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A wheel having moment of inertia 2 kg-m² about its geometrical 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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$\frac{π}{15} N - m$

$\frac{π}{12} N - m$

$\frac{π}{18} N - m$

$\frac{2 π}{15} N - m$

answer is A.

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Detailed Solution

We have, $ω_{2} = ω_{1} + α t$

$\Rightarrow ω_{1} = ω_{2} - α t$

Here, $ω_{1} = 0$

and $ω_{2} = 60 rpm$

$= \frac{60 \times 2 π}{60} {rads}^{- 1}$

$\Rightarrow 0 = 2 π \times \frac{60}{60} - α \cdot 60$

or $α = \frac{2 π}{60}$

$∴$ Torque, $τ = I α = 2 \times \frac{2 π}{60} = \frac{π}{15} N - m$

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