The moment of inertia of a wheel rotating abo | vedclass

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(B) Given: Moment of inertia $I = 3 	imes 10^2 \ kg \ m^2$,Retarding torque $	au = 6.9 	imes 10^2 \ N \ m$,Initial angular speed $\omega_0 = 4.6 \

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$2$

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(B) Given: Moment of inertia $I = 3 	imes 10^2 \ kg \ m^2$,Retarding torque $	au = 6.9 	imes 10^2 \ N \ m$,Initial angular speed $\omega_0 = 4.6 \ rad \ s^{-1}$,Final angular speed $\omega = 0 \ rad \ s^{-1}$.Using the relation $	au = I \alpha$,we find the angular acceleration $\alpha$:$\alpha = \frac{	au}{I} = \frac{6.9 	imes 10^2}{3 	imes 10^2} = 2.3 \ rad \ s^{-2}$.Since it is a retarding torque,$\alpha = -2.3 \ rad \ s^{-2}$.Using the equation of motion $\omega = \omega_0 + \alpha t$:$0 = 4.6 + (-2.3)t$$2.3t = 4.6$$t = \frac{4.6}{2.3} = 2 \ s$.Thus,the wheel will come to rest in $2 \ s$.

The moment of inertia of a wheel rotating about an axis is $3 \times 10^2 \ kg \ m^2$ and it has a constant angular speed of $4.6 \ rad \ s^{-1}$. If a retarding torque of $6.9 \times 10^2 \ N \ m$ is applied to the wheel,in how many seconds will the wheel come to rest?

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