Conservation of Angular MOmentum
Posted on 2006-04-23
A 55kg runner runs around the endge of a horizontal turn table mounted on a vertical frictionless axis through its center. The runners velocity relative to the earth is 2.8m/s the turntable is rotating in the opposite direction with magnitude angular velocity 0.20rad/sec relative to the earth. The radius is 3.0m and the moment of inertia about the axis of rotation is 80 kg m^2. Find the final angular velocuty of the system if the runner comes to rest relative to the turn table.
I was thinking that I would start by finding the combined moment of inertia.
I_t = I + I_r = 80kg m^2 + (55kg)(3.0m)^2 = 575kg m^2
So the initial angular momentum is equal to the final angular momentum.
L_1 = L_2
But I guess I have to figure out his velocity relative to the turn table to be able to apply conservation of angular momentum. All I know is that the tangential accleration of the turn table is equal to ra (a is alpha).
So we know he has acceleration, and his final velocity relative to the turn table is zero, but how do I figure out his velocity relative to the turn table? I was thinking Galieliean Transformations of Velocitys, but does that apply with angular and linear velocities?
I have to figure this out.