Q.
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 $\mathbf{A}$- the disc is not free to rotate about its centre and case $\mathbf{B}$ – 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
(d) $\tau_{A} = \tau_{B} = (mg\frac{L}{2}\sin{\theta + M\ gL\sin{\theta)}}$
= Restoring torque about point $O$.
In case $A$, moment of inertia will be more. Hence, angular acceleration $(\alpha = \frac{\tau}{I})$ will be less. Therefore angular frequency will be less. Note Question is difficult because this type of SHM is rarely.
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.