Question

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. 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. When rod-disc system is lowest position total kinetic energy for case A = total kinetic energy for 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

Answer 1

Answer: (B), (C)

The correct options are
B Restoring torque in case A = Restoring torque in case B
C Angular frequency for case A > Angular frequency for case B

$\text{Given:-}$ A metal rod of length L and mass m

$\text{Solution:-}$ Torque is same for both the cases.

${\tau }_A={\tau }_B=2\pi \sqrt{\frac{I}{(M+m)gd}}$

Here,$I_A=\frac{m{L}^3}{3}+(\frac{M{R}^2}{2}+M{L}^2)$

When the disc is free to rotate about its centre, it will not rotate w.r.t. ground. It will act as a point mass, so moment of inertia is given by

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

So,

$I_A>I_B\Rightarrow T_A>T_B\Rightarrow {\omega }_A<{\omega }_B$

$\text{Hence the correct option is B and C}$