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7 years ago

28 Points
```										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

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All the best AQmrit Pal !!!

Regards,
```
7 years ago
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