If we did all things we are capable of, we would literally astound ourselves.

– Thomas A. Edison

Monday, May 28, 2007

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) 2MR22 (b) MR22 (c) 2MRAω2

(d) MRAω2 (e) MRAω2/2

You may be remembering the expression for maximum acceleration of a simple harmonic motion: amax = 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 (αmax) of the cylinder is given by

αmax = amax/R = ω2A/R.

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

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

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

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

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