Conservation of Angular Momentum of a man

[andyrk]

A man of mass m₁ is standing at the edge of a disc of mass m₂ which is pivoted about its CM. The man starts waling along the rim of the disc with a velocity μ_o. Find the angular velocity acquired by the disc by the motion of the man along the rim?
I couldn't do this on my own so I turned to the solutions for the help. There it is written that since no external torque is present on the (man+disc) system therefore we can apply conservation of angular momentum about the axis of rotation, i.e the CM.
But after that it is written that:

ζ_man+ζ_disc=0​

Why is this?

 

[Doc Al]

The total angular momentum is zero.

 

[andyrk]

Then if there is no net torque how is the disc rotating? Since it has acquired some angular velocity wouldn't the disc possesses some torque? Also I have one more doubt..As the man moves along the rim he pushes the disc backward with some force so that he moves forward. So the disc rotates. But the force with which he pushed the disc becomes the force required for rotating the disc by giving the necessary torque. So, won't the disc possesses some angular acceleration rather than some angular velocity? Since if torque exists, angular acceleration has to exist.

 

[Doc Al]

Then if there is no net torque how is the disc rotating? Since it has acquired some angular velocity wouldn't the disc possesses some torque?

There's no net torque on the system as a whole, thus the total angular momentum remains constant. Of course the two parts--man and disk--exert torques on each other and thus their individual angular momenta change.

Also I have one more doubt..As the man moves along the rim he pushes the disc backward with some force so that he moves forward. So the disc rotates. But the force with which he pushed the disc becomes the force required for rotating the disc by giving the necessary torque. So, won't the disc possesses some angular acceleration rather than some angular velocity? Since if torque exists, angular acceleration has to exist.

Certainly while they exert torque producing forces on each other they will experience an angular acceleration. At some point those forces cease and they just continue to move as they do.

 

[andyrk]

Certainly while they exert torque producing forces on each other they will experience an angular acceleration. At some point those forces cease and they just continue to move as they do.

Why would the force exerted by the man on the disc ever cease? If he keeps walking then he keeps pushing so angular acceleration always stays there. Right?

 

[Doc Al]

Why would the force exerted by the man on the disc ever cease? If he keeps walking then he keeps pushing so angular acceleration always stays there. Right?

If he's walking at a constant rate, the average force he exerts must be zero.

 

[andyrk]

If he's walking at a constant rate, the average force he exerts must be zero.

Maybe I am not aware about the statement you said just now. Ok I agree since he is moving with constant speed there wouldn't be any force on him or as you say average force. But if we don't take into consideration average force and analyse man's motion instantaneously we will find that a force is existing that is acting on the man. Also why would the average force be 0? I mean it has to be 0 but how will it be 0?

 

[thecommexokid]

Certainly while they exert torque producing forces on each other they will experience an angular acceleration. At some point those forces cease and they just continue to move as they do.

I don't think I agree with this statement. With every step, the man exerts a force on the disk, causing it to spin, and the disk exerts an equal and opposite force on the man, propelling him forward around the circle. If the man ever stopped walking, the disk would also come to a stop, because together they have a net 0 angular momentum.

 

[Doc Al]

I don't think I agree with this statement. With every step, the man exerts a force on the disk, causing it to spin, and the disk exerts an equal and opposite force on the man, propelling him forward around the circle. If the man ever stopped walking, the disk would also come to a stop, because together they have a net 0 angular momentum.

Not sure why you think I would disagree. Whenever the man exerts any force on the disk, the disk will of course exert an equal and opposite force on the man. When he first starts walking, there must be an average net force accelerating him forward (and the disk backward). And when the man stops walking there must also be an average net force slowing him down.

Once the man gets to his desired walking speed, the average force exerted on the disk (in that tangential direction) must be zero. If not, the disk and man will keep increasing speed!

 

[thecommexokid]

Once the man gets to his desired walking speed, the average force exerted on the disk (in that tangential direction) must be zero. If not, the disk and man will keep increasing speed!

What was vaguely puzzling me until this conversation forced me to think about it more carefully is just how walking works in general. If we zoom in on a single step, when the man's weight is on his back foot there is a net force pushing the man forward (and the disk backward), and as he puts down his front foot there is a net force pushing the man backward (and the disk forward). So as you say, the average force over a whole step is 0. I was focused entirely on the back foot pushing the disk backward, and not thinking about the front foot coming down and pushing it forward.

Hopefully this helps you, too, andyrk. As you pointed out, there are instantaneous forces happening. But they average to 0 over a whole step. If it helps, think about the man taking his steps really slowly, and pausing each time he puts his foot down.

 

[haruspex]

If it helps, think about the man taking his steps really slowly, and pausing each time he puts his foot down.

Alternatively, assume that as he puts a foot down it is already moving backwards relative to the man (which is how one does walk), and in particular such that it is not moving relative to the disk. Thus there are no tangential forces.

