A wheel of moment of inertia 10 kg cdot m^2 | vedclass

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(B) Given: Moment of inertia $I = 10 \ kg \cdot m^2$,initial angular speed $N_1 = 10 \ rpm$.The initial angular velocity is $\omega_1 = \frac{2 \pi N_

Vedclass · Accepted Answer

$131.4$

Vedclass · Answer

(B) Given: Moment of inertia $I = 10 \ kg \cdot m^2$,initial angular speed $N_1 = 10 \ rpm$.The initial angular velocity is $\omega_1 = \frac{2 \pi N_1}{60} = \frac{2 \pi 	imes 10}{60} = \frac{\pi}{3} \ rad/s$.The final angular velocity is $\omega_2 = 5 \omega_1 = 5 	imes \frac{\pi}{3} = \frac{5 \pi}{3} \ rad/s$.According to the work-energy theorem,the work done is equal to the change in rotational kinetic energy:$W = \Delta K = \frac{1}{2} I (\omega_2^2 - \omega_1^2)$.Substituting the values: $W = \frac{1}{2} 	imes 10 	imes \left[ (\frac{5 \pi}{3})^2 - (\frac{\pi}{3})^2 ight]$.$W = 5 	imes \left( \frac{25 \pi^2}{9} - \frac{\pi^2}{9} ight) = 5 	imes \frac{24 \pi^2}{9} = 5 	imes \frac{8 \pi^2}{3} = \frac{40 \pi^2}{3}$.Using $\pi^2 \approx 9.8696$,$W = \frac{40 	imes 9.8696}{3} \approx 131.59 \ J$. Given the options,the closest value is $131.4 \ J$.

$A$ wheel of moment of inertia $10 \ kg \cdot m^2$ is rotating at $10$ rotations per minute. The work done in increasing its speed to $5$ times its initial value will be .......... $J$.

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