# If a pendulum is on a rotating table, will a torque be generated?

Here is the set up. Very simple. A flat (i.e. horizontal table, there is no gravity) and rounded table that spins on its axis (through the center of the table). A spring mass system is now put on the table inside a slot along the radius of the table.

One end of the spring is attached to the origin of the table, and the other end is the small bob mass. Lets assume the relaxed length of the spring is half way along the radius.

So the above is the physical set up. Here is a picture, showing the spring with a little bit of tension in it.

So, the pendulum can only move along the radius, as it is inside a slot in the table. Now I spin the table, say clockwise with some initial angular speed, and at the same time release the bob so it also starts to vibrate back and forth on the radius as the table rotates.

What I am not sure about is that whether a torque be generated due to the reaction of the bob on the edge of the slot as the disk spins or not? This reaction force has an arm length from the center, hence a torque will be present.

In otherwords, will this system's angular momentum be constant or not? If there is no torque generated, then the angular momentum is ofcourse constant.

Thanks

Edit May be I should explain why I thought at first there might be a torque generated. When the bob is rotating along with the table, it will have 2 components of accelerations. One component is along the radius direction, which is (x'' - x (theta')^2) where x is the bob coordinate along the radius, and theta(t) is the angle of rotation. In the above, the term x(*theta')^2 is the centripetal acceleration and x'' is the acceleration of the bob along the radius.

There is also an acceleration perpendicular to that, which is (x theta'' + 2 x' theta'). In the above, the second term is the coriolis acceleration and the first term is the standard Euler acceleration. I made this diagram

Now, whenever I see something accelerating, then there must be a force along that direction. Right? This is the force that I took as generating a torque on the disk (actually it is the reaction to this force, ie. back to the disk).

This is why I thought there is a torque.

But when I did the same derivation using Lagrangian method L=T-V, I got different equation of motion. (no torque). Hence I thought to ask the experts here.

update 4 I have finally I think sorted this whole thing out!. I have updated the derivation. Solved this problem using Newton and Lagrangian methods, and both give the same equation of motions, and verified that the angular momentum remain constant. I have a PDF and HTML page and the applet here

http://12000.org/my_notes/mma_demos/slot_on_disk/index.htm

This is not as easy as it first seemed. There is actually a torque involved. Due to Coriolis force. This torque however changes in value such that the angular momentum remain constant. The critical thing for me to see was that the spring in infinitly rigid against rotation. This was mentioned below by Shaktyai. This is critical, as it means a slot is not even needed to keep the spring stright, and hence there is no Euler acceleration term perpendicular to the radial direction. The only side acceleration is the Coriolis one. This does generate a torque. I am not sure what this torque is called ( Coriolis torque?).

It all now fits nice. I removed the slot from the applet since it is not needed, and updated it. If someone can verify it for me it will be appericated. The applet runs in the browser. Written in Mathematica.

Wiki has a nice illustration of the Coriolis effect here http://en.wikipedia.org/wiki/Coriolis_effect

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If there is no external torque, then the angular momentum of the system will be conserved. This seems to be the case in the system you're describing, assuming no frictional forces.

The spring will contribute to the moment of inertia of the system as a whole(*), but once you have the whole system rotating with a given angular momentum, it should continue to do so.

(*) For one thing, as the system rotates faster the effective equilibrium position of the spring should move further from the center due to the centrifugal force. So some energy that could otherwise be going towards speeding up the rotation is instead being stored in the potential energy of the spring.

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Thanks. I had a feeling that might be the case. So this means the action/reaction between the bob itself and the table along the edge of the slot basically cancels out. Since the table "pushes" the bob to rotate, and the bob "pushes" back to the table, there is zero net force on the edge. – Nasser Aug 15 '12 at 22:05
Why do you believe it would continue to rotate at a constant angular velocity? Conservation of angular momentum is not the same as conservation of angular velocity. – AdamRedwine Aug 15 '12 at 22:06
@AdamRedwine, you're right, I should have said "constant angular momentum". I'll edit. – Tim Goodman Aug 15 '12 at 22:07
@NasserM.Abbasi, Yes, essentially it's the same as why you can't lift yourself up by your suspenders. As Newton's 3rd law says, as you pull on them they pull back on you in the same amount. But as Adam rightly mentioned, that just means the momentum is conserved, not the velocity. (Angular momentum is angular velocity times moment of inertia, so as the moment of inertia changes (by moving the mass distribution closer or further from the axis) the rotation speed will change. – Tim Goodman Aug 15 '12 at 22:21
@NasserM.Abbasi, Yes if I'd seen your edit (mentioning Coriolis forces and Lagrangian dynamics) I would not have assumed I might need to tell you the definition of angular momentum. You never know what level of background one is coming from. :) – Tim Goodman Aug 15 '12 at 22:38

Without outside interference, the angular momentum will be constant owing to the Law of Conservation of Momentum. The angular velocity, however, will change. As the mass moves toward the outside of the wheel, the rate of rotation will slow; as the mass moves back to the center, the rate of rotation will increase. This is just like what happens when an ice skater pulls in her arms while spinning.

Angular momentum question like this are always tricky. I find it helpful to imagine being in a space station and conceptualizing the difference between doing cartwheels and having the space station rotate around me. In the first case, the astronaut would feel a force pulling out from the center of the axis of rotation; in the second case, no force would be felt.

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+1 I think maybe this is what the asker was really wondering. I focused on the question "Is the angular momentum constant", but of course the angular velocity can change even as angular momentum is conserved. – Tim Goodman Aug 15 '12 at 22:10

There will be Coriolis force terms causing torques between the pendulum bob and whatever is constraining it to stay on the line.

The angular momentum for the whole system must be conserved, but the crucial observation is that the moment of inertia of the bob, and thus of the whole system, is not constant in time, and that therefore the angular velocity of the system (which geometrically must be the same for both the table and the bob) cannot be constant.

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"There will be Coriolis force terms causing torques". Well. That is what I thought first. But in the above discussion, it was said that due to action/reaction, the Coriolis force has no net effect on the disk. Hence there is no torque on the disk present. I am confused now :) – Nasser Aug 16 '12 at 10:09
I went over this again, and when I add a torque, the angular momentum is no longer conserved. I am not able to get a solution where there is a torque AND also having the angular momentum constant. – Nasser Aug 16 '12 at 11:45
There is no net torque on the whole system. The bob exerts a torque on the disk and vice versa. These torques should exactly cancel out and leave the total angular momentum unchanged. – Emilio Pisanty Aug 16 '12 at 16:16
There is a torque actually. It is due to Coriolis force. It is not an external applied torque. Not sure what it is called. But it is there. Yes, the angular momentum does remain constance. Verified from simulation. – Nasser Aug 17 '12 at 16:56

Your problem lies the system being ill defined:

You study the bob as being a mass point linked to a massless spring rotating in the plane. The spring is supposed to be infinitively rigid in the direction perpendicular to its axis (physically the slot forces the spring to only move along its axis). The only force applied on the bob are : weight, table reaction, spring force, and reaction of the slot. Any torque du to the coriolis force is balanced by the reaction of the slot.

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Any torque du to the coriolis force is balanced by the reaction of the slot. Yes. That is how the simulation is currently implemented. No torque. My main question was, if there will be a torque generated on the disk by the bob or not. It looks like most here agree that there is no net torque present. thanks – Nasser Aug 16 '12 at 14:42
The force on the disk is applied through the axis of rotation (by the other end of the spring) and can't have a torque. – Shaktyai Aug 16 '12 at 14:48