A uniform solid cylinder of mass 5 kg and radius 0.1 m is reting on a horizontal platform ( parllel to the x-y plane) and is free to rotate about its axis the x-direction and given by x=0.2 cos (10t)m. If there is no slipping, then maximum torque action on the cylinder during its motion is

Options

A. `(Momega^(2)AR)/3`
B. `(Momega^(2)AR)/2`
C. `2/3xxMomega^(2)AR`
D. The situation is not possible

Answer

B

Explanation

The cylinder is having a fixed axis, so it can perform rotational motion only.

`fR=(MR^(2))/2alphaimpliesf=(MR)/2alpha`
For no slipping,
`Ralpha=` acceleration of the platform
`(2f)/M=omega^(2)Acosomegat`
For the maximum torque `f` would maximum and `f_(max)=Momega^(2)A/2`
So the maximum torque,
`tau_(max)=f_(max)xxR-=(M^(2)omega^(2)AR)/2`