# Final angular velocity of a merry-go-round?

## Homework Statement

A child (mc = 36 kg) is playing on a merry-go-round (mm = 225 kg, R = 2.9m) that is initially at rest. The child then jumps off in a direction tangent to the edge of the merry-go-round. The child has a speed of 5.0 m/s just before she lands on the ground. What is the magnitude of the final angular velocity of the merry-go-round?

## Homework Equations

L = I*ω
Li,c + Li,m = Lf,c + Lf,m
I = (1/2)mmR2 for the merry-go-round.
I = mcR2 for the child.

## The Attempt at a Solution

The initial angular momentum of the system is 0 so Li,c + Li,m = 0.
So Lf,c + Lf,m = 0
(1/2)mmR2 * ωf + mcR2 * ωf = 0
(1/2)mmR2 * ωf = - mcR2 * ωf

However when I plug the values for mass of the merry-go-round and the child as well as the radius for the merry-go-round I cannot seem to find a way to isolate ωf to one side of the equation, I just keep getting 0 rad/s.

jbriggs444
Homework Helper
The angular velocity of the child is not very well specified since the child is not in circular motion around the selected axis of rotation. But that turns out not to matter. The angular momentum of the child can be calculated more directly. Angular momentum for a non-rotating body moving in a straight line is the product of momentum times the perpendicular offset from the selected axis of rotation.

The angular velocity of the child is not very well specified since the child is not in circular motion around the selected axis of rotation. But that turns out not to matter. The angular momentum of the child can be calculated more directly. Angular momentum for a non-rotating body moving in a straight line is the product of momentum times the perpendicular offset from the selected axis of rotation.

So I can find the final angular momentum for the child, and then plug it into the equation for the conservation of momentum to solve for the final angular velocity of the merry-go-round? So the momentum for him would be mass*velocity (momentum) and then I'm not sure what you mean by the perpendicular offset for the selected axis of rotation. The axis of rotation for the merry-go-round goes straight through the middle, and the child jumps off the edge in a direction tangent to that edge.

PeroK
Homework Helper
Gold Member
2020 Award

## Homework Statement

A child (mc = 36 kg) is playing on a merry-go-round (mm = 225 kg, R = 2.9m) that is initially at rest. The child then jumps off in a direction tangent to the edge of the merry-go-round. The child has a speed of 5.0 m/s just before she lands on the ground. What is the magnitude of the final angular velocity of the merry-go-round?

## Homework Equations

L = I*ω
Li,c + Li,m = Lf,c + Lf,m
I = (1/2)mmR2 for the merry-go-round.
I = mcR2 for the child.

## The Attempt at a Solution

The initial angular momentum of the system is 0 so Li,c + Li,m = 0.
So Lf,c + Lf,m = 0
(1/2)mmR2 * ωf + mcR2 * ωf = 0
(1/2)mmR2 * ωf = - mcR2 * ωf

However when I plug the values for mass of the merry-go-round and the child as well as the radius for the merry-go-round I cannot seem to find a way to isolate ωf to one side of the equation, I just keep getting 0 rad/s.

You could use conservation of angular momentum, but you can also think of the child applying an impulse (torque) when it jumps off. As the child is not rotating or in a circular orbit, I think it's confusing to consider its angular velocity. There is, of course, a more fundamental definition/equation for angular momentum that it would be better to use.

To understand the angular momentum in this case, you could consider the following:

A particle moves at a constant velocity ##v## from the point (0, 1) along the line ##y=1##. Calculate its angular momentum about the origin and check that this is conserved as it follows its linear path.

jbriggs444