 

[andyrk]

I got even more confused. The total net force of the action reaction pair is zero but that doesn't matter! Right? Why won't the force being imparted on the disc not cause angular acceleration? The man is walking with a constant rate from the starting. But then here arises another query. A force causes acceleration. When we walk on a rough surface it is the roughness of the surface that makes us go forward. Like think of a rough surface as having many imperfections in it. When we walk we push the earth. Zooming in, we see that parts of our shoes/boots interlock with the imperfection. We now apply force to these imperfections which in turn exert a force on us, so we move forward. But now as a force was given to us we must have accelerated (certainly if we started walking from rest). But once we are already walking, you mean to say that The ground doesn't apply any force to us? And if it does, we don't accelerate? Why?? That means that any kind of walking would make the person accelerate! And the man is not moving forward because of the average force, its the whole (disc + man) system that is moving with the help of average force. The man is moving with the help of the force exerted by the disc. Also if the man stops walking, why would the disc also stop turning? It would keep rotating with the angular velocity it had acquired upto the point when the men stopped walking? Angular momentum would have been conserved only if no external torque would have been present. The man himself is creating an external torque. In fact the average force(zero) is not making anything move! The disc will only stop if there is some torque provided that opposes its present rotating motion! #Confused. :/

 

[haruspex]

I got even more confused. The total net force of the action reaction pair is zero but that doesn't matter! Right? Why won't the force being imparted on the disc not cause angular acceleration? The man is walking with a constant rate from the starting. But then here arises another query. A force causes acceleration. When we walk on a rough surface it is the roughness of the surface that makes us go forward.

Divide the man's walk into three phases:
1. Accelerates from rest to cruising speed. In this phase, the man applies a tangential force to the disk. The disk applies the same force forward on him. The man accelerates and acquires angular momentum about the centre of the disk. The torque the man exerts on the disk gives the disk equal and opposite angular momentum.
2. The man maintains a steady tangential speed. To do this he only has to exert a radially outward force on the disk (to keep going in a circle; he would have to exert the same force if standing still on a rotating disk). He exerts no tangential force. So he does not speed up any more, and neither does the disk. The reaction from the disk is the centripetal force on the man.
3. The reverse of 1. The man exerts a forward (in his sense) tangential force, bringing himself and disk to a stop.
If it bothers you to think about walking at a steady speed without applying a force in the direction of travel, you can model the walking as a series of complete cycles of the 1-2-3 pattern above.

 

[andyrk]

To do this he only has to exert a radially outward force on the disk

3. The reverse of 1. The man exerts a forward (in his sense) tangential force, bringing himself and disk to a stop.
(

Why would the man exert a forward force on the disc if he needs to stop? And ok, going by your way this creates an anticlockwise torque as compared to the motion of the disc when the man had started moving or was continuously moving. So the disc would only stop if the the tangential force the man applied on the disc while stopping for a very short time creates an angular velocity equal to that gained by the disc in the starting as was asked in the question. How do we know that the stopping force he exerted equals the starting force he exerted and was applied for the same time?

What I think makes me come to your explanation:

It is true that every body executing uniform circular motion has a tendency to move out of it tangentially. Take for example a car turning on a curved path. As the car traverses the curved path it has a tendency to move out tangentially so it tries to pull the road beneath it with itself. So the road also pulls the car in the form of frictional force. So the car continues to be in circular motion. Is this the same case with the man?

 

[haruspex]

How do we know that the stopping force he exerted equals the starting force he exerted and was applied for the same time?

The force need not be equal and opposite, and the time for which it is applied need not be the same, but the change in angular momentum, ∫Fr.dt, must be equal and opposite to bring himself to rest, and so it will also bring the disk to rest.

 

[andyrk]

If it bothers you to think about walking at a steady speed without applying a force in the direction of travel, you can model the walking as a series of complete cycles of the 1-2-3 pattern above.

I tired but couldn't. Like in the case of walking in a circle, he exerts a radially outward force on the disc and thus the disc exerts a centripetal force on the man so that he moves in a circle at a constant speed.

So what happens in the case when the man wants to move at a steady speed while walking on a straight path/road?

Also, if there is a centripetal force on the man while moving on the rim of the disc why doesn't the man go towards the centre of the disc?

 

[andyrk]

but the change in angular momentum, ∫Fr.dt, must be equal and opposite to bring himself to rest, and so it will also bring the disk to rest.

Why should it be equal and opposite? The angular momentum?

 

[andyrk]

2. The man maintains a steady tangential speed. To do this he only has to exert a radially outward force on the disk (to keep going in a circle; he would have to exert the same force if standing still on a rotating disk). He exerts no tangential force. So he does not speed up any more, and neither does the disk. The reaction from the disk is the centripetal force on the man.

If he exerts no tangential force but is moving tangentially then how is this possible? The man moves outwards radially rather than tangentially if he wants to carry on witht he speed he has gained till now? So why doesn't he fall off the disc as he is no more moving along the rim of
the disc but radially outward. How can the man simultaneously move forward also while trying to push the disc radially outwards? You said that the disc exerts a centripetal force on the man which is the frictional force (assuming that the surface of the disc is rough). So the frictional force opposes relative sliding between the surfaces. So that means that the man wants to move radially outward? And at the same time you say that he move along the rim? HOW?

 

[haruspex]

So what happens in the case when the man wants to move at a steady speed while walking on a straight path/road?

Same deal. If there were no air resistance then you would not need to generate any forward force to keep moving in a straight line at a steady speed. Most of your effort goes into vertical forces to compensate for losses in compressing shoes, skin, muscles..

Also, if there is a centripetal force on the man while moving on the rim of the disc why doesn't the man go towards the centre of the disc?

The whole point about centripetal force is that you need it just to stay at a constant distance from the centre. To approach the centre you would need to generate some extra radial force.

Why should it be equal and opposite? The angular momentum?

The man, mass m, stays at distance r from the centre. If he accelerated to speed v (relative to the ground) he did so by exerting a force, possibly varying, over some time t, such that mv = |∫F(t).dt|. His angular momentum about the centre is then mvr = |∫F(t)r.dt|. If the M of I of the disc is I and the disk is now rotating at rate ω then we have Iω = ∫F(t)r.dt. Bringing the man to a stop, relative to the ground, requires an equal and opposite change in angular momentum.