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Question
Two wheels of the moment of inertia 4 kgm2 rotate side by side at the rate of 120 rev/min and 240 rev/min respectively in the opposite directions. If now both the wheels are coupled by means of a weightless shaft so that both the wheels rotate with a common angular speed. Calculate the new speed of rotation.
Solution
I1 = I2 = I = 4 kg m2 , n1 = 120 r.p.m, n2 = 240 r.p.m.
Initially, the angular velocities of the two wheels are `vecω_1` and `vecω_2` and, therefore the angular momentum `vecL_1` and `vecL_2` are in opposite directions.
∴ The magnitude of the total initial angular momentum, L = -L1 + L2
∴ L = -Iω1 + Iω2 .................(i)
After coupling on the same shaft, the total moment of inertia is 2I.
Let, ω = 2πn be the common angular speed.
∴ The magnitude of the total final angular momentum L′ = 2Iω ….(ii)
By the principle of conservation of angular momentum, L = L′
∴ Equating equation (i) and (ii), we have
I(ω2 − ω1) = 2Iω
∴ 2 × 2πn = 2π(n2 - n1)
∴ 2n = n2 − n1
∴ n = `(n_2 - n_1)/2 = (240 - 120)/2` = 60 r.p.m.
The new speed of rotation of the wheels would be 60 r.p.m.
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