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Rohan Singh Grade:
        

http://www.goiit.com/posts/downloadAttach/21768.htm

7 years ago

Answers : (1)

Chetan Mandayam Nayakar
312 Points
										

I will give a sketch of the solution because the detailed solution involves complicated algebra. It is just "brute force", and straightforward


a)moment of inertia about axis of rotation =(1/2)(MR2+mr2+2mR2)=I


at equilibrium position, kinetic energy =decrease in potential energy=mgR(1-cosθ)=(1/2)Iω2=(1/2)Iv2/R2, upon doing the simplification you will arrive at the answer given


b)Torque =T =I*alpha=mgRθ=-I(ω^2)θ,ω2= mgR/I, T=2π/ω

6 years ago
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