JEE Main 2021 · Physics · Rotational Motion · medium

A uniform rod of mass M and length L can rotate freely in a vertical plane about a smooth horizontal pivot at one end. It is released from the…

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A uniform rod of mass

M

and length

L

can rotate freely in a vertical plane about a smooth horizontal pivot at one end. It is released from the horizontal position. When the rod passes the vertical position, the magnitude of the force exerted by the pivot on the rod is

A. $\dfrac{Mg}{2}$
B. $\dfrac{3Mg}{2}$
C. $3Mg$
D. $\dfrac{5Mg}{2}$

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Answer: (D) $\dfrac{5Mg}{2}$

Energy conservation:

Mg\dfrac{L}{2}=\dfrac12\cdot\dfrac{ML^2}{3}\,\omega^2\Rightarrow \omega^2=\dfrac{3g}{L}

. At the vertical position the torque of gravity about the pivot is zero, so the angular acceleration is zero and the centre of mass has only centripetal acceleration

\omega^2 L/2

directed upward. Hence

N-Mg=M\omega^2\dfrac L2=\dfrac{3Mg}{2}

, giving

N=\dfrac{5Mg}{2}

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