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Question:

A massless rod of length 2l has equal point masses attached to its two ends as shown in the figure. The rod is rotating about its axis passing through its center and making an angle α with the axis. The magnitude of the change of momentum of the rod, i.e., (|dL/dt|), equals?

mω²sin²θ

2mω²sinθcosθ

mv²sin²θ

m(l/2)ωsinθcosθ

Solution:

Correct option is B. mω²sin²θ
We know that the rate of change of angular momentum is equal to the torque acting on the system
⇒|dL/dt| = τ
From the figure, the force on the mass m is given as:
F = mrω² (Due to the centripetal force)
r = lsinα
⇒F = m * lsinα * ω²
⇒τ = r⊥ * F
r⊥ = lcosα
⇒τ = lcosα * m * lsinαω²
⇒τ = ml²ω²sinαcosα
Total torque due to both masses is given by
τnet = τ₁ + τ₂ = ml²ω²sinαcosα + ml²ω²sinαcosα
⇒τnet = 2ml²ω²sinαcosα = ml²ω²sin2α
∴|dL/dt| = τnet = ml²ω²sin2α