# Where's the torque when you retract your arms on a spinning chair?

1. Jan 3, 2010

### DocZaius

If you spin on a chair with your arms and legs out and then pull them in, your rotation accelerates. What torque is responsible for that acceleration? It seems to me all the movement is towards the axis of rotation and should not be responsible for any torque. In fact, take an idealized body that can reduce its radius with movements entirely towards the axis of rotation. Where does the torque come from?

I know that the moment of inertia has changed, and from that point of view it makes sense that the angular velocity would increase. I was just trying to look at it from a free body diagram perspective.

2. Jan 3, 2010

### rcgldr

The path that your arms take when being retracted is a inwards spiral. Part of that force is in the direction of travel of that spiral path, so the speed of your arms increases. In order to keep your arms orientation constant relative to yourself and the chair, you need to apply a torque, coexistant with an equal and opposing torque from your arms that increases your rate of rotation. Image of an example spiral path showing the line perpendicular to the path, and the radial line that is not perpendicular to the path.

3. Jan 3, 2010

### DocZaius

Thank you very much!

4. Jan 3, 2010

### Cleonis

As Jeff reid points out with his diagram:
The trajectory of your feet and fists is an inward spiral. The general case is contraction of a rotating system. At each point in time a rotating system will contract when the actual centripetal force is larger than the required centripetal force. Over the course of the contraction the centripetal force is doing https://www.physicsforums.com/library.php?do=view_item&itemid=75".

If a trajectory is circular then the centripetal force is not doing work. When the trajectory is a spiral then the work that is being done is causing rotational acceleration.

Peculiarly, quite a few authors write statements such as: "The person on the chair starts rotating faster because his moment of inertia decreases." I believe that line of reasoning violates causality. The causal factor is the centripetal force. During contraction the centripetal force causes angular acceleration. The contracted system has a smaller moment of inertia.

Cleonis

Last edited by a moderator: Apr 24, 2017
5. Jan 3, 2010

### rcgldr

The centripetal force only does work when there's a component of force in the direction of the spiral path. In the case of a object pulled by a string that wraps around a post, the path is an involute of circle, and no work is done because the force is always perpendicular to the spiral path. Angular momentum doesn't appear to be conserved, until you take into account that a torque force is applied to the post, imparting a angular acceleration to whatever the post is attached to (like the earth, in which case the angular acceleration is tiny), in which case angular momentum is conserved again when you include whatever the post is attached to as part of a closed system.

6. Jan 3, 2010

### diazona

There is no torque, in the ideal case. As the idealized body reduces its radius, its moment of inertia decreases and you would observe its angular velocity $\vec{\omega}$ increasing in such a way as to keep the angular momentum $\vec{L} = \mathbf{I}\vec{\omega}$ constant. And since
$$\vec{\tau} = \frac{\mathrm{d}\vec{L}}{\mathrm{d}t}$$
if the angular momentum stays constant, there is no torque.

There would, of course, be internal forces involved. But the moments of all those forces about the axis of rotation would cancel out, so that the net torque involved is zero.

7. Jan 3, 2010

### Cleonis

Indeed you point out an important difference.

The simplest case is a circumnavigating mass subject to pull from a massless cord, causing it to spiral inward. If the force is at all times a central force then:
- The centripetal force is doing work
- Angular momentum of the circumnavigating object is conserved.
- As the rotating system contracts the tangential velocity increases all the time.

In the case of the cord wrapping around a post:
- Pull of the cord on the object is at all times perpendicular to the velocity: the force is not doing work.
- Angular momentum of the circumnavigating object is not conserved
- The tangential velocity stays the same.

Cleonis

8. Jan 3, 2010

### D H

Staff Emeritus
No.

Quite a few authors say that because they are correct.

Can you lift yourself by your bootstraps? No. The reason is because pulling on your bootstraps is an internal force. An external force is required to change your momentum. Similarly, an external torque is required to change your angular momentum. In the ideal case, a person on a frictionless spinning chair, flailing ones arms about (or pulling them in) does not result in an external torque.

Look at the rotational equation of motion:

$$\frac{D\vec L}{dt} = \frac{d}{dt}\left(\mathbf I \vec {\omega}\right) = \vec{\tau}$$

No torque is required to change the angular velocity. In fact, if there is zero torque (as it is in the ideal case), angular velocity *must* increase to keep the angular momentum constant when the moment of inertia is decreasing.

A causal factor is indeed needed to change the moment of inertia. Work is being done here! However, it is mistaken to think of this as necessitating a torque. Torque is change in angular momentum, not angular velocity. The kinetic energy of the system is

$$T = \frac 1 2 \left(\vec L \cdot \vec {\omega}\right)$$

Differentiating wrt time,

$$\frac {dT}{dt} = \frac 1 2 \left(\frac{d\vec L}{dt}\cdot \vec{\omega} + \vec L \cdot \frac{d\vec{\omega}}{dt}\right)$$

With no external torque, this reduces to

$$\frac {dT}{dt} = \frac 1 2 \vec L \cdot \frac{d\vec{\omega}}{dt}$$

In the simple freshman physics case of rotation about an eigenaxis of the system, angular momentum is parallel to angular velocity and angular acceleration. The above vector equations become scalar equations.

