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Two identical uniform discs of mass m and radius r are arranged as shown in the figure. If α is the angular acceleration of the lower disc and acm is acceleration of centre of mass of the lower disc, then relation among acm, α and r is

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Two identical uniform discs

The net torque about a point is,

\(T_{net} = I\alpha\)

Where \(I\) is the moment of inertia and α is the angular acceleration.

The net torque of the upper disc of mass \(m\) and radius \(r\) is,

\(Tr = \frac{mr^2}{2} \alpha_1 \quad....(i)\)

For the lower disc,

\(Tr = \frac{mr^2}{2} \alpha \quad....(ii)\)

From (i) and (ii),

\(\alpha_1 = \alpha \quad....(iii)\)

As the string does not get slack,

Acceleration of point A is equal to the acceleration of point B.

\(r\alpha_1 = a_{cm} - r\alpha\)

From equation (iii),

\(a_{cm} = 2r\alpha\)

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