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Angular momentum and kinetic energy

by ash1262
Tags: angular, energy, kinetic, momentum
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ash1262
#1
Dec4-13, 11:33 PM
P: 12
A boy sitting on a rotating chair (ignore friction) with his hands stretched outwards, pulls in his hands thus reducing the moment of inertia of the chair-boy system. The angular velocity will increase to conserve the angular momentum.
The kinetic energy of the system will also increase (as kinetic energy = (1/2)Iωω).
What I can not understand is where does the added energy come from?
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Drakkith
#2
Dec4-13, 11:54 PM
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I is the object's moment of inertia. (r2dm) As the radius decreases the momentum is conserved. When the boy pulls in his arms, the angular velocity increases, but the radius decreases, which reduces I. The two effects equal out so that both momentum and energy are conserved.
Nugatory
#3
Dec5-13, 12:13 AM
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Quote Quote by ash1262 View Post
What I can not understand is where does the added energy come from?
The boy had to pull his arms in against the centrifugal force. The added energy came from his muscles doing this work.

ash1262
#4
Dec5-13, 02:33 AM
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Angular momentum and kinetic energy

Thanks Nugatory.
When the boy pulls out his hand again the energy is restored to the previous level. Where does the extra energy go? Friction of the muscles?
sophiecentaur
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Dec5-13, 04:16 AM
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Quote Quote by ash1262 View Post
Thanks Nugatory.
When the boy pulls out his hand again the energy is restored to the previous level. Where does the extra energy go? Friction of the muscles?
Basically yes. If two masses, the equivalent of his arms, were allowed to fly out on springs, the springs would hold potential energy. But, without some loss mechanism, there would be oscillation and the process would continue in-out-in-out for ever. It would be another of those 'paradoxical' situations like connecting capacitors in parallel.
ash1262
#6
Dec7-13, 09:30 AM
P: 12
Thanks PF Patron.
I was trying to calculate the work done by the centripetal force when a stone rotating at the end of a string is brought in from initial radius R to final radius R/2, by integrating [∫mω^2/x^3 r^4 dx] from r to r/2. The magnitude is the same as the gain in kinetic energy, but the sign is not correct. What I want to ask is that in such an integration, do we take dx as positive or negative?
Nugatory
#7
Dec7-13, 09:46 AM
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Quote Quote by ash1262 View Post
The magnitude is the same as the gain in kinetic energy, but the sign is not correct.
If you go back and check all your signs carefully, you'll find where you got one wrong. Most likely you chose the wrong sign for the force. If x is increasing in the outwards direction, then the centripetal force is acting in the direction of -x.
sophiecentaur
#8
Dec7-13, 09:54 AM
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You are moving (definite integration) from a big r to a small r and the force is negative. -X- =+
ash1262
#9
Dec10-13, 02:13 AM
P: 12
In the same lines, a truck moving horizontally with velocity 'v' is loaded with a mass 'm', the mass being lowered vertically. The truck with the mass will move at a lower velocity to conserve the momentum. Again the kinetic energy will drop. Where does the kinetic energy go? I could not detect any work done here. Please help.
sophiecentaur
#10
Dec10-13, 03:21 AM
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This is a new model, you're discussing now, is it (flat Earth with uniform g?)?I don't understand how your model leads to the truck slowing down as the mass goes lower. The two motions are at right angles to each other, as I understand it so they can be discussed separately. The g force is an external force so conservation doesn't apply (you can discuss a common centre of mass of Earth and truck if you want to but the Earth's share of any KE gained is vanishingly small.

If this mass is not in free fall, there will be work done on a braking system. Otherwise, the falling mass will hit the bottom and bounce back with no loss of mechanical energy.
ash1262
#11
Dec10-13, 08:02 AM
P: 12
The truck (mass M) is moving with velocity v horizontally on earth (considered an inertial frame) when a stationary mass (m) is lowered (loaded) slowly on it vertically. The mass (m) had zero momentum in the horizontal direction before it was lowered on the truck and zero kinetic energy (assume). As there is no horizontal force applied on the truck (ignoring friction), the momentum in the horizontal direction will be conserved, and so the velocity of the truck and the mass system (now moving together) will reduce to v'. [v'=Mv/(M+m)]
The initial kinetic energy of the system (truck and mass) was 1/2 M v^2, and the final is 1/2 (M+m) v'^2. The final K.E is less the the initial.
Where does the energy go?
Doc Al
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Dec10-13, 08:44 AM
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Quote Quote by ash1262 View Post
The initial kinetic energy of the system (truck and mass) was 1/2 M v^2, and the final is 1/2 (M+m) v'^2. The final K.E is less the the initial.
Where does the energy go?
Think of the mass and the truck as undergoing an inelastic collision. The "missing" energy ends up mainly as internal energy. (Things heat up a bit.)
sophiecentaur
#13
Dec10-13, 12:24 PM
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Quote Quote by ash1262 View Post
The truck (mass M) is moving with velocity v horizontally on earth (considered an inertial frame) when a stationary mass (m) is lowered (loaded) slowly on it vertically. The mass (m) had zero momentum in the horizontal direction before it was lowered on the truck and zero kinetic energy (assume). As there is no horizontal force applied on the truck (ignoring friction), the momentum in the horizontal direction will be conserved, and so the velocity of the truck and the mass system (now moving together) will reduce to v'. [v'=Mv/(M+m)]
The initial kinetic energy of the system (truck and mass) was 1/2 M v^2, and the final is 1/2 (M+m) v'^2. The final K.E is less the the initial.
Where does the energy go?
There is a horizontal force (impulse) when the mass starts to touch the truck and momentum is shared by M and m (conservation). The difference in KE is always going to be there if they end up at the same speed (energy conservation includes heat / sound energy).
A general principle in this sort of problem is usually that there is some inelastic collision somewhere in the process. You just have to spot it.
redoopi
#14
Jan3-14, 06:38 AM
P: 10
Quote Quote by ash1262 View Post
Thanks Nugatory.
When the boy pulls out his hand again the energy is restored to the previous level. Where does the extra energy go? Friction of the muscles?
Quote Quote by sophiecentaur View Post
Basically yes. If two masses, the equivalent of his arms, were allowed to fly out on springs, the springs would hold potential energy. But, without some loss mechanism, there would be oscillation and the process would continue in-out-in-out for ever. It would be another of those 'paradoxical' situations like connecting capacitors in parallel.
I don't understand... Let me try to analyze the situation in this way:

