Solved: Spinning Situation: What Happens to the Merry-Go-Round?

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When a person stops the spinning bicycle wheel on a stationary merry-go-round, the conservation of angular momentum dictates that the merry-go-round must begin to rotate. Since the wheel was spinning counterclockwise, the merry-go-round will also rotate counterclockwise to conserve the total angular momentum of the system. The absence of external torque allows this conservation to occur. Therefore, the correct outcome is that the merry-go-round begins to rotate counterclockwise as observed from above. This illustrates the fundamental principle of angular momentum in rigid body dynamics.
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Homework Statement



Suppose you are standing on the center of a merry-go-round that is at rest. You are holding a spinning bicycle wheel over your head so that its rotation axis is pointing upward. The wheel is rotating counterclockwise when observed from above.

For this problem, neglect any air resistance or friction between the merry-go-round and its foundation.

Suppose you now grab the edge of the wheel with your hand, stopping it from spinning.
What happens to the merry-go-round?

A. It remains at rest.
B. It begins to rotate counterclockwise (as observed from above).
C. It begins to rotate clockwise (as observed from above).

Homework Equations



??

The Attempt at a Solution



the merry go round cannot stay at rest because the angular momentum from the spinning bicycle wheel cannot disappear.

the merry go round will spin counterclockwise (as observed from above) because that the direction the wheel was spinning?
 

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Hello,

I hope you are familiar with the condition that is required for the conservation of angular momentum. In addition, recall that there is a simple equation describing the relationship between angular velocity and angular momentum. (Remember, angular velocity is a vector, and the sign conventions will determine the direction of revolution.)

Regards,
Horatio
 
We will assume this system is a rigid body rotating around a symmetry axies.

In this case L = I \omega (all vectors) .

L is conversed as the external torque on the system is 0.

Hence to keep the same angular momentum vector as the wheel comes to a stop, you and the merry go-ahead must start spinning in the same direction ie counter clock wise.
 
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