Does Angular Momentum Change When We Drop Weights From A Rotating Turntable?

In summary, the man's angular velocity should remain constant when he drops the weights outside the turntable due to the constant distance between the weights and the axis of rotation. The torque exerted by the weights on the man can be taken with respect to any point on the axis of rotation without changing the angular momentum. However, for a symmetric body, the total angular momentum is parallel to the axis of rotation while the individual angular momenta of the particles are at an angle with respect to the axis.
  • #1
breez
65
0
A man, holding a weight in each hand, stands at the center of a horizontal frictionless rotating turntable. The effect of the weights is to double the rotational inertia of the system. As he is rotating, the man opens his hands and drops the two weights. They fall outside the turntable. What happens to the man's angular velocity?

The answer is that his angular velocity should stay relatively constant. I am confused to why this occurs. This is my reasoning for this:

As the weights are in free fall, they still are roughly same distance from the axis of rotation. Thus I is constant, and hence the nothing in the system changes. Angular velocity is still constant.

However, once the weights fall on the ground, the external friction force exerts a net torque on the weights, stopping them. However, since the system is not closed, the angular momentum need not be conserved, so the man's angular velocity remains unchanged.
 
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  • #2
Do you have a question? A cleaner way to see the effect of dropping the weights is to ask: What torque is exerted by the weights on the man when he drops them?
 
  • #3
Oh I see. There is no net torque acting on the man when he drops them.
 
  • #4
Oh, and on a tangential question about angular momentum...

I was reading about precession of a gyroscope in my physics text and it seems to take the angular momentum with respect to the axis, but the torque with respect to a point on the axis. I know that for torque and angular momentum to be coherent, we should always take both with respect to the same origin. So I am assuming it is reasonable to take torque with respect to any point on the axis of rotation for which the angular momentum of a rigid body in simple rotation is taken in respect to?

This seems to make sense, as in my text, the proof of the relation, [tex]||L_z|| = I\omega[/tex], with respect to an arbitrary point on the axis of rotation seems to reduce to the situation of treating the situation as finding L for each particle making up the rigid body, with the point on the axis perpendicular to the particle. Since in simple rotation, the velocity and radius are on the same plane as any differential cross section of the rigid body which the point particle lies on, each individual L must be along the axis, perpendicular to the cross section. That would mean L is independent of any specific point on the axis, but rather only the axis itself. Therefore, as long as torque is taken with respect to a point on the axis, the two quantities are compatible. Is my reasoning correct?
 
  • #5
Doc Al where are youuuuu
 
  • #6
I'm here, I'm here. (Sorry for the delayed response.)

If you're saying what I think you're saying, then I agree. Let me rephrase part of what you said. For a body rotating about an axis of symmetry, the total angular momentum of the body is the same about any point on the axis. So, you can analyze torque about any point on that axis without redefining the angular momentum. (I think that's what you meant about the quantities being "compatible".)

So far, so good. But I will pick a nit with this paragraph:
breez said:
This seems to make sense, as in my text, the proof of the relation, [tex]||L_z|| = I\omega[/tex], with respect to an arbitrary point on the axis of rotation seems to reduce to the situation of treating the situation as finding L for each particle making up the rigid body, with the point on the axis perpendicular to the particle. Since in simple rotation, the velocity and radius are on the same plane as any differential cross section of the rigid body which the point particle lies on, each individual L must be along the axis, perpendicular to the cross section. That would mean L is independent of any specific point on the axis, but rather only the axis itself. Therefore, as long as torque is taken with respect to a point on the axis, the two quantities are compatible. Is my reasoning correct?
Just to be clear: It is not true that the angular momentum of each individual particle is along the axis of rotation. (Consider [itex]\vec{L} = \vec{r} \times \vec{p}[/itex] applied to a particle of the body. That angular momentum vector is at an angle with respect to the axis of rotation.) It's only the sum of the angular momenta that's parallel to the axis. That's why [itex]L = I \omega[/itex] is only true for symmetric bodies. (But [itex]L_z = I \omega[/itex] always holds.)
 

FAQ: Does Angular Momentum Change When We Drop Weights From A Rotating Turntable?

What is Angular Momentum Scenario?

Angular Momentum Scenario is a concept in physics that describes the rotational motion of an object around an axis. It is a measure of the quantity of rotation an object possesses.

What factors affect angular momentum?

The two main factors that affect angular momentum are the mass of the object and its velocity. The more mass an object has and the faster it is moving, the greater its angular momentum will be.

How is angular momentum calculated?

Angular momentum is calculated by multiplying the mass of an object by its velocity and its distance from the axis of rotation. This can be represented by the equation L = mvr, where L is angular momentum, m is mass, v is velocity, and r is the distance from the axis of rotation.

What is the conservation of angular momentum?

The conservation of angular momentum states that the total angular momentum of a system remains constant unless acted upon by an external torque. This means that the angular momentum of a system cannot be created or destroyed, only transferred or redistributed within the system.

What are some real-life examples of angular momentum?

Some real-life examples of angular momentum include the spinning of a top, the rotation of a planet around its axis, the orbit of a satellite around a planet, and the motion of a spinning figure skater. Additionally, any object that is rotating or moving in a circular motion exhibits angular momentum.

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