A wheel having moment of inertia $2 k g - m^{2}$ about its vertical axis, rotates at the rate of $60 r p m$ about this axis. The torque which can stop the wheels rotation in one minute would be

\frac{2 π}{15} N - m

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\frac{π}{12} N - m

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\frac{π}{15} N - m

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\frac{π}{18} N - m

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Solution

The correct option is C $\frac{π}{15} N - m$
$α = \frac{2 π (n_{2} - n_{1})}{t} == \frac{- 2 π}{60} = \frac{- π}{30} r a d / s^{2}$
$=∴ r = I α = \frac{2 \times π}{30} = \frac{π}{15} N - m$

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