A wheel of moment of inertia 10 kg-m^2 is rot | vedclass

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Question

Given Moment of inertia, $I=10 k g-m^{2},$ speed $N=10 r p m$

$\Rightarrow \omega_{1}=\frac{2 \pi N}{60}=\frac{2 \pi 	imes 10}{60}=\frac{\pi}{3}$

Vedclass · Accepted Answer

$131.4$

Vedclass · Answer

Given Moment of inertia, $I=10 k g-m^{2},$ speed $N=10 r p m$

$\Rightarrow \omega_{1}=\frac{2 \pi N}{60}=\frac{2 \pi 	imes 10}{60}=\frac{\pi}{3}$

$\Rightarrow \omega_{2}=5 \omega_{1}=\frac{5 \pi}{3}$

$	herefore$ work done $=\frac{1}{2} I\left(\omega_{2}^{2}-\omega_{1}^{2}ight)=\frac{1}{2} 	imes 10 	imes\left(\frac{25 \pi^{2}}{9}-\frac{\pi^{2}}{9}ight)=131.4 J$

A wheel of moment of inertia $10\ kg-m^2$ 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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