# A uniform disc of radius R and mass M is free to rotate only about its axis.  A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure.  The body is released from rest.  Then the acceleration of the body is : Option 1) Option 2) Option 3) Option 4)

As we learnt in

Analogue of second law of motion for pure rotation -

$\vec{\tau }=I\, \alpha$

- wherein

Torque equation can be applied only about two points

(i) centre of motion.

(ii) point which has zero velocity/acceleration.

Short trick

$a=\frac{forward\ force-backward\ force}{total\ mass}$

$a=\frac{mg-0}{m+\frac{I}{R^{2}}}\ \; \Rightarrow\ \;a=\frac{mg}{m+\frac{1}{2}\frac{MR^{2}}{R^{2}}}$

$a=\frac{2mg}{2m+M}\ \; \; \;\left [ \because \;I_{c}=\frac{1}{2}mR^{2} \right ]$

Option 1)

This is the correct oiption.

Option 2)

This is an incorrect option.

Option 3)

This is an incorrect option.

Option 4)

This is an incorrect option.

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