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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

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Solution

The correct option is **A** 3Kθ2∘l

Finding position of COM

r1=mm+m2l=2l3

r2=l−r1=l3

When rod is tilted by θ0,

Centripetal force on m is

Fc=mv2l/3

⇒v=Aω⇒v=(l3θ∘)ω

ω is frequency of torsional pendulum.

⇒ω=√KI

Tension in the rod is

T=mv2l/3=3mv2l=3ml(l3θ∘ω)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

Finding position of COM

r1=mm+m2l=2l3

r2=l−r1=l3

When rod is tilted by θ0,

Centripetal force on m is

Fc=mv2l/3

⇒v=Aω⇒v=(l3θ∘)ω

ω is frequency of torsional pendulum.

⇒ω=√KI

Tension in the rod is

T=mv2l/3=3mv2l=3ml(l3θ∘ω)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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