A metal rod of length L and mass m is pivoted | vedclass

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Question

(A,D) The restoring torque $	au$ for a small angular displacement $	heta$ is given by $	au = -(mg \cdot \frac{L}{2} + Mg \cdot L) \sin 	heta \appr

Vedclass · Accepted Answer

$(A)$ Restoring torque in case $A =$ Restoring torque in case $B$

Vedclass · Answer

(A,D) The restoring torque $	au$ for a small angular displacement $	heta$ is given by $	au = -(mg \cdot \frac{L}{2} + Mg \cdot L) \sin 	heta \approx -(mg \cdot \frac{L}{2} + Mg \cdot L) 	heta$. Since the restoring torque depends only on the masses and their positions relative to the pivot,it is the same in both cases $A$ and $B$. Thus,statement $(A)$ is true.
The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{|	au|/	heta}{I_{pivot}}}$.
In case $A$,the disc is fixed to the rod,so it rotates with the rod. The moment of inertia $I_A = I_{rod} + I_{disc, pivot} = \frac{mL^2}{3} + (\frac{MR^2}{2} + ML^2) = \frac{mL^2}{3} + \frac{MR^2}{2} + ML^2$.
In case $B$,the disc is free to rotate,so it does not rotate about its own center. The moment of inertia $I_B = I_{rod} + I_{disc, CM} = \frac{mL^2}{3} + ML^2$.
Since $I_A > I_B$,and the restoring torque is the same,the angular frequency $\omega_A < \omega_B$. Thus,statement $(D)$ is true.

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