JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Rotational Mechanics

A uniform rod of length '

l

' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed

ω

the rod makes an angle

θ

with it (see figure). To find

θ

equate the rate of change of angular momentum (direction going into the paper)

\frac{ml ^{2}}{12} ω^{2} sin θ cos θ

about the centre of mass (CM) to the torque provided by the horizontal and vertical forces

F_{H}

and

F_{V}

about the CM. The value of

θ

is then such that:

Answer: (C)

Solution

For an element of the spinning rod at distance

x

from the pivot, the outward inertial effect is

\frac{m}{ℓ} d x ω^{2} x sin θ

, and its torque about the pivot is this times the lever arm

x cos θ

. Integrating along the rod,

τ_{cf} = \int_{0}^{ℓ} \frac{m}{ℓ} ω^{2} x^{2} sin θ cos θ d x = \frac{m ω ^{2} ℓ ^{2} sin θ cos θ}{3}

. Gravity acting at the centre of mass supplies the balancing torque

m g \frac{ℓ}{2} sin θ

. Setting them equal,

m g \frac{ℓ}{2} sin θ = \frac{m ω ^{2} ℓ ^{2}}{3} sin θ cos θ

, the

sin θ

cancels and

cos θ = \frac{3 g}{2 ℓ ω ^{2}}

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option