Question 770: Centre of Mass and Rotational Motion

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 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?

Accepted Answer

We have
Restoring torque = Force of gravity on disc and rod which is same in both cases.
$$\bullet$$ Case A : The moment of inertia is
$${I_A} = {{M{R^2}} \over 2} + M{L^2} + {{m{l^3}} \over 3}$$
Therefore,
$${	au _A} =  - {I_A}\omega _A^2	heta  =  - \left( {{{M{R^2}} \over 2} + M{L^2} + {{m{l^3}} \over 3}} ight)\omega _A^2	heta $$
$$\bullet$$ Case B : The moment of inertia is
$${I_B} = {{m{l^3}} \over 3} + M{L^2}$$
Therefore,
$${	au _B} =  - {I_B}\omega _B^2	heta  =  - \left( {{{m{l^3}} \over 3} + M{L^2}} ight)\omega _B^2	heta $$
since $${	au _A} = {	au _B} \Rightarrow {\omega _A} < {\omega _B}$$.

Answer

Restoring torque in case A = Restoring torque in case B.

Answer

Restoring torque in case A < Restoring torque in case B.

Answer

Angular frequency for case A > Angular frequency for case B.

Answer

Angular frequency for case A < Angular frequency for case B.

Centre of Mass and Rotational Motion

Choose the correct answer: