Discussion Forum

The uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad/s about its own axis, is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are genty placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis.  The new angular velocity (in rad s^-1) of the system is

Answer:

Explanation:

$I_{1}\omega_{1}=I_{2}\omega_{2}$

 $\therefore$                   $\omega_{2}=\left(\frac{I_{1}}{I_{2}}\right)\omega_{1}$

  $\left[\frac{\frac{1}{2}MR^{2}}{\frac{1}{2}MR^{2}+2(mr)^{2}}\right]\omega_{1}$

   = $\left[\frac{50(0.4)^{2}}{50(0.4)^{2}+8\times(6.25)\times(0.2)^{2}}\right](10)$

   = 8 rad/ s