Angular velocity question

[John O' Meara]

A solid uniform disk of mass m and radius R is pivoted about a horizontal axis through its center, and a small body of mass m is attached to the rim of the disk. If the disk is released from rest with the small body at the end of a horizontal radius, find the angular velocity when the body is at the bottom.
Loss of P.E., = gain in K.E. Therefore

m*g*R = .5*I*w^2 + .5*m*v^2.

Where I = rotational inertia and w = angular velocity.

w=2*(g*R - 2*v)/R^2)^2.
Is this correct, if it isn't why or where? Thanks very much.

[John O' Meara]

Note: I= .5*m*R^2

[John O' Meara]

I was woundering what is your ideas on above. Many thanks.

[Doc Al]

Since the disk and the small mass constitute a single object that ends up rotating at some angular speed, you need to find the rotational inertia of that composite object. The final KE will be .5*I*w^2, where I is the total rotational inertia.