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A solid cylinder of mass m and radius r ...

A solid cylinder of mass m and radius r rolls down with angular velocity `omega` on an, inclined plane of angle of inclination `theta` and height h. Rotational kinetic energy of the cylinder when it reaches the foot of the inclined plane is

A

`1/2mgh`

B

`1/3mr^2omega^2`

C

`1/2mr^2omega^2`

D

`1/4mr^2omega^2`

Text Solution

Verified by Experts

The correct Answer is:
A
Rotational kinetic energy `K.E_("rot")=1/2Iomega^2=1/2 ((mr^2)/2)omega^2=1/4 mr^2omega^2=1/4 mv^2`
but `v=sqrt(2gh) :. K.E_("rot")=1/4 2mgh =1/2 mgh`
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