JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Rotational Mechanics

A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle of

3 0^{\circ}

fro the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad

s^{- 1}

) will be (

g = 10 ms^{- 2}

Answer: (B)

Solution

Use energy conservation about the pivot. Released from

3 0^{\circ}

above the horizontal, the centre of mass (at

\frac{ℓ}{2}

from the pivot) drops by

h = \frac{ℓ}{2} sin 3 0^{\circ} = \frac{ℓ}{4}

as the rod swings down to the horizontal. The lost gravitational energy becomes rotational kinetic energy:

m g \frac{ℓ}{4} = \frac{1}{2} I ω^{2}

with

I = \frac{m ℓ ^{2}}{3}

for a rod about its end. Solving,

ω = \frac{3 g}{2 ℓ} = \frac{3 \times 10}{2 \times 0.5} = 30 rad/s

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option