Question

A solid cylinder is released from rest . If the string does not slip relative to cylinder then find relative acceleration of A w.r.t B just after it is released....

Answer 1

...........



Dear Student,

Please find below the solution to the asked query:

The moment of inertia of the cylinder I=12mr2Let us consider the velocity of the block after it falls through height  h is v. Then equatingkinetic energy and potential energy we get12Iω2+12mv2=mgh⇒12mr22(vr)2+12mv2=mgh⇒32mv2=mgh⇒v=2gh3‾‾‾‾√Now if a is the acceleration then we haveh=12at2⇒t2=2haand v=at⇒t=va=2gh3a2‾‾‾‾√⇒t2=2gh3a2then we have2ha=2gh3a2⇒a=g3a) acceleration is g3b) velocity after it falls through a distance h is 2gh3‾‾‾‾√

Hope this information will clear your doubts about rotational motion.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards