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sohan singh singh Grade: 12

7 years ago

Answers : (1)

Askiitians Expert Soumyajit IIT-Kharagpur
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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Soumyajit Das IIT Kharagpur

7 years ago
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