Question

A uniform cylinder is released from rest from top of an inclined plane of inclination theta and length l if cylinder rolls without slipping then what is the speed of cylinder at the bottom of plane

Answer 1

Here is your answer:

Explanation:

Gain in the Kinetic Energy = loss in potential energy

1/2 Iω2 + 1/2 mV2 = mgl Sinθ

1/2 [1/2 mR2 ] ω2 + 1/2 mV2 - 3/4 mv2 = mgl Sinθ

V =  √ 4gl Sinθ/3

Answer 2

Explanation:

Gain in  KE= loss in PE

$\frac{1}{2} I w^{2}+\frac{1}{2} m v^{2}=m g \ell \sin \theta$$\frac{1}{2}\times\frac{1}{2} m R^{2}\right] w^{2}+\frac{1}{2} m v^{2}-\frac{3}{4} m v^{2}=m g \ell \sin \theta$$v=\sqrt{\frac{4 g \ell \sin \theta}{3}}$