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A grindstone has a moment of inertia of $6 kg m^{2}$ . A constant torque is applied and the grindstone is found to have a speed of $150 rpm$ , $10$ seconds after starting from rest. The torque is

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$4 π Nm$

$3 π Nm$

$3 N m$

$\frac{π}{3} N m$

answer is A.

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Detailed Solution

Here, $I = 6 kg m^{2}, t = 10 s, ω_{0} = 0$

$ω = 150 rpm = \frac{150}{60} rps = \frac{5}{2} rps$

$ω = 2 π v = 2 π \times \frac{5}{2} = 5 π {rads}^{- 1}$

$α = \frac{ω - ω_{0}}{t} = \frac{5 π - 0}{10} = \frac{π}{2} {rads}^{- 2} ∴ Torque, τ = I α = 6 \times \frac{π}{2} = 3 π Nm$

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