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6.System of Particles and Rotational Motion
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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.......... $J$
A
$100$
B
$131.4$
C
$13.4$
D
$0.1341$
Solution
Given Moment of inertia, $I=10 k g-m^{2},$ speed $N=10 r p m$
$\Rightarrow \omega_{1}=\frac{2 \pi N}{60}=\frac{2 \pi \times 10}{60}=\frac{\pi}{3}$
$\Rightarrow \omega_{2}=5 \omega_{1}=\frac{5 \pi}{3}$
$\therefore$ work done $=\frac{1}{2} I\left(\omega_{2}^{2}-\omega_{1}^{2}\right)=\frac{1}{2} \times 10 \times\left(\frac{25 \pi^{2}}{9}-\frac{\pi^{2}}{9}\right)=131.4 J$
Standard 11
Physics
