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course PHY241
11/29 9
Rotational dynamics with magnets on strapBasically, after a little setup and practice, you are going to spin the strap 15 times, using a rubber band chain. The system doesn't behave all that consistently, so don't go to a lot of trouble trying to be very precise. But do take some care to use the length of the rubber band chain for each of your 'real' trials. Setup and data collection should take between 15 and 30 minutes.
You will then answer some questions. The amount of calculation necessary is not extensive, and most of the calculations are of the same type you have done many times already in the course. Do be sure to use units throughout your calculations. The main challenge will be interpretation of the questions in terms of the system.
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You will accelerate the strap using a chain of rubber bands, as demonstrated in class.
Assume for now that the restoring force of your chain is
F = - k x,
where x is the length of the chain in excess of its length when supporting a single suspended domino, and k is
k = (1/2 N / cm) / (number of rubber bands in chain).
For example, suppose you have a chain of 7 rubber bands, which has length 44 cm when supporting the weight of a domino. When the chain is 50 cm long, it exerts a force, which is calculated as follows:
For this chain k = ( 1/2 N / cm ) / 7, since there are 7 rubber bands. We get k = .07 N / cm, approx..
The chain is 6 cm longer than it was when supporting a domino, so x = 6 cm.
The force exerted at this length is therefore F = - k x = -.07 N / cm * 6 cm = -.42 N.
Forces for other lengths can be calculated using F = - k x = -.07 N / cm * x. Just remember that x is the length in excess of the 44 cm basic length.
Loop the last rubber band in your chain around the edge of the strap in such a way that when released the strap rotates in the clockwise direction. Most likely this will cause the nut to spin in such a way that it descends the threaded rod.
Extend the rubber band chain far enough that when the strap is released it will rotate through several revolutions before coming to rest. However make it spin it fast enough to be dangerous, and make sure it's spinning slowly enough that you can count its revolutions.
Find a chain length that accomplishes the above goals.
What is the length?
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Unstretched rubberband is 34cm.
Stretched rubberband is 40cm.
40cm-34cm=x, x=6cm
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You will use this chain length for every trial in your experiment.
Advice:
For each of the trials below, if the nut does rotate, return it to its original position between trials.
If you wish you can lubricate the system with WD-40 or something similar. This is entirely optional. If you don't have anything like WD-40 handy, don't worry about it. Your instructor doesn't yet know whether this will yield better results, or worse results, than not lubricating the system.
Using this chain length for every trial, do five trials, releasing the system and counting the half-revolutions. It should take several seconds for the system to come to rest, so you should also be able to look at the second hand or digital display of a clock or wristwatch when you release the strap, and when it comes to rest. You should be able to remember the two clock positions and the number of half-revolutions long enough to write them down.
Report your data for these five trials:
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13, 8
16, 10
12, 8
9, 7
15, 10
This set was just the strap itself, with no magnets. The first number represents the half-revolutions and the second number represents the seconds the 2nd hand on the clock made until it came to rest.
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Now repeat, but this time with a magnet halfway between the axis and each end of the strap. Again do five trials.
Report your data for these five trials:
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9, 6
10, 7
8, 6
7, 5
6, 5
magnets halfway between axis and the end
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Repeat once more, with the magnets near the ends of the strap. Again do five trials.
Report your data for these five trials:
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13, 13
9.5, 10
13, 13
9.5, 11
9, 11
magnets on the ends, laying longways
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Based on your data:
What is the median number of half-revolutions through which the strap rotated for each set of five trials?
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1st set: 13
2nd set: 8
3rd set: 9.5
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What is the median time required for the strap to come to rest for each set of five trials?
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1st set: 8 seconds
2nd set: 6 seconds
3rd set: 11 seconds
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Based on your median values, what is the average angular velocity of the strap for each set of five trials, calculated in half-revolutions / second?
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1st set: 1.625 half-rev/sec
2nd set: 1.333 half-rev/sec
3rd set: 0.8636 half-rev/sec
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Assuming uniform angular acceleration, what is the initial angular velocity of the strap for each set of five trials, calculated in half-revolutions / second?
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1st set: 3.25 half-rev/sec
2nd set: 2.66 half-rev/sec
3rd set: 1.727 half-rev/sec
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Assuming uniform angular acceleration, what is the acceleration of the strap for each set of five trials, calculated in half-revolutions / second^2?
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1st set:-0.406 half-rev/sec^2
2nd set: -0.435 half-rev/sec^2
3rd set: -0.156 half-rev/sec^2
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The moment of inertia of the strap is about 1/12 m L^2, where L is its length (about 30 cm) and m its mass (about 65 grams). The moment of inertia of a magnet whose center is at distance r from the axis is m r^2, where m is its mass (about 50 grams).
What therefore is the moment of inertia of the strap (plus magnets if present) in each of your trials?
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1st set: 4875g*cm^2 [1/12(65)(30^2)=4875]
2nd set: 7687.5g*cm^2 [4875+(50(7.5^2))]
3rd set: 13325g*cm^2 [4875+(50(13^2))]
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The angular kinetic energy of an object whose moment of inertia is I is given by the formula KE = 1/2 * I * omega^2, where omega is its angular velocity. Since a half-revolution corresponds to pi radians of angular distance, the angular velocity in rad / sec is pi times the angular velocity in half-rev / sec.
What is the angular KE of the strap corresponding to each set of five trials, based on the initial angular velocity you previously calculated?
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63525 g*cm^2*rad/s [1/2(4875)(1.625*pi)^2]
67408 g*cm^2*rad/s [1/2(7687.5)(1.333*pi)^2]
49041 g*cm^2*rad/s [1/2(13325)(0.8636*pi)^2]
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What average force was exerted by the rubber band system during each trial? This result should be the same for all trials, if you followed the instruction to use the same chain length for each trial.
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-0.75N
[F=-ks, k=0.5/4 rubberbands=0.125, F=-0.125(40-34)=-0.75N]
This would be the force of maximum magnitude. The force changes linearly from 0 to magnitude .75 N, so the average force .75 N / 2 = .38 N, approx..
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Through what distance do you estimate the rubber band chain exerted its force? Again a single result will probably apply to all trials.
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About how much work did the rubber band chain therefore do during each trial? Once more a single result will probably apply to all trials.
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-0.75N(6cm)=-4.5N*cm
Should get half of this. A Newton is 100 000 g cm/s^2, so your result, to compare with those you got above, would be about 200 000 g cm/s^2.
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What is the average torque on the system during each trial? (recall that torque = force * moment-arm; in this case the moment-arm is the distance from the axis to the point where the rubber band exerted its force)
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-11.25N*cm
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Through what fraction of a radian do you estimate the system traveled between release and the instant the rubber band chain went slack?
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6 cm, on a circle of radius 15 cm, is 6/15 rad = .4 rad
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What therefore is the product of average torque and angular displacement, for this interval?
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(-11.25N*cm)*(6cm)=-67.5N*cm^2
The product would be ( 11.25 N * cm ) * .4 rad = 4.5 N cm^2, consistent with your previous result.
You should actually get half of this; the average torque would be half the max torque.
This should reinforce that ave. torque * angular displacement = work.
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