\aligned \dot{\omega} &= -\,\frac{\dot I}{I} \omega \\ \dot T &= \frac 1 2 \dot I \omega^2 \endaligned

Work must be being done to accomplish that increase in kinetic energy. Just because work is being done does not mean that a torque must exist.

9. Jan 4, 2010

### rcgldr

In this particular case, internal torques are required to increase the angular momentum of the persons body and chair that are not moving "inwards", and to decrease the angular momentum of the arms and hands that are moving "inwards". The overall angular momentum of this system is preserved, but the components of this system experience internal torques and changes in angular momentum.

10. Jan 4, 2010

### D H

Staff Emeritus
Let's reduce this to the simplest case, Jeff. Consider a massless rod rotating about its axis. The ends of the rod are connected to some massive structure via frictionless couplings. Two point masses are connected to the rod by equal length flexible massless strings. The strings are connected to the rod via a pair of tiny frictionless pulleys. A motor controls the length of the strings. The entire system is in empty space; no gravity.

The rod+masses subsystem is set into rotation and the masses are slowly pulled toward the rod. There is no external torque on this subsystem. Because the forces needed to draw the masses inward are purely radial, there are no internal torques, either. Yet the angular velocity still increases.

11. Jan 4, 2010

### rcgldr

I agree, I was only pointing out the case where the rod is not massless, with the analogy of the person and chair being the rod with mass. The first diagram I posted would be similar to the case of the massless rod, all of the mass is in the moving object, and I'm assuming a massless string.

12. Jan 4, 2010

### Cleonis

Obviously I agree there is no external torque.
In my replies in this thread I have not mentioned torque; your disagreement is not with me.

As Jeff Reid points out, in the case of a person on a swiveling chair you can opt to track the torso and the limbs separately, and then there are internal torques to evaluate.

Cleonis

Last edited: Jan 4, 2010
13. Jan 4, 2010

### D H

Staff Emeritus
Whether the rod is massless is irrelevant. It just makes the calculation of the inertia tensor a bit simpler. Let's go back to that post.

This does not necessarily mean a torque is exerted. Think of it in terms of a comet orbiting the Sun. Comets have highly elliptical orbits. The force is only normal to the ellipse at the apsides. Yet there is no torque on the comet. Gravitation is a central force.

Torque is $$\vec r \times \vec F$$. It has nothing to do per se with velocity, which (locally) defines the path being followed.

This is correct. Work is a different concept than torque. Work must be performed to decrease the moment of inertia of a constant mass spinning object. Just because work is being performed does not mean a torque is being applied.

14. Jan 4, 2010

### blitz.km

I am sure I can justify what many authors write.

The angular momentum can be conserved because we do not have any external force on the system.

Angular momentum= I.w
I: Moment of inertia
w: Angular velocity

As the person folds his hands, the moment of inertia decreases.. because now more mass is concentrated near the axis of rotation.
As the angular momentum has to remain constant, the angular velocity increases.

15. Jan 4, 2010

### Cleonis

My perspective is tracing causality.
My demand is that the sequence must be tracable from cause to effect.

Now, in the case of reasoning like "the angular momentum has to remain constant", the metaphor is vague. What is the cause?

Of course the calculation that is based on conservation of angular momentum will produce the correct result. But the original poster is asking why the person in the chair starts spinning faster. The request is to help understand.

Instead of using conservation of angular momentum you can calculate the increase in angular velocity by evaluating how much work the centripetal force is doing. By looking at the direction of velocity and the direction of force one gets a visceral understanding of what physics is taking place.

Last edited: Jan 4, 2010
16. Jan 4, 2010

### D H

Staff Emeritus
Why is the concept of work, which does not work without the concept of conservation of energy, more fundamental than conservation of angular momentum?

17. Jan 4, 2010

### Cleonis

I have made no statements about one principle being "more fundamental" than another; your disagreement is not with me.

Work deals with energy conversion; as a force is causing acceleration one form of energy is converted to another form. Work is associated with proceeding in time from cause to effect. Conservation of angular momentum on the other hand, is strongly associated with spatial symmetry.

In the end it all converges: it is possible to derive conservation of angular momentum from the work/energy-theorem, but in doing so one of the postulates is symmetry of the laws of dynamics for all oriëntations in space.

18. Jan 4, 2010

### DocZaius

I think that the reason I have trouble understanding how this system could have had an angular acceleration applied to it without an external torque is that this corresponding situation is different in the linear case.

It seems that linearly, a mass cannot simply change its value as the moment of inertia does here. If it does, it does so by ejecting matter and we treat the resulting bodies seperately (but we remember they were together when considering conservation of linear momentum).

So linearly, if I see the whole body having undergone linear acceleration, it is safe to say that an external force was applied to it, while rotationally that is not necessarily the case (the whole body could have changed its moment of inertia entirely internally (say that 3 times fast)).

Let me know if/how I am wrong!

Last edited: Jan 4, 2010
19. Jan 5, 2010

### blitz.km

Well, can you explain this:
"An object would have constant linear momentum until we apply an external force on it."
Even this must be vague, because according to you we can not apply this to solve any query.

20. Jan 5, 2010

### A.T.

I still don't get why a force is a better "cause" than conservation of angular momentum.

Both are just assumptions based on experience. We assume that there is a force, if we see acceleration. And we assume that angular momentum is conserved. Isn't it just that we are more used to say "a force caused something"?