Real life problem about angular momentum conservation

[Krushnaraj Pandya]

Homework Statement

suppose you're sitting on a rotating stool holding a 2kg mass in each outstretched hand, if you suddenly drop the masses, will your angular velocity increase, decrease or remain the same?

Homework Equations

dL/dt=net torque
when net torque is 0, L=constant=Iw
therefore, I1w1=I2w2

The Attempt at a Solution

Since by dropping the masses, moment of inertia will decrease, to keep the angular momentum same w must increase. But my textbook says it'll remain the same. Why?

 

[Nathanael]

The Attempt at a Solution

Since by dropping the masses, moment of inertia will decrease, to keep the angular momentum same w must increase. But my textbook says it'll remain the same. Why?

You are right, moment of inertia decreases. But why do you think angular momentum should stay the same? When you let go, do the weights fall straight down or go flying off?

 

[Krushnaraj Pandya]

You are right, moment of inertia decreases. But why do you think angular momentum should stay the same? When you let go, do the weights fall straight down or go flying off?

I think there is a serious flaw in my understanding of when angular momentum is conserved...If we take the man and the weights as a system, there isn't any net external torque in the horizontal direction- doesn't that mean angular momentum is conserved in the horizontal direction?? and so I1w1=I2w2, they will go flying off surely but I don't understand how that relates here. I am having trouble visualizing and developing intuition on angular momentum in general, I would be grateful if you could shed some light on that

 

[Nathanael]

I think there is a serious flaw in my understanding of when angular momentum is conserved...If we take the man and the weights as a system, there isn't any net external torque in the horizontal direction- doesn't that mean angular momentum is conserved in the horizontal direction?? and so I1w1=I2w2, they will go flying off surely but I don't understand how that relates here. I am having trouble visualizing and developing intuition on angular momentum in general, I would be grateful if you could shed some light on that

You are correct that the angular momentum is conserved if we consider the entire system, weights included... but you are ignoring the angular momentum of the weights after they go flying off. If you want to consider only the person as your system then the weights apply external forces.

Good night(/day?) I’ll be back tomorrow(/today?)

 

[Krushnaraj Pandya]

You are correct that the angular momentum is conserved if we consider the entire system, weights included... but you are ignoring the angular momentum of the weights after they go flying off. If you want to consider only the person as your system then the weights apply external forces.

Good night(/day?) I’ll be back tomorrow(/today?)

umm...will the force applied by the weights be the opposite of the centripetal force? what'll be the mathematics of this situation- that ought to make things clearer

 

[Nathanael]

I was actually thinking about what I said and wanted to edit it. When I said that (“if the person is your system the weights are external forces”) I meant it on a more general note. In this particular problem no forces are involved because we assume that we just let go. You could have other scenarios where the weights are thrown.

If you just let go, then there’s no transfer of angular momentum between the person and the weights. The angular momentum of the person never changes and neither does their moment of inertia.

 

[jbriggs444]

A external torque is the only way that one can change the angular momentum of a closed system. But if one considers the system consisting of the man and the weights currently in his hands, one does not have a closed system. One has an open system -- a system in which mass is allowed to enter or leave. The weights start in his hands, inside the system. The weights end flying freely, outside the system.

If one is trying to account for angular momentum in an open system, one must consider the angular momentum carried in or out of the system when mass enters or leaves. Or, one can draw the lines around the system differently so that it is closed, as @Nathanael suggests in #6 above.

 

[Merlin3189]

Consider the initial setup as two (or 3) objects, the man and the weights, each having some angular momentum. The only forces between these objects - the centripetal forces and their reactions - are perpendicular to the axis of rotation, acting through it and cannot alter the angular momentum.
Allowing the bodies to separate by removing these forces, affects the angular momentum of neither (or none.)

The moment of inertia of the man has not changed (unless he allows his arms to pull in after releasing the objects *) and he still has the same AM he had before.
The weights are more interesting as they have the same angular momentum, but an ever increasing moment of inertia and an ever decreasing angular velocity.

* This may be a bit more messy than I was thinking. Presumably the weights were stretching his arms a little and when he let's go, the arms shrink back a bit. On the other hand he has to bend his fingers outwards in order to release the weights. Maybe more blood can flow out into his hands and arms now that the fingers are not clenched and the muscles not tensed.

To get round all this mess, perhaps he should have the masses attached to the ends of a pole, whose centre is pivoted on the axle of his stool. The pole provides the centripetal forces and he just holds the pole lightly at the centre, in order to claim that he is attached and to hold a light locking mechanism clamping the weights to the pole. When he releases this mechanism by loosening his grip very close to the axis of rotation, the weights can fly off and the pole rotate independently.
I expect he'll still move a bit, but maybe we've minimised it. And since he and the weights never applied any significant force to each other, maybe it's easier to see that they are independant of his AM all along..

 

[Krushnaraj Pandya]

I was actually thinking about what I said and wanted to edit it. When I said that (“if the person is your system the weights are external forces”) I meant it on a more general note. In this particular problem no forces are involved because we assume that we just let go. You could have other scenarios where the weights are thrown.

