Simple Harmonic Motion Ques 57
- 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: 57.$(a, d)$
Solution:
Formula:
Compound Pendulum / Physical Pendulum:
- $\tau _A=\tau _B=(m g \frac{L}{2} \sin \theta+M g L \sin \theta)$
$=$ 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.
BETA
