A wheel having a moment of inertia of 4 ,kg c | vedclass

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(B) Given: Moment of inertia $I = 4 \,kg \cdot m^2$,initial angular velocity $\omega_0 = 240 \,rpm$,time $t = 1 \,min = 60 \,s$,and final angular velo

Vedclass · Accepted Answer

$\frac{8 \pi}{15}$

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(B) Given: Moment of inertia $I = 4 \,kg \cdot m^2$,initial angular velocity $\omega_0 = 240 \,rpm$,time $t = 1 \,min = 60 \,s$,and final angular velocity $\omega = 0 \,rad/s$.First,convert the angular velocity from $rpm$ to $rad/s$:$\omega_0 = 240 	imes \frac{2 \pi}{60} = 8 \pi \,rad/s$.Using the equation of rotational motion $\omega = \omega_0 - \alpha t$,where $\alpha$ is the angular deceleration:$0 = 8 \pi - \alpha(60) \implies \alpha = \frac{8 \pi}{60} = \frac{2 \pi}{15} \,rad/s^2$.The torque $	au$ required is given by $	au = I \alpha$:$	au = 4 	imes \frac{2 \pi}{15} = \frac{8 \pi}{15} \,N \cdot m$.

$A$ wheel having a moment of inertia of $4 \,kg \cdot m^2$ about its symmetrical axis rotates at a rate of $240 \,rpm$ about it. The torque required to stop the rotation of the wheel in one minute is ............ $N \cdot m$.

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