Q101 / 150 · MHT CET 2023 (13 May Shift 2) · Physics

A thin uniform rod $A B$ of mass ' $\mathrm{m}$ ' and length ' $l$ ' is hinged at one end A to the ground level. Initially the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its end B strikes the ground is ( $g$ = acceleration due to gravity)

  • A
    $\sqrt{\frac{g}{l}}$
  • B
    $\sqrt{\frac{\mathrm{mg}}{l}}$
  • Correct answer
    C
    $\sqrt{\frac{3 \mathrm{~g}}{l}}$
  • D
    $\sqrt{\frac{\mathrm{mg}}{3 l}}$

Solution

We know, Loss in P.E = Gain in rotational K.E. When the centre of mass of the rod falls through a distance $\frac{\mathrm{L}}{2}$, Loss in P.E $=m g \frac{L}{2}$ ...(i) and, Gain in Rotational K.E $=\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2}\left[\frac{\mathrm{ML}^2}{3}\right] \omega^2$ ...(ii) Equating (i) and (ii) $$ \begin{aligned} & \frac{\mathrm{MgL}}{2}=\frac{1}{2} \frac{\mathrm{ML}^2}{3} \omega^2 \\ & \omega^2=\frac{3 \mathrm{~g}}{\mathrm{~L}} \\ & \omega=\sqrt{\frac{3 \mathrm{~g}}{\mathrm{~L}}}=\sqrt{\frac{3 \mathrm{~g}}{l}} \quad \ldots(\because \text { here } \mathrm{L}=l) \end{aligned} $$ $\ldots(\because$ here $\mathrm{L}=l)$

Hence, the correct answer is option: 'C'.

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