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  • How to solve this problem- 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 fr

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)

    \frac{2mg}{2m+M}

  • Option 2)

    \frac{2Mg}{2m+M}

  • Option 3)

    \frac{2mg}{2M+m}

  • Option 4)

    \frac{2Mg}{2M+m}

 

Answers (1)

best_answer

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)

\frac{2mg}{2m+M}

This is the correct oiption.

Option 2)

\frac{2Mg}{2m+M}

This is an incorrect option.

Option 3)

\frac{2mg}{2M+m}

This is an incorrect option.

Option 4)

\frac{2Mg}{2M+m}

This is an incorrect option.

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