A solid cylinder of mass 2 kg and radius 4 cm rotating about its axis at the rate of 3 rpm . The torque required to stop after 2π revolutions is: - Physics MCQ NEET 2019 2026 | ExamTest.live

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A solid cylinder of mass 2 kg and radius 4 cm rotating about its axis at the rate of 3 rpm . The torque required to stop after 2 π revolutions is:

Solution & Step-by-step Explanation

Initial angular velocity ω_{0} = 3 rpm = 3 \times \frac{2 π}{60} = 0.1 π rad/s .Angular displacement Δ θ = 2 π rev = 2 π \times 2 π = 4 π^{2} rad .Final angular velocity ω = 0 .Using ω^{2} = ω_{0}^{2} + 2 α Δ θ : 0 = (0.1 π)^{2} + 2 α (4 π^{2}) ⟹ α = - \frac{0.01 π ^{2}}{8 π ^{2}} = - 1.25 \times 1 0^{- 3} rad/s^{2} .Moment of inertia of solid cylinder I = \frac{1}{2} M R^{2} = \frac{1}{2} \times 2 \times (0.04)^{2} = 16 \times 1 0^{- 4} kg m^{2} .Torque τ = I α = (16 \times 1 0^{- 4}) \times (1.25 \times 1 0^{- 3}) = 20 \times 1 0^{- 7} = 2 \times 1 0^{- 6} N m .

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A solid cylinder of mass 2 kg and radius 4 cm rotating about its axis at the rate of 3 rpm . The torque required to stop after 2 π revolutions is:

2 \times 1 0^{- 6} N m

2 \times 1 0^{- 3} N m

12 \times 1 0^{- 4} N m

2 \times 1 0^{6} N m

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