Simple Harmonic Motion 5 Question 17

22. A metal rod of length $L$ and mass $m$ is pivoted at one end. A thin disc 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 $\boldsymbol{A}$ - the disc is not free to rotate about its centre and case $\boldsymbol{B}$ - 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?

(2011)

(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 $B$

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Answer:

Correct Answer: 22. $(a, c)$

Solution:

  1. $\tau _A=\tau _B=m g \frac{L}{2} \sin \theta+M g L \sin \theta$

$=$ Restoring torque about point $O$.

In case $A$, moment of inertia will be more. Hence, angular acceleration $(\alpha=\tau / I)$ will be less. Therefore, angular frequency will be less.



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