A wheel having moment of inertia of 2 kg-m² about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel's rotation in one minute would be

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$\frac{2 π}{15} N - m$

$\frac{π}{12} N - m$

$\frac{π}{15} N - m$

$\frac{π}{18} N - m$

answer is C.

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Detailed Solution

Given that
$\begin{array}{l} I = 2 kg - m^{2}, ω_{0} = (60 / 60) \times 2 πrad / \sec \\ ω = 0 and t = 60 \sec \\ Now, torque λ = Iα = I (\frac{ω_{0} - ω}{t}) \\ ∴ λ = 2 (\frac{2 π - 0}{60}) = \frac{π}{15} N - m \end{array}$

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