Ignore the translational K.E. of the hands (e.g. the boy is spinning very fast, and he pulls his hands very slowly).

When the boy pulls in his hands:
decrease in P.E. + work done by muscles to pull hands in + work done by friction when pulling hands in (negative) = increase in rotational K.E.

When the boy pulls out his hands:
decrease in rotational K.E. + work done by muscles to pull hands out + work done by friction when pulling hands out (negative) = increase in P.E.

Assume that the boy pulls his hands such that:
(1) decrease in P.E. = increase in P.E. (i.e. the hands return to the same position before pulling in and after pulling out), and
(2) increase in rotational K.E. = decrease in rotational K.E. (i.e. angular speed is the same before pulling in and after pulling out)

As a result,

work done by muscles to pull hands in + work done by muscles to pull hands out = - work done by friction when pulling hands in (negative) - work done by friction when pulling hands out (negative)

This seems to make sense. But what if we replace the boy by a "perfect machine" that has no friction of its "muscles"? Clearly the work done by friction is zero, but we can still supply energy to the machine to make "work done by muscles to pull hands in" positive. In this case, "work done by muscles to pull hands out" must be negative. But what does it mean? I guess that once we don't supply energy to the hands, the hands will move to a position further away from the position before pulling in, with the extra P.E. and rotational K.E. gained equals the energy supplied to the hands, so that negative work must be done when pulling out if we have to keep it at the position before pulling in (similar to the case that we do negative work to stop a moving ball). Is it correct? But if it is, why would there be oscillations? and how to determine the ratio of the extra P.E. gained to the extra rotational K.E. gained?
sophiecentaur
#15
Jan3-14, 09:58 AM
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However much energy is 'lost' during the cycle of hands out - hands in - hands out etc., it has to be made up from the muscles or the motor in the ideal machine. If the machine runs out of energy supply then the distance that the arm can move back will be less. If you had a mass on a spring on the arm, the oscillations would just die down due to friction.
The only thing you can rely on is for the Angular Momentum to be conserved (ignoring any friction forces, coupling the turntable to the Earth)
redoopi
#16
Jan3-14, 07:26 PM
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Quote Quote by redoopi View Post
I don't understand... Let me try to analyze the situation in this way:

Ignore the translational K.E. of the hands (e.g. the boy is spinning very fast, and he pulls his hands very slowly).

When the boy pulls in his hands:
decrease in P.E. + work done by muscles to pull hands in + work done by friction when pulling hands in (negative) = increase in rotational K.E.

When the boy pulls out his hands:
decrease in rotational K.E. + work done by muscles to pull hands out + work done by friction when pulling hands out (negative) = increase in P.E.

Assume that the boy pulls his hands such that:
(1) decrease in P.E. = increase in P.E. (i.e. the hands return to the same position before pulling in and after pulling out), and
(2) increase in rotational K.E. = decrease in rotational K.E. (i.e. angular speed is the same before pulling in and after pulling out)

As a result,

work done by muscles to pull hands in + work done by muscles to pull hands out = - work done by friction when pulling hands in (negative) - work done by friction when pulling hands out (negative)

This seems to make sense. But what if we replace the boy by a "perfect machine" that has no friction of its "muscles"? Clearly the work done by friction is zero, but we can still supply energy to the machine to make "work done by muscles to pull hands in" positive. In this case, "work done by muscles to pull hands out" must be negative. But what does it mean? I guess that once we don't supply energy to the hands, the hands will move to a position further away from the position before pulling in, with the extra P.E. and rotational K.E. gained equals the energy supplied to the hands, so that negative work must be done when pulling out if we have to keep it at the position before pulling in (similar to the case that we do negative work to stop a moving ball). Is it correct? But if it is, why would there be oscillations? and how to determine the ratio of the extra P.E. gained to the extra rotational K.E. gained?
OK. Let me try to answer my own question. If there is no friction, the "hands" will continue to move out until tension of arms holds it. If we think of the "hands" as two masses sliding along a frictionless infinitely-long arm at opposite ends, then the two masses will continue to move out until gravitational force holds them. In both cases, loss in K.E. = gain in P.E.

If friction exists, when the masses move out, loss in K.E. + work done by friction (negative) = gain in P.E., so as the masses move out, the system will rotate at a rate slower than when there is no friction, and the masses will stop moving out when static friction holds the masses.

If energy is supplied to move the masses out, it will reduce the effect of friction.

(Note: I am learning the subject, so this is just my guess...)


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