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IIT JEE Advanced 2011 Paper 1

Section: Physics

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Question : 8 of 69

Marks: +1, -0

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 statements(s) is(are) true?

[JEE Adv 2011 P1]

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

Solution:

We have

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

• Case A: The moment of inertia is

I_A

=

\frac{MR^2}{2} + ML^2 + \frac{ml^3}{3}

Therefore,

\tau_A

=

-I_A \omega_A^2 \theta

=

-\left(\frac{MR^2}{2} + ML^2 + \frac{ml^3}{3}\right) \omega_A^2 \theta

• Case B: The moment of inertia is

I_B

=

\frac{ml^3}{3} + ML^2

Therefore,

\tau_B

=

-I_B \omega_B^2 \theta

=

-\left(\frac{ml^3}{3} + ML^2\right) \omega_B^2 \theta

since

\tau_A

=

\tau_B

⇒

\omega_A

<

\omega_B

.

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