A solid cylinder of mass M and radius R rolls without slipping on an inclined plane of length L and height h . What is the speed of its centre of mass when the cylinder reaches its bottom?

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Quantrex Academy · Accepted Answer

Correct answer: C. (4)/(3) g h

Here K.E. of centre of mass when cylinder reached the bottom =(1)/(2) m v^2+(1)/(2) I ω^2 (i) Here I=(1)/(2) M K^2 and ω=(V)/(R) aligned & KE=(1)/(2) m v^2+(1)/(2) M K^2 · (V^2)/(r^2) & =(1)/(2) m v^2[1+(K^2)/(R^2)] aligned But for solid cylinder K= R 2 or (K^2)/(R^2)=(1)/(2) K E=(3)/(4) m v^2 PE of the solid cylinder at height h is PE=m g h From eq. (ii) and (iii) aligned & m g h=(3)/(4) m v^2 & v= (4)/(3) g h aligned

A solid cylinder of mass M and radius R rolls without slipping on...

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