Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

Grade:

1 Answers

Chetan Mandayam Nayakar
312 Points
10 years ago

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

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free