JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Elasticity & SHM

Two masses

m

and

\frac{m}{2}

are connected at the two ends of a massless rigid rod of length

ℓ

. The rod is suspended by a thin wire of torsional constant k at the centre of mass of the rodmass system (see figure). Because of torsional constant k , the restoring torque is

τ = k θ

for angular displacement

θ

. If the rod is rotated by

θ_{0}

and released, the tension in it when it passes through its mean position will be

(A) $\frac{3 k θ _{0} ^{2}}{ℓ}$
(B) $\frac{2 k θ _{0} ^{2}}{ℓ}$
(C) $\frac{k θ _{0} ^{2}}{ℓ}$ ✓ Correct
(D) $\frac{k θ _{0} ^{2}}{2 ℓ}$

Answer: (C)

Solution

First locate the centre of mass, which is where the wire attaches. Measuring from mass

m

x_{c m} = \frac{\frac{m}{2} ℓ}{\frac{3 m}{2}} = \frac{ℓ}{3}

, so

m

sits

\frac{ℓ}{3}

from the axis and

\frac{m}{2}

sits

\frac{2 ℓ}{3}

away. The moment of inertia is

I = m (\frac{ℓ}{3})^{2} + \frac{m}{2} (\frac{2 ℓ}{3})^{2} = \frac{m ℓ ^{2}}{3}

. The system swings as a torsional oscillator with

ω^{2} = \frac{k}{I}

, and at the mean position its angular speed is

\dot{θ} = θ_{0} ω

. The string tension supplies the centripetal force on

m

T = m \dot{θ}^{2} \cdot \frac{ℓ}{3} = m θ_{0}^{2} \frac{k}{I} \cdot \frac{ℓ}{3} = \frac{k θ _{0}^{2}}{ℓ}

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Elasticity & SHM

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