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Dear Amrit Pal,
Ans:- The every part of your problem is not clear, what is K ?
Ok fine I am solving it for a cylinder and calculating angle A. you calculate thr rest.
Now when the body was in horizontal positon, the center of mass was at a height R from the horizontal level. But,
when it starts falling through the incline then this becomes RcosA
Hence the change in PE=MgR(1-cosA)
This is totally converted into the translational and rotational KE
Hence MgR(1-cosA)=1/2 M v² + 1/2 I w²
Now when the body starts loosing the contact with the plane, then the normal reaction should be 0 hence,
MgcosA=Mw²R (eq for centipetal ACC)
again for non sleepping we have v=wr
Then putting this values in Eq 1 we get,
MgR(1-cosA)=1/2 M gR cosA+1/4 MgR cosA
solving we get,
Cos A=4/7
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best AQmrit Pal !!!
Regards,
Askiitians ExpertsSoumyajit Das IIT Kharagpur
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