Data:
“Conservation of Angular Momentum”
Adam Capriola
Experiment Performed: 10/16/09
Report Due and Handed in: 10/30/09
Partners: Danielle Teitelman and Christine Loomis
Purpose
To compare the moments of inertia calculated using two different methods, and to verify that angular momentum is conserved in an interaction between a rotating disk and a ring dropped onto the disk.
Hypothesis
If a weighted ring is added to the disk, the moment of inertia will be the same as the disk without the weighted ring. The angular momentum before the ring is dropped on the disk during part two will be greater than the angular momentum after the ring is dropped.
Labeled Diagrams
See attached sheet.
Data
Part 1
Mass of disk (M): 1.500 kg Radius of disk (R): 0.114 m
Radius of shaft (r): 0.006 m Mass of ring (m): 1.420 kg
Inner radius of ring (R1): 0.054 m Outer radius of ring (R2): 0.064 m
Disk Alone
|Force of Kinetic|Angular acceleration|Final angular |Tension T in |T - fk |Net torque τ = |Moment of inertia|Moment of inertia I =τ/α |
|friction fk (N) |α |velocity ω |string |(N) |r(T- fk) (Nm) |I = ½ MR2 |=r(T- fk)/α |
| |(rad/s2) |(rad/s) |(N) | | |(kgm2) |(kgm2) |
|0.394 |1.732 |16.36 |3.52 |3.13 | 0.019 |0.010 |0.011 |
Disk plus ring
|Force of Kinetic|Angular acceleration|Final angular |Tension T in |T - fk |Net torque τ = |Moment of inertia|Moment of inertia I =τ/α |
|friction fk (N) |α |velocity ω |string |(N) |r(T- fk) (Nm) |I = ½ MR2 + ½ |=r(T- fk)/α |
| |(rad/s2) |(rad/s) |(N) | | |m(R12 + R22) |(kgm2) |
| | | | | | |(kgm2) | |
|0.443 |1.126 |14.52 |3.33 |2.89 |0.017 |0.015 |0.015 |
Part 2
|Angular velocity before|Angular velocity after |Moment of inertia of |Moment of inertia plus |Angular momentum before |Angular momentum before |
|ring is dropped (ωi) |ring is dropped (ωi) |disk |ring |ring is dropped (L = Iiωi)|ring is dropped (L = Ifωf)|
|(rad/s) |(rad/s) |(I = ½ MR2) |(I = ½ MR2 + ½ m(R12 + |(kgm2/s) |(kgm2/s) |
| | |(kgm2) |R22)) | | |
| | | |(kgm2) | | |
|16.27 |9.681 |0.010 |0.015 |0.163 |0.145 |
Graphs
Part One (Disk Alone)
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Part One (Disk Plus Ring)
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Part Two
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Questions
Part 1
1. In your data table in Part 1, you have two values for the moment of inertia. One is found from the theoretical equation for moment of inertia that is introduced in the Theory section and other is an experimental value obtained using Newton’s 2nd law for rotational motion, τ = Iα, in conjunction with the definition of torque, τ =rF. How well do your two values agree with each other? What is the percent difference? Which do you think is likely a better way to calculate a value for moment of inertia?
The values are extremely close, as the percent difference for the disk alone is 9.5% and the disk plus the ring is 0% (they are of equal value). I think the better way to calculate the moment of inertia is to use I = ½ MR2, as it is a more elegant equations that takes into account less variables. The other equation takes more variables into account, mainly for calculation torque, which I feel leads to increased error.
Part 2
1. How do your values for the angular momentum before and after the ring is dropped onto the disk compare? What is the percent difference?
The angular momentum before the ring is dropped onto the disk is greater than the angular momentum after the ring is dropped onto the disk. The percent difference is 11.7%.
2. Does there appear to be an inverse relationship between moment of inertia and angular velocity?
No, there appears to be a direct relationship between moment of inertia and angular velocity. As the angular velocity decreased, so did the moment of inertia.
3. How well do your results support the theory of conservation of momentum? What are the limitations of the experimental setup?
The results somewhat support the theory of conservation of momentum. The percent difference is 11.7%, which I suppose isn’t a huge discrepancy, but it could be better. The limitations of the experimental setup were that it is difficult to drop the ring on the spinning disk perfectly. We were able to drop the ring into the grooves of the disk, but there was still some wiggle room in those grooves. The ring would need to fit in the grooves like a puzzle piece in order to be positioned dead center to yield the least amount of error.
Conclusion
Lab Summarized
During the first part of the lab, the moments of inertia for a spinning disk with and without a weighted ring on top were calculated using two different methods. The force coercing the disk to spin was a 300 g weight attached to the shaft of the disk using a string a pulley system. The weight was allowed to free fall and the resulting graph of velocity versus time was used to find the final angular velocity by taking the mean of the segment after which the string had completely unraveled from the shaft. The angular acceleration was found from the slope of this graph up to that point.
These values, along with the force of kinetic friction, found by determining the minimum force needed to get the disk spinning, were used to find the moment of inertia. The moment of inertia was also calculated a second way, using the radii and masses of the disk and ring.
The second part of the experiment was performed much like part one of the experiment using the disk alone, only this time shortly after the string had unraveled, the ring was dropped onto the spinning disk. Using the angular velocities and moment of inertias, determined much like they were in part one, the angular moments before and after the ring were dropped were calculated and compared.
The percent differences between the two different calculations of the moments of inertia in part one were quite low. Using the disk alone, the percent difference was 9.5% and with the disk plus the ring, the percent difference as 0%. Under perfect conditions, the values should have been equal. The angular acceleration and final velocity for the disk along were greater than that of the measurement for the disk plus the ring, which could be expected, due to the extra mass. The percent difference between the angular momentums before and after the ring was dropped in part two was 11.7%. They should have been equal under ideal conditions.
As stated previously in the questions, some of this error is most likely due from the ring not being placed dead center around the spinning disk. If the disk was dropped at and angle and did not make complete contact with the disk at the same instant, this could have also caused error. If the shaft was not properly lubricated, this would have caused error throughout the experiment. Lastly, if the string ever caught a snag while unraveling, this would have also contributed to the error.
Equations
I =τ/α
I = ½ MR2
L = Ifωf
Iiωi = Ifωf
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