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IIT-JEE 2011 Paper 1 Offline

MCQ (More than One Correct Answer)

A metal rod of length L and mass m is pivoted at one end. A thin disk of mass M and radius R ( < L) is attached at its centre to the free end of the rod. Consider two ways the disc is attached : (case A). The disc is not free to rotate about its centre and (case B) the disc is free to rotate about its centre. The rod-disc system performs SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is(are) true?

A
Restoring torque in case A = Restoring torque in case B.
B
Restoring torque in case A < Restoring torque in case B.
C
Angular frequency for case A > Angular frequency for case B.
D
Angular frequency for case A < Angular frequency for case B.

Explanation

We have

Restoring torque = Force of gravity on disc and rod which is same in both cases.

$$\bullet$$ Case A : The moment of inertia is

$${I_A} = {{M{R^2}} \over 2} + M{L^2} + {{m{l^3}} \over 3}$$

Therefore,

$${\tau _A} = - {I_A}\omega _A^2\theta = - \left( {{{M{R^2}} \over 2} + M{L^2} + {{m{l^3}} \over 3}} \right)\omega _A^2\theta $$

$$\bullet$$ Case B : The moment of inertia is

$${I_B} = {{m{l^3}} \over 3} + M{L^2}$$

Therefore,

$${\tau _B} = - {I_B}\omega _B^2\theta = - \left( {{{m{l^3}} \over 3} + M{L^2}} \right)\omega _B^2\theta $$

since $${\tau _A} = {\tau _B} \Rightarrow {\omega _A} < {\omega _B}$$.

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