A rod AB is free to rotate in a vertical plan | vedclass

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(B) Let $L$ be the length of the rod and $m$ be its mass. The moment of inertia of the rod about $A$ is $I = \frac{mL^2}{3}$.When the rod falls from t

Vedclass · Accepted Answer

the coefficient of restitution between the rod and the peg is $\frac{1}{\sqrt{2}}$

Vedclass · Answer

(B) Let $L$ be the length of the rod and $m$ be its mass. The moment of inertia of the rod about $A$ is $I = \frac{mL^2}{3}$.When the rod falls from the unstable vertical position to the stable vertical position,the change in potential energy is converted into rotational kinetic energy: $mg(L) = \frac{1}{2} I \omega^2$.Substituting $I = \frac{mL^2}{3}$,we get $mgL = \frac{1}{2} (\frac{mL^2}{3}) \omega^2$,which simplifies to $\omega = \sqrt{\frac{6g}{L}}$.The velocity of end $B$ just before impact is $v = \omega L = \sqrt{6gL}$.After the collision,the rod swings up to a horizontal position where it comes to rest. The rotational kinetic energy just after the collision is converted into potential energy: $\frac{1}{2} I \omega'^2 = mg(\frac{L}{2})$.Substituting $I = \frac{mL^2}{3}$,we get $\frac{1}{2} (\frac{mL^2}{3}) \omega'^2 = \frac{mgL}{2}$,which simplifies to $\omega' = \sqrt{\frac{3g}{L}}$.The velocity of end $B$ just after impact is $v' = \omega' L = \sqrt{3gL}$.The coefficient of restitution $e$ is given by $e = \frac{v'}{v} = \frac{\sqrt{3gL}}{\sqrt{6gL}} = \frac{1}{\sqrt{2}}$.

Quantities	Their numerical values (in $SI$ units)
a. Angular acceleration of smaller wheel	$1$. $5/3$
b. Torque on the larger wheel	$2$. $100/3$
c. Angular acceleration of larger wheel	$3$. $5/2$

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