If I = 50, kg-m^2,then how much torque (in N- | vedclass

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(A) Given: Moment of inertia $I = 50\, kg-m^2$,initial angular velocity $\omega_i = 20\, rad/s$,final angular velocity $\omega_f = 0\, rad/s$,and time

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$100$

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(A) Given: Moment of inertia $I = 50\, kg-m^2$,initial angular velocity $\omega_i = 20\, rad/s$,final angular velocity $\omega_f = 0\, rad/s$,and time $t = 10\, s$.Angular acceleration $\alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 20}{10} = -2\, rad/s^2$.The torque required is given by $	au = I \alpha$.$	au = 50 	imes (-2) = -100\, N-m$.The magnitude of the torque applied is $100\, N-m$ in the direction opposite to the angular velocity.

If $I = 50\, kg-m^2$,then how much torque (in $N-m$) will be applied to stop it in $10\, s$? Its initial angular speed is $20\, rad/s$.

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