Question

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

Answer 1

Answer:

Acceleration of a rolling body down on inclined plane given as

a=1+R2K2gsinθ

For a solid cylinder

I=MK2=2MR2

∴R2K2=21

∴a=1+21gsinθ=32gsinθ

Also v2=u02+2as

=(0)2+2×32gsinθ×L

34g×Lh×L=34gh

v=34gh

Answer 2

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Work-energy theorem,        Mgh=21Mv2+21Iw2

For solid cylinder,  I=21MR2

Also   v=Rw       (due to pure rolling)

Thus     Mgh=21Mv2+21×21MR2w2

$\implies  v= \sqrt{\dfrac{4}{3}gh}$