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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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All the best AQmrit Pal !!!


 



Regards,


Askiitians Experts
Soumyajit Das IIT Kharagpur

7 years ago
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