TTU Dynamics: Find normal/angular acceleration of a point on a merry go round

[Green Lantern]

Homework Statement

A boy mass mb and a girl mg are riding on a frictionless merry go round mr that has mass moment of inertia I = 1/2(mr)R^2. The boy runs back and forth from center to edge and completes a circuit in 2 seconds. If the initial speed is ωo, what is the acceleration the girl experiences on the outer edge R?

Homework Equations

I = 1/2(m)r^2
F = ma
M = dH/dt
M = F cross r

The Attempt at a Solution

I have as my diagram a circle that I assume spins clockwise. The plane of the circle is the y-z plane, so I have all weights pointing in the -i direction. There is 1 degree of freedom. I can apply F = ma right away and get:

F = -(mr + mb + mg)gi

I am unsure of how to go about tackling the rest of the problem. I have to assume the merry go round is a circular plate. I assume the girl and boy can be treated as point masses, since no mass moment of inertia is given for them. My knowledge of angular dynamics is rusty. What am I not seeing here?

 

[anathema]

Firstly, it is important to note that the boy's motion will cause the merry go round to rotate about its center of mass. This means that the forces acting on the system will have both linear and rotational components.

To find the acceleration experienced by the girl on the outer edge, we can use the equation M = dH/dt, where M is the net torque acting on the system and H is the angular momentum. In this case, the net torque is caused by the boy's motion, and can be calculated using the equation M = F cross r, where F is the force exerted by the boy and r is the distance from the center of mass to the point where the force is applied (in this case, the edge of the merry go round).

Once you have calculated the net torque, you can use the equation M = Iα to find the angular acceleration α of the merry go round. This can then be related to the linear acceleration a of the girl on the outer edge using the equation a = Rα.

I hope this helps in solving the problem. Good luck!