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Question

Two masses m and 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 rod - mass system as shown in the figure.
Because of torsional constant K, the restoring torque is τ=Kθ for angular displacement θ.
If the rod is rotated by θ and released so that rod oscillates. The tension in the rod, when it passes through its mean position will be


A
3Kθ2l
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B
2Kθ2l
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C
Kθ2l
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D
Kθ22l
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Solution

The correct option is A 3Kθ2l
Finding position of COM


r1=mm+m2l=2l3
r2=lr1=l3
When rod is tilted by θ0,


Centripetal force on m is
Fc=mv2l/3
v=Aωv=(l3θ)ω
ω is frequency of torsional pendulum.


ω=KI

Tension in the rod is
T=mv2l/3=3mv2l=3ml(l3θω)2
T=mθ20×l3×KI
Moment of inertia about COM is
I=m2(2l3)2+m(l3)2=m24l29+ml29=ml23
Hence, T=mθ20×l3×K(3)ml2=Kθ20l

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