devarshi-dt-logo

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 (see figure). Because of torsional constant k, the restoring torque is τ=kθ for angular displacement θ. If the rod is rotated by θ0 and released, the tension in it when it passes through its mean position will be:

kθ0²/2l

3kθ0²/l

2kθ0²/l

kθ0²/l

Solution:

ω=√(k/I)
ω=√(3km/l²)(Ref. image 1)
ω=ωθ0=average velocity
T=mω²r1
T=mω²l/3=mω²θ0²/l/3=m(3km/l²)(θ0²/l/3)=kθ0²/l
I=μl²/2=m(m/2)/(m+m/2)l²/2=ml²/6=(ml²/3)
(Ref. image 2)
r1/r2=1/2 =>r1=l/3