Here is a question which checks your understanding of basic things in angular motion and simple harmonic motion:

**A solid cylinder of mass M and radius R is resting on a horizontal platform which is parallel to the XZ plane. The cylinder can rotate freely about its own axis, which is along the X-direction. The platform is given a linear simple harmonic motion of angular frequency ‘ω’ and amplitude ‘A’ in the Z-direction. If there is no slipping between the cylinder and the platform, the maximum torque acting on the cylinder is**

**(a) 2MR ^{2}Aω^{2} (b) MR^{2}Aω^{2} (c) 2MRAω^{2}**

**(d) MRAω ^{2} (e) MRAω^{2}/2 **

You may be remembering the expression for maximum acceleration of a simple harmonic motion: a_{max }= ω^{2}A

[ If you don’t remember the above expression, you may use the simplest form of simple harmonic motion of amplitude ‘A’ and angular frequency ‘ω’ and differentiate it twice:

z = A sin ωt (We write the displacement as ‘z’ since it is in the Z-direction).

a = d^{2}z/dt^{2} = –ω^{2}A^{ }sin ωt

Therefore, the maximum acceleration is ω^{2}A].

The maximum angular acceleration (α_{max}) of the cylinder is given by

α_{max} = a_{max}/R = ω^{2}A/R.

The maximum torque (τ_{max}) on the cylinder is given by

τ_{max} = α_{max}I, where ‘I’ is the moment of inertia of the cylinder about its own axis (which is equal to MR^{2}/2).

Therefore, maximum torque τ_{max} = (ω^{2}A/R) (MR^{2}/2) = **MRAω ^{2}/2**

You will find an interesting MCQ on rolling at physicsplus: MCQ on Rolling Bodies** **