Question

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She throws a rock (mass m) in a horizontal direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is v. Calculate (a) the angular speed of the merry-go- round and (b) the linear speed of the girl after the rock is thrown.

Answer 1

a)
`\omega=(-mvR)/(I+MR^2)`

b)
`v=(-mvR^2)/(I+MR^2)`

Step-by-step explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

`L_1=0`

Later, after the girl throws the rock, the angular momentum will be:

`L_2=(I_M+I_g)\omega +L_r`

where:

`I` is the moment of inertia of the merry-go-round

`I_g=MR^2` is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

`\omega` is the angular speed of the merry-go-round and the girl

`L_r=mvR` is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

`L_1=L_2`

So we find:

`0=(I+I_g)\omega +mvR\\\omega=(-mvR)/(I+MR^2)`

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

`v=\omega r`

where

`\omega` is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

`\omega=(-mvR)/(I+MR^2)` is the angular speed

`r=R` is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

`v=\omega R=(-mvR^2)/(I+MR^2)`