I Does a person standing up on a swing in oscillatory motion do any work?

[brochesspro]

Also, I wonder if a system can have work done to it due to its internal forces, as it is generally said that the work done due to internal forces is zero or at least I have learnt that(cannot trust my knowledge anymore, lol). I got this question in my mind because of a situation in another question, to be honest, but it is related to work and is conceptual so I guess it is alright if I post it here.

If a man sitting on a swing in oscillatory motion stands up in the middle of its journey stands up, is he said to have done any work? Since the angular velocity of the swing changes, especially if he stands when the swing is in its bottommost position, in this specific case we can apply conservation of angular momentum and confirm the change in angular velocity. So has the man done any work on himself? Since he is also a part of the swing? I think this is the only question I have left related to work.

Here is the complete question for reference:
A person of mass $M$ is sitting on a swing of length $L$ and swinging with an angular amplitude $\theta_0$. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance $l$ is($l<<L$):

 

[hutchphd]

His legs do an amount of work mgl (of course extending his legs will shift the body locus of the center of mass making a small complication for calculation). But that's why we can pump ourselves in a swing. I am assuming l is vertical motion when he pumps and l<<L.

 

[jbriggs444]

Also, I wonder if a system can have work done to it due to its internal forces, as it is generally said that the work done due to internal forces is zero or at least I have learnt that(cannot trust my knowledge anymore, lol).

The "net work" done by internal forces is zero.

Newton's third law assures us of this. Suppose that we have a third law pair of forces: $\vec{F_{ab}}$ and $\vec{F_{ba}}$. Newton's third law assures us that $\vec{F_{ba}} + \vec{F_{ab}} = 0$.

Multiply the sum of those two forces by the displacement of the center of mass. Still zero. Add all of your other internal forces to the accounting. All of them will have internal third law partners. Still zero "net work"

This holds, even if the center of mass is moving. The key is that the sum of all internal forces is zero. We multiply them all by the same displacement.The "real work" done by internal forces can be non-zero.

Edit: changing the "real work" example to something not involving rotation.

Suppose that we have a piston in a cylinder and an explosive mixture of gasoline and air. The spark plug ignites the mixture. The piston is driven one way and the cylinder the other. The total kinetic energy of the assembly has increased. "real work" has been done. The force of the [negligibly massive] gasses on the one end has done "real work" on the cylinder and on the other end has done "real work" on the piston.

Real work is done even though the net force was zero. It will turn out not to matter how fast the center of mass is moving. The total real work done by all of the internal forces will be an invariant that does not depend on the choice of [inertial] frame. The same total no matter what reference frame one chooses to adopt.

Maybe you put a bunch of these together and drive counter-rotating flywheels with the resulting engine. Surely there is an increase in kinetic energy in there somewhere.

 

[jbriggs444]

A person of mass $M$ is sitting on a swing of length $L$ and swinging with an angular amplitude $\theta_0$. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance $l$ is($l<<L$):

"The work done by him"...

Sounds like a simple question. But it is not. What sort of work are we talking about? Work done by exactly what? Work done on exactly what?

I have two ideas on how to proceed. One based on how the energy of the system changes as a result of the action. Another based on how much force his legs exert on his body and the displacement during that force application.

However, this question belongs in its own thread. I am reporting this post in order to help that happen.

 

[brochesspro]

The real work is the sum of the work done by the forces at one end of her muscles and the work done by the forces at the other ends. The displacements at the two ends are different. So even though the sum of the forces nets to zero, the sum of the works does not.

I do not understand this part, could you clarify it a bit?

His legs do an amount of work mgl (of course extending his legs will shift the body locus of the center of mass making a small complication for calculation). But that's why we can pump ourselves in a swing. I am assuming l is vertical motion when he pumps and l<<L.

Not only that, there is an additional factor of $1 + \theta_0^2$ due to the change in the kinetic energy of the swing. Here is the solution if needed.
https://www.toppr.com/ask/question/a-person-of-mass-m-is-sitting-on-a-swing/

 

[jbriggs444]

I do not understand this part, could you clarify it a bit?

So we have a muscle. For instance, in your calf (gastrocnemius). The bottom end pulls upward on the tendon that attaches on the back of your ankle. The top end pulls downward on an attachment point on the tibia (or femur?) near the knee. [I'm not an orthopedic surgeon].

The forces applied at the two attachment points are nearly equal and opposite. But the two attachment points displace by different amounts. The ankle attachment point moves up. The knee attachment point stays in place.

The total work done is non-zero. As one would hope -- the purpose of muscles is to do work.

 

[brochesspro]

"The work done by him"...

Sounds like a simple question. But it is not. What sort of work are we talking about? Work done by exactly what? Work done on exactly what?

I know the question is vague in that sense, I thought it was on the swing and the man as a system as there is nothing else mentioned and that is why I asked the question on internal forces.

However, this question belongs in its own thread. I am reporting this post in order to help that happen.

I thought so, I will create a thread tomorrow if needed. I will go to bed after this post since it is pretty late for me.

Note that I gave you a simpler problem, the rod I described was around a fixed pivot and the CoM does not travel a straight line down. Anyway, you can choose any point you want.

Thanks for the info and the example, though I do not know what the purpose of that example was.

For your problem, why CoM does not move at constant acceleration. For that to happen the normal force must be smaller than mg and constant. Now, why should the normal force be constant?

I know, it does not have to, the whole reason for this thread to exist was that I could not prove that the acceleration was not constant.
I guess that is enough from me for today. Have a nice day, guys.