Work & Energy of disc on conveyor belt

[jjiimmyy101]

Problem: The coefficient of kinetic friction between the 100 lb disk and the surface of the conveyor belt is 0.2. If the conveyor belt is moving with a speed of 6 ft/s when the disk is placed in contact with it determine the number of revolutions the disk makes before it reaches a constant angular velocity.

So this is what I have so far:

At state 1: T1 = 1/2 * m * V^2 + 1/2 * Ig * $$\omega^2$$


Work done from state 1 to state 2: U = Moment * $$\theta$$

Moment = friction *(0.5) = (0.2)*(100)*(0.5)

At State 2: I get stuck right here. Don't I have to know how far the disk travels to be able to calculate velcocity at state 2 and so on.

Any pointers would be appreciated.

 

[jjiimmyy101]

anyone?

 

[arildno]

Okay:
1) The kinetic friction will stop acting once the CONTACT POINT on the disk has achieved the velocity of the conveyor belt (no relative motion!)
2) Use Newton's 2.law to determine the velocity of the center of mass as a function of time.
3) Calculate the torque of the frictional force with respect to the center of mass, and gain an expression for the disk's angular velocity as a function of time.
4) The contact point velocity is given by the rigid body motion formula; use this to determine the time at which the contact point achieves the conveyor belt velocity.
5) Use this time to determine the number of revolutions.