A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of 1200 rpm. It is brought to rest in 10 s by applying a constant torque. The

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Correct answer: D. 0.0128π Nm, 100

Initial angular velocity ω_0 = 1200 rpm = (1200 × 2π)/(60) = 40π rad/s Final angular velocity ω = 0 Time t = 10 s Angular acceleration α = (ω - ω_0)/(t) = (0 - 40π)/(10) = -4π rad/s^2 Moment of inertia of the solid sphere I = (2)/(5)MR^2 = (2)/(5) × 5 × (0.04)^2 = 0.0032 kg m^2 Torque = I|α| = 0.0032 × 4π = 0.0128π Nm Angle rotated θ = (ω_0 + ω)/(2)t = (40π + 0)/(2) × 10 = 200π rad Number of rotations N = (θ)/(2π) = (200π)/(2π) = 100 Answer: 0.0128π Nm, 100

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