A wheel having a moment of inertia 2 , kg cdo | vedclass

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(D) Given: Moment of inertia $I = 2 \, kg \cdot m^{2}$,initial angular velocity $\omega_{i} = 60 \, rpm = 60 	imes \frac{2\pi}{60} \, rad/s = 2\pi \,

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$\frac{\pi}{15} \, Nm$

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(D) Given: Moment of inertia $I = 2 \, kg \cdot m^{2}$,initial angular velocity $\omega_{i} = 60 \, rpm = 60 	imes \frac{2\pi}{60} \, rad/s = 2\pi \, rad/s$,final angular velocity $\omega_{f} = 0$,and time $\Delta t = 1 \, minute = 60 \, s$.Using the impulse-momentum theorem for rotation,the torque $	au$ required is given by $	au \Delta t = \Delta L = I(\omega_{f} - \omega_{i})$.Taking the magnitude,$	au 	imes 60 = I 	imes \omega_{i}$.Substituting the values: $	au 	imes 60 = 2 	imes 2\pi$.$	au = \frac{4\pi}{60} = \frac{\pi}{15} \, Nm$.

$A$ wheel having a moment of inertia $2 \, kg \cdot m^{2}$ 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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