A Question (MCQ) on Rolling

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) 2MR²Aω² (b) MR²Aω² (c) 2MRAω²

(d) MRAω² (e) MRAω²/2

You may be remembering the expression for maximum acceleration of a simple harmonic motion: a_max = ω²A

[ 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 = d²z/dt² = –ω²A sin ωt

Therefore, the maximum acceleration is ω²A].

The maximum angular acceleration (α_max) of the cylinder is given by

α_max = a_max/R = ω²A/R.

The maximum torque (τ_max) on the cylinder is given by

τ_max = α_maxI, where ‘I’ is the moment of inertia of the cylinder about its own axis (which is equal to MR²/2).

Therefore, maximum torque τ_max = (ω²A/R) (MR²/2) = MRAω²/2

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