Suppose a cylindrical platform with moment of inertia 0.005kg m2 is freely rotating at angular speed 10 rad/s. A disc initially, rotationally at rest with mass 0.6kg and radius 0.1 m is dropped onto the platform so that their centers coincide. The kinetic frictional coefficient between the faces of the platform and disk is 0.25. Eventually both rotate with a common speed. Calculate the final speed.

Question

Also, how much energy is lost in the collision?
Also, assuming that all the energy is lost to friction and that the torque due to friction is 1/2FkR, determine the relative slide angle of the disk over the platform.
Finally, how much time does it take for the disk to reach the final speed?

Answer 1

The final anglular velocity can be obtained by simply using conservation of angular momentum, and will not depend upon the friction coefficient. The moment of inertia of the dropped disc is
(1/2)mr^2 = 0.003 kg/m^2.
10*I1 = wfinal*(I1 + I2)
wfinal = 10*(.005)/(.008) = 6.25 rad/s

Once you have that final angular velocity w, the loss of rotational kinetic energy is easily calculated.

Energy is converted to heat at a rate (frictional torque)*(angular velocity)
Use that relationship and the KE loss to compute the time required to stop slipping.