JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Elasticity & SHM

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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 center 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 center and (case B) the disc is free to rotate about its center. 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✓ Correct
(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.✓ Correct

Answer: (A), (D)

Solution

The restoring torque depends only on the gravitational forces and their moment arms, which are identical in both cases:

τ = m g \frac{L}{2} sin θ + M g L sin θ

So option A is correct.

The moment of inertia differs:

Case A: $I_{A} = \frac{m L ^{2}}{3} + \frac{M R ^{2}}{2} + M L^{2}$ (disc cannot rotate about its center)
Case B: $I_{B} = \frac{m L ^{2}}{3} + M L^{2}$ (disc free to rotate, its rotational inertia about its center doesn't contribute)

Since

I_{A} > I_{B}

and

ω = \frac{τ / θ}{I}

, we have

ω_{A} < ω_{B}

(option D correct).

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · One or more correct options

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JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Elasticity & SHM

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