A thin uniform rod of mass 1 kg and length 1 | vedclass

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Question

(A) By the principle of conservation of energy,the potential energy lost by the rod is equal to the rotational kinetic energy gained by it.Initial pot

Vedclass · Accepted Answer

$\omega = \sqrt{30} \ rad \ s^{-1}$

Vedclass · Answer

(A) By the principle of conservation of energy,the potential energy lost by the rod is equal to the rotational kinetic energy gained by it.Initial potential energy of the rod (taking the center of mass at height $L/2$) is $U_i = mg(L/2)$.Final rotational kinetic energy when the rod hits the ground is $K_f = \frac{1}{2} I \omega^2$.Here,$I$ is the moment of inertia of the rod about the hinged end,given by $I = \frac{mL^2}{3}$.Equating the two: $mg \frac{L}{2} = \frac{1}{2} (\frac{mL^2}{3}) \omega^2$.Simplifying the equation: $g = \frac{L}{3} \omega^2$,which gives $\omega = \sqrt{\frac{3g}{L}}$.Substituting the given values $g = 10 \ m \ s^{-2}$ and $L = 1 \ m$:$\omega = \sqrt{\frac{3 	imes 10}{1}} = \sqrt{30} \ rad \ s^{-1}$.

$A$ thin uniform rod of mass $1 \ kg$ and length $1 \ m$ is hinged at one end to the ground. It originally stands vertically and is allowed to fall to the ground. If the rod hits the ground with angular speed $\omega$,then the correct statement is (Assume $g = 10 \ m \ s^{-2}$):

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