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Say there a child, that stands on the edge of an AT REST merry go round. When they jump off, the child has a linear velocity, and the merry go round begins to turns.

They say there is no net torque on the merry go round/child system and angular momentum is conserved.

I understand the torque applied to the merry go round from the child(force of child x radius of merry go round), but how does the merry go round provide a torque to the child. Doesn't it just apply a force?

This is the same issues as a child jumping on a merry go round that is coming in straight/tangentially to the edge.

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  • $\begingroup$ Objects moving in a straight line can have non-zero angular momentum. physics.stackexchange.com/questions/458404/… $\endgroup$
    – BowlOfRed
    Sep 2, 2020 at 15:23
  • $\begingroup$ I understand that. But they say the net torque on the system is zero, since angular momentum is conserved. If the child gains angular momentum, then is there a torque on the child? The child is providing a torque to the merry go round, we know this for sure. The only thing applied the child is the reaction force, which is where i am confused $\endgroup$ Sep 2, 2020 at 15:28
  • $\begingroup$ I might have answered my own question. There is no OUTSIDE external torque. Therefore no ang. momentum added from outside the system. However, the child exerts a torque on the merry go round, reducing or giving it negative ang momentum. And because of this, the merry go round produces a force on the edge back to the child, causing them to fly off with tangent velocity and gain ang. momentum. Is this correct BowlOFRed? $\endgroup$ Sep 2, 2020 at 15:36

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how does the merry go round provide a torque to the child.

It doesn't.

Let's move the experiment into a weightless environment: We start with the child clinging to the edge of a disk, and the angular momentum of the child-disk system is zero. Then, the child pushes off, on a line that is tangent to the edge of the disk, and which lies in the plane of the disk.

If you look at the disk in isolation, you'd see the disk rotating about its own center of mass: It has angular momentum. If you look at the child in isolation, assuming they pushed off in just the right way, you could see no rotation. (I.e., you could see that the child, in isolation has no angular momentum.)

The key though, is that the child and disk now have linear motion relative to each other, on a line that does not pass through the center of mass of the child-disk system.

I don't know the math, but that's where the missing angular momentum went. The off-center linear motion of the child and disk contributes to the angular momentum of the system, and that contribution will be equal and opposite to the angular momentum that you'd see if you looked just at the disk in isolation.

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  • $\begingroup$ Then why do they say the merry go round/child system is zero net torque? I know the momentum is conserved, but if i were to draw out forces and torques, the merry go round would have a force/torque on it, but the child has just force on it. $\endgroup$ Sep 2, 2020 at 15:30
  • $\begingroup$ @KevinCSpeltz, ...no net torque on the whole system. "Net torque" refers to a torque coming from outside the system. There's nothing outside of the system exerting torque in my weightless version of the experiment. I imagine that having the merry-go-round connected to the Earth by an axle changes the story somewhat, but maybe that can be hand-waved away. I haven't thought too hard about that. $\endgroup$ Sep 2, 2020 at 15:35
  • $\begingroup$ i just realized that. I was thinking too hard and responded similar to BowlOFRed. So externally nothing is going on. But internally, the child exerts a torque, causing it to spin, giving a negative angular momentum to the merry go round. To keep ang momentum constant, the merry go round applies a force to the child(newtons 3rd law), causing child to accelerate linearly, which in turn, increases the angular momentum, keeping the system balanced. Is this correct? $\endgroup$ Sep 2, 2020 at 15:38
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I understand the torque applied to the merry go round from the child(force of child x radius of merry go round), but how does the merry go round provide a torque to the child. Doesn't it just apply a force?

Torques are "just forces", but interpreted about a specific point or axis. If we have an axis of consideration (probably the platform axis is convenient), then a force applied off-axis is also a torque.

So any torque the child applies to the platform, the opposite torque is applied to the child (even though the child doesn't start rotating). The sum of torques and the sum of angular momentum remains zero.

Why doesn't the child rotate though if they are experiencing a torque?

Unbalanced torques create a change in angular momentum. Rotations are only one form of angular momentum. Motion that is not collinear with the axis is also a form of angular momentum.

Does this change the work done by the child also, since now they are accelerating linear and yet gaining angular momentum.

No. The work done is the same. Just because we account for the motion as having angular momentum in this case doesn't change the energy transfer.

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  • $\begingroup$ So essentially the force being applied by the platform on the child is due to the torque about the axis of rotation....so no it doesn't cause the child to rotate, but still converses angular momentum. Correct? $\endgroup$ Sep 2, 2020 at 17:19
  • $\begingroup$ Also, was my earlier comment above right? I was trying to make sure i understood the external torques...all torques comes from inside the system. I was over thinking that. If the system is closed, nothing from outside is adding to it or taking away from it $\endgroup$ Sep 2, 2020 at 17:20
  • $\begingroup$ Yes correct. I thought your original question was how that could be if after the jump the child had zero and the platform had non-zero angular momentum. It is all internal and the way it balances is that the linear motion of the child is still a non-zero angular momentum. Each portion has its own equal and opposite angular momentum that was created by equal and opposite torques. $\endgroup$
    – BowlOfRed
    Sep 2, 2020 at 18:42
  • $\begingroup$ I guess my original questions is how can the force on the child, which comes from the edge of the disk/merry go round, be seen as a torque. Relative to the child, there is no torque arm or radius. But what you are saying is that from the axis of rotation of the disk, at that radius/distance, that torque is shown as a force. So that is how? $\endgroup$ Sep 2, 2020 at 19:36
  • $\begingroup$ All torques (and angular momenta) have to be relative to a point or axis. For the system to be consistent, that point has to be the same in all cases. If we use the axis of the platform, then the force on the child and the platform are at the same distance from the axis and forms a torque. This torque causes the platform to rotate and the child to move. Both can be calculated as forms of angular momentum. $\endgroup$
    – BowlOfRed
    Sep 2, 2020 at 22:11
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If the child jumps off tangentially, he pushes backward on the merry-go-round and it pushes him forward. Both forces act at a distance (R) from the reference point at the axis of the merry-go-round and both exert a torque. The impulsive torque acting on the child changes his angular momentum relative to the reference point: FRΔt = mRΔv. For the merry-go-round, the impulsive torque also changes the angular momentum (in the opposite direction): FRΔt = IΔω, where I is the rotational inertia.

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