Answer :
a) [tex]\omega=\frac{-mvR}{I+MR^2}[/tex]
b) [tex]v=\frac{-mvR^2}{I+MR^2}[/tex]
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:
[tex]L_1=0[/tex]
Later, after the girl throws the rock, the angular momentum will be:
[tex]L_2=(I_M+I_g)\omega +L_r[/tex]
where:
[tex]I[/tex] is the moment of inertia of the merry-go-round
[tex]I_g=MR^2[/tex] 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
[tex]\omega[/tex] is the angular speed of the merry-go-round and the girl
[tex]L_r=mvR[/tex] 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,
[tex]L_1=L_2[/tex]
So we find:
[tex]0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}[/tex]
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
[tex]v=\omega r[/tex]
where
[tex]\omega[/tex] is the angular speed
r is the distance of the body from the axis of rotation
In this problem, for the girl, we have:
[tex]\omega=\frac{-mvR}{I+MR^2}[/tex] is the angular speed
[tex]r=R[/tex] is the distance of the girl from the axis of rotation
Therefore, her linear speed is:
[tex]v=\omega R=\frac{-mvR^2}{I+MR^2}[/tex]