JEE Main · 2017 · 02 Apr 2017 · Offline · Rotational Mechanics

A slender uniform rod of mass M and length

l

is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle

θ

with the vertical is:

(A) $\frac{3 g}{2 l} cos θ$
(B) $\frac{2 g}{3 l} cos θ$
(C) $\frac{3 g}{2 l} sin θ$ ✓ Correct
(D) $\frac{2 g}{3 l} sin θ$

Answer: (C)

Solution

The rod turns under its own weight acting at the centre, a distance

\frac{l}{2}

from the pivot, giving a torque

M g \frac{l}{2} sin θ

when the rod is

θ

from the vertical. Its moment of inertia about the end is

\frac{M l ^{2}}{3}

, so the angular acceleration is

α = \frac{M g \frac{l}{2} sin θ}{\frac{M l ^{2}}{3}} = \frac{3 g}{2 l} sin θ

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option

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JEE Main · 2017 · 02 Apr 2017 · Offline · Rotational Mechanics

Solution

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