A flywheel having moment of inertia 4 kgm² about its vertical axis, rotates at the rate of 120 rpm about the axis. The minimum constant torque which can stop the wheel’s rotation in two minutes would be

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$\frac{4 π}{15} N - m$

$\frac{3 π}{15} N - m$

$\frac{2 π}{15} N - m$

$\frac{π}{15} N - m$

answer is C.

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Detailed Solution

$ω_{0} = 120 \times \frac{2 π}{60} = 4 πrad / s & α = \frac{ω_{0}}{t} = \frac{4 π}{120} = \frac{π}{30} rad / s^{2}; τ = Iα = 4 \cdot \frac{π}{30} = \frac{2 π}{15} Nm$

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