A flywheel of mass 25 ,kg has a radius of 0.2 | vedclass

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(B) Given: Mass $M = 25 \,kg$, Radius $R = 0.2 \,m$, Initial angular velocity $\omega_i = 240 \,rpm = \frac{240 	imes 2 \pi}{60} \,rad/s = 8 \pi \,ra

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$0.4 \pi \,Nm$

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(B) Given: Mass $M = 25 \,kg$, Radius $R = 0.2 \,m$, Initial angular velocity $\omega_i = 240 \,rpm = \frac{240 	imes 2 \pi}{60} \,rad/s = 8 \pi \,rad/s$, Time $t = 20 \,s$, Final angular velocity $\omega_f = 0 \,rad/s$.First, calculate the angular deceleration $\alpha$:$\alpha = \frac{\omega_i - \omega_f}{t} = \frac{8 \pi - 0}{20} = \frac{8 \pi}{20} = 0.4 \pi \,rad/s^2$.Next, calculate the moment of inertia $I$ of the flywheel (assuming it is a disc):$I = M R^2 = 25 	imes (0.2)^2 = 25 	imes 0.04 = 1 \,kg \cdot m^2$.Finally, calculate the torque $	au$:$	au = I \alpha = 1 	imes 0.4 \pi = 0.4 \pi \,Nm$.

$A$ flywheel of mass $25 \,kg$ has a radius of $0.2 \,m$. It is rotating at $240 \,rpm$. What is the torque necessary to bring it to rest in $20 \,s$?

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