The angular momentum of a body changes by $80 \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}$ when its angular velocity changes from $20 \mathrm{rad} / \mathrm{s}$ to $40 \mathrm{rad} / \mathrm{s}$. Find the change in its kinetic energy of rotation.
Q 101.1
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Data : $\omega_1=20 \mathrm{rad} / \mathrm{s}, \omega_2=40 \mathrm{rad} / \mathrm{s}$
If $\mathrm{I}$ is the $\mathrm{MI}$ of the body, its initial angular momentum is $\mathrm{I} \omega_1$, and final angular momentum is $\mathrm{I}_2$.
Change in angular momentum
$
\begin{aligned}
& =1 \omega_2-1 \omega_1\left(\omega_2-\omega_1\right) \\
& \therefore 80=1(40-20) \\
& \therefore I=4 \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}
$
$
\begin{aligned}
\text { Change in KE } & =\frