experiment run for many cycles, we can see the amplitude of the oscillations slowly decay
(Miranda E-R’s team), clear evidence of gradual energy loss.
Possible sources of energy loss: Friction between the two disks: the bottom disk should
remain still or spin slowly at a constant rate, but it appears to spin up when the top disk spins
up, and may slow down when the falling mass rises.
In some experiments, there is a jog in the x(t) and v(t) data for the falling mass when it hits
the bottom. Matt E. noted that appears to be when the string is positioned such that it slips
from side to side in the little opening on the disk.
Note that the string is wound around a little disk on top of the big disk, with lever arm = rtop
Data: Raw data for frequencies and speeds are in the attached Excel chart (rot.xls). The
physical characteristics for parts of the apparatus are tabulated below. Aman D., Chris C.
and Andrew Y. measured these data (with some guidance from Zita). Sizes are uncertain to
about r = 0.1 mm, and masses to about m = 1g.
Falling mass: m = 26.5 g
disk: M = 1.353 kg
R = Rdisk = 127.5/2 mm
Rinner hole= 15.65 mm
Little disk on top of big disk:
r = rp = rpulley = 25.3/2 mm
rgroove = 24.85/2 mm
rtop = 25.2/2 mm
rinner hole= 15.6/2 mm
hole depth = d = 10 mm
mtop = 0.015 kg = 15 g
mpulley = 25.9 g (with
inner cylinder removed
5. Calculations:
a. Frequency: The disk frequency data were recorded in radiation counts due to a software
setup peculiarity. Radiation counts measure the number of black stripes per second to pass
the optical sensor (f ‘ in stripes/sec). We can convert this to actual frequency ( f in rev/s or 
in rad/s) by noting that each black stripe is 1 mm wide, separated by 1 mm white stripe, so
one count is 2 mm of arclength.
R
Circumference C= 2R =  127.5 mm ~ 400 mm
Number of stripes / revolution = N = 400 mm/rev / (2 mm/stripe) = 200 stripes/rev
(I also estimated this by observing rotations, and counted stripes by hand – consistent.)
f(rev/s) = f ‘ (stripes/sec) / N (stripes/rev)
We calculated f in the Excel column next to counts/sec by dividing by 200.
Then the angular frequency (rad/sec) = 2f (rev/sec).
b. Energies: The moments of inertia above contribute to the rotational kinetic energies
calculated in the Excel Sheet. The angular frequency of the pulley (p) is related to that of
the disk () by the linear velocity (v) of the mass (and the string):
v = rtop  = rgroove p
2
Kdisk=½ Idisk 
Ktop=½ Itop 2
Kpulley=½ Ipulley p2
For the hanging mass, the translational kinetic energy is Kt=½ m v2 and the
gravitational potential energy is U = mgh where h=0 at the lowest value of x - call it xmin. So
U = mg(x-xmin).
c. Moments of inertia: For a disk or cylinder of mass M and radius R with uniform mass
distribution, the moment of inertia I=½ MR2. From this, we must subtract the moments of
inertia due to the 10 mm deep hole in the big disk, and the non-rotating central core of the
pulley.
(i) Big disk: Idisk = Isolid + Ihole where Isolid = ½ MsolidR2 and Ihole = ½ Mhole R2hole.
I’m going to approximate Idisk = Isolid =½ MsolidR2, assuming the hole is filled. The actual
moment of inertia is slightly greater, since there is less mass near the center. My estimate
will be off by approximately M hole  Volumehole  radiushole  depthhole
M solid
radiusdisk  thicknessdisk
Volumesolid

We did not measure the thickness of the disk – was it about twice the hole depth?
Idisk ≈ ½ MR2 = 0.002749 kg*m2
(I values are calculated in the Excel sheet, rot.xls)
(ii) Little disk on top: Itop = ½ mtop r2top = 1.19E-06 kg*m2
(iii) Pulley: Ipulley = ½ mpulley (r2p – r2inner) = 1.28E-06 kg*m2
Giancoli Fig.10-21.d (p.252)
d.(i) Angular momenta of disks and pulley: L = I, and the angular frequency  varies
with time, as tabulated in the Excel data sheet. L of the disk and top are up, and L of the
pulley is parallel to the pulley axis, so we cannot add these to try to conserve total angular
momentum.
(ii) Angular momentum of falling mass: L = mrv where r = the distance from the axis
of the big disk to the pulley where the mass is falling down. Direction is parallel to L of the
pulley, perpendicular to L of the disk. (See Giancoli Ch.11 # 25.)
6. Analyses and results
Here are the energy data from (rot.xls). These look quite reasonable.
Energies vs time
Energy(J)
8.00E-02
7.00E-02
6.00E-02
U=mgh
Kt=1/2 mv^2
5.00E-02
4.00E-02
3.00E-02
2.00E-02
1.00E-02
0.00E+00
-1.00E-02 0
Kdisk
Ktop
Kpulley
Ktot
Etot
5
10
15
20
25
30
35
Time (s)
The total energy (aqua + signs on top) is slowly decreasing, bounce by bounce, but
reasonably tops out at the maximum of both the potential energy U (blue diamonds) and the
total kinetic energy (brown dots). Ktot rises as U falls, and vice versa. The kinetic energy is
almost entirely due to the big spinning disk (yellow). The hanging mass (Kt, pink)
contributes little to the total kinetic energy because it is not very massive (ditto for the pulley
and the little top disk). The changing potential energy of the hanging mass drives the system.
Energy drops from to 6.57 x10-2 J to 5.97x10-2 J over one cycle, from t=5.5 to 16.7 s. Energy
loss rate = 5.36 x10-4 J/s, over 9% per cycle. That’s a considerable loss, if accurate. We
could calculate uncertainties or average the data over 3 time intervals, if ambitious.
7. Discussion: The data and calculations test our predictions and answer our questions. Our
predictions regarding energy are pretty well borne out. The bumps at the top of each peak in
kinetic energy may be due to the effect noted by Matt in Sec.4 above, or to actual jostling of
the disk each time the hanging mass is jerked upward at maximum speed, or to some other
effect. Angular momentum could also be plotted, and uncertainties could be calculated.
Energy is generally conserved as expected, except for the 9% friction loss each cycle.
Sample data from rot.xls
Rotational Dynamics lab - week 6 - Physical Systems - Feb.2006 - E.J. Zita
Data are from Aman Dhillon, Chris Casanovas, and Andrew Yu
object
big disk
top disk
pulley
1.353
1.50E-02
2.59E-02
63.75
1.26E+01
1.27E+01
7.80E+00
0.002749
1.19E-06
1.28E-06
mass (kg)
radius
(mm)
r inner
I =1/2
MR^2
1m=
1.00E+03
time (s)
0.1
0.2
0.3
mm
Falling
mass
Position
(m)
h
6.67E-01
6.67E-01
6.66E-01
Spinning disk
pi=
Velocity(m/s)
Acceleration
(m/s^2)
counts
f'
(stripes/sec)
f=f ' /
200.00
(rev/sec)
-4.73E-03
-7.87E-03
-1.03E-02
-2.93E-02
-2.48E-02
-1.95E-02
2
2
2
20
20
20
0.10
0.10
0.10
3.14
w=2pi*f
(rad/s)
0.63
0.63
0.63