# Angular momentum conservation on a merry-go-round

• kirito
kirito
Homework Statement
A girl Sarah, with mass m, runs toward a playground merry-go- round, which is initially at rest, and jumps on at its edge. Sarah's velocity is tangent to the circular merry-go-round.
Sarah and the merry-go-round then spin together with a constant angular velocity . The merry-go-round has a mass M, a radius R, and has the form of a uniform solid
disk.
Relevant Equations
Li=Lf

I know that after Sarah jumps the system has an angular momentum since its turning , before she jumps the marry go round was at rest and Sarah had a linear momentum and that linear momentum can be viewed an angular momentum in respect to the vertical axis of rotation in the center of the marry go

We do not have to concern ourselves with vertical forces, such as the force of gravity or the normal force applied to the merry-go-round by the ground, because vertical forces give no torque about a vertical axis of rotation.
1.i tried to check this out by identifying all the forces in play
and got the following forces normal marry go on Sarah and Sarah on marry go gravitational force pulling Sarah and gravitational force pulling the array go normal force between marry go and ground which is equal to marry go + Sarahs wight but opposite in direction , then friction on marry go and friction on Sarahs ,
the internal forces will have opposite toques canceling out but I am not sure how to convince myself that the normal from the ground will cancel the torque of the normal on the marry go since would not the force be applied on a distance R from the axis of rotation or will it be applied at the Center of Mass at the axis of rotation?
and is it always the case that vertical forces give no torque

However, the turntable does not accelerate to the right. This is because there is a horizontal force applied on the turntable by whatever the turntable’s axis is connected to, which we can consider to be the Earth. As shown in Figure, the Sarah/merry-go-round system has a net external force acting on it at this point, which is why the linear momentum is not conserved
2.I don't really understand what isF from ground here and so I don't understand what it means for it to imply that linear momentum is not conserved,

kirito said:
I am not sure how to convince myself that the normal from the ground will cancel the torque of the normal on the marry go since would not the force be applied on a distance R from the axis of rotation or will it be applied at the Center of Mass at the axis of rotation?
Consider how the merry-go-round is supported by the ground. It is not supported like a spinning top on a point that is free to both tilt and rotate. It is supported on a vertical axle that is free to rotate but not to tilt.

If you like, you can think of it as having two bearings, one above the other. A torque about a horizontal axis will be resisted by a horizontal force one way by the upper bearing and an equal horizontal force the other way by the bottom bearing.

Yes, you are correct that a spinning top will wobble if pushed downward on one side. The torque will not be resisted by the normal force and will result in precession.

MatinSAR
jbriggs444 said:
Consider how the merry-go-round is supported by the ground. It is not supported like a spinning top on a point that is free to both tilt and rotate. It is supported on a vertical axle that is free to rotate but not to tilt.

If you like, you can think of it as having two bearings, one above the other. A torque about a horizontal axis will be resisted by a horizontal force one way by the upper bearing and an equal horizontal force the other way by the bottom bearing.

Yes, you are correct that a spinning top will wobble if pushed downward on one side. The torque will not be resisted by the normal force and will result in precession.
sorry would you be able to use different terms to explain this , I tried to translate the new words such as two bearings precession, yet it is still a bit hard to digest I appreciate the help

kirito said:
sorry would you be able to use different terms to explain this , I tried to translate the new words such as two bearings precession, yet it is still a bit hard to digest I appreciate the help
Ignore the paragraph about precession. It is not needed.

I played on many merry-go-rounds as a child.

There is a vertical axle under the center. The ride is free to spin around that vertical axis.

The ride is not free to tilt. The axle is held vertical. It is sturdy. A child or an adult can stand on the rim without tipping it over. There is no power source. Children pull on the pipes (stanchions) to start it spinning. They can then climb on.

As the oldest child, I would usually work on pushing while my brother and sister climbed on and held on for dear life. In one memorable incident, our foster brother climbed on without due respect for the rotation speed. He did not brace well and wound up face down, head outward. Half of the groove outside the merry-go-round was then contaminated with regurgitated breakfast.

Over time, most of these merry-go-rounds would develop a tilt. Rain and time would soften the support and allow for settling in one direction or another. In another memorable incident, I was solo, standing upright in the center, walking uphill as if on a treadmill. The merry-go-round was accelerating slowly beneath me. Belatedly, I realized that I had no exit strategy. I lacked the skill to walk backward on the opposite side at the requisite speed. The ensuing collision was a painful learning experience.

kirito
jbriggs444 said:
Over time, most of these merry-go-rounds would develop a tilt. Rain and time would soften the support and allow for settling in one direction or another.
I remember these as having a slight designed tilt to the axis. This allowed the budding physicists in the cohort to "pump" them to impressiive speeds by leaning out at the top and moving toward the axle center axle at the bottom.: repeat as needed. The speed was limited only by the inability to provide sufficient centripetal force at the low point to pull ones posterior back inwards at high rotational speed. This was pretty damned fast as I recall. I don't think most kids understood the technique: probably a good thing that greatly reduced trips to the ER. Nausea was indeed a constant threat, but I loved those things.

MatinSAR
kirito said:
2.I don't really understand what isF from ground here and so I don't understand what it means for it to imply that linear momentum is not conserved,
I don’t understand what pictures a) and b) are for. Perhaps there are more parts to the question. I assume you are asking about picture c).
If the merry-go-round were sitting on ice instead of being mounted on an axle then when Sarah jumps on the whole system would slide to the right. Since it does not, the mount must exert an impulse in the opposite direction to Sarah's. As that is external to the system, linear momentum of the system is not conserved.
But F does not exert a torque about that axis, so, if you take angular momentum of the system about that axis, that is conserved.

MatinSAR and kirito

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