If you just let go, then there’s no transfer of angular momentum between the person and the weights. The angular momentum of the person never changes and neither does their moment of inertia.

so you meant to say the weights don't apply any net external toque if we just let go right? ok, I understand that
*After reading the answers below, I understand that MI of man is always the same but initially when we consider the entire system it is MI of man + weights...later for the man it is I(man) and I(weight) for the weights but the total initial AM=I(man+weights)w is equal to final AM=r1 x p1 + r2 x p2 + I(man)*w. and neither w nor I of man changes but I of weights increase and w decreases but ultimately their individual AM remains same. Am I right?

 

[Krushnaraj Pandya]

A external torque is the only way that one can change the angular momentum of a closed system. But if one considers the system consisting of the man and the weights currently in his hands, one does not have a closed system. One has an open system -- a system in which mass is allowed to enter or leave. The weights start in his hands, inside the system. The weights end flying freely, outside the system.

If one is trying to account for angular momentum in an open system, one must consider the angular momentum carried in or out of the system when mass enters or leaves. Or, one can draw the lines around the system differently so that it is closed, as @Nathanael suggests in #6 above.

my view of this situation mathematically is that initial AM is (I(man)+2mr^2)*w and final AM is addition of AM's of the weights+the man. how do I know from this whether w of person changes or not? will the weights and person have the same w after they fly off?

 

[Krushnaraj Pandya]

Consider the initial setup as two (or 3) objects, the man and the weights, each having some angular momentum. The only forces between these objects - the centripetal forces and their reactions - are perpendicular to the axis of rotation, acting through it and cannot alter the angular momentum.
Allowing the bodies to separate by removing these forces, affects the angular momentum of neither (or none.)

The moment of inertia of the man has not changed (unless he allows his arms to pull in after releasing the objects *) and he still has the same AM he had before.
The weights are more interesting as they have the same angular momentum, but an ever increasing moment of inertia and an ever decreasing angular velocity.

* This may be a bit more messy than I was thinking. Presumably the weights were stretching his arms a little and when he let's go, the arms shrink back a bit. On the other hand he has to bend his fingers outwards in order to release the weights. Maybe more blood can flow out into his hands and arms now that the fingers are not clenched and the muscles not tensed.

To get round all this mess, perhaps he should have the masses attached to the ends of a pole, whose centre is pivoted on the axle of his stool. The pole provides the centripetal forces and he just holds the pole lightly at the centre, in order to claim that he is attached and to hold a light locking mechanism clamping the weights to the pole. When he releases this mechanism by loosening his grip very close to the axis of rotation, the weights can fly off and the pole rotate independently.
I expect he'll still move a bit, but maybe we've minimised it. And since he and the weights never applied any significant force to each other, maybe it's easier to see that they are independant of his AM all along..

I understood this the most clearly. From what I understand now, there is no torque on the system from the forces involved and therefore no change in total AM of any bodies involved since all forces pass through the axis of rotation. I also get that the MI of the weights keeps increasing and their w decreasing...so all AM's of every body is conserved? correct? Now I also understand why the weights flying off was off significance- thank you for writing such a long explanation.
*This would be messy in real life, and you must be a pretty good engineer to have thought of these things (I admit I was quite surprised to see you studied experimental psychology and not mechanical engineering or something related) but it isn't of much importance here- this question was in my high school textbook so it is probably just an ideal case to improve my concepts

 

[jbriggs444]

The act of opening one's hands does not change one's angular momentum.

The formula for moment of inertia as ##L = \omega\ I## applies for rigid bodies rotating around a fixed point. Two weights flying in straight lines do not form a single rotating rigid body. Their total angular momentum will be computed not as ##L = \omega\ I## but as ##L = \vec{r_1} \times \vec{p_1} + \vec{r_2} \times \vec{p_2}##

 

[Krushnaraj Pandya]

The act of opening one's hands does not change one's angular momentum.

The formula for moment of inertia as ##L = \omega\ I## applies for rigid bodies rotating around a fixed point. Two weights flying in straight lines do not form a single rotating rigid body. Their total angular momentum will be computed not as ##L = \omega\ I## but as ##L = \vec{r_1} \times \vec{p_1} + \vec{r_2} \times \vec{p_2}##

Got it! so at first AM is total I of man+weights into w since they were one single body (of sorts), later it is L=r1 x p1 + r2 x p2 + I(of man)w, am I right?

 

[jbriggs444]

Got it! so at first AM is total I of man+weights into w since they were one single body (of sorts), later it is L=r1 x p1 + r2 x p2 + I(of man)w, am I right?

Yes, that is a correct statement of angular momentum conservation.

 

[Krushnaraj Pandya]

Yes, that is a correct statement of angular momentum conservation.

Can I also say that initial AM of a weight is I(weight)*w= final AM=r1 x p1?

 

[jbriggs444]

Can I also say that initial AM of a weight is I(weight)*w= final AM=r1 x p1?

Yes. If you work the arithmetic, you will note that the two are equal.

 

[Krushnaraj Pandya]

Yes. If you work the arithmetic, you will note that the two are equal.

Thank you! all my doubts are resolved and my understanding of AM is much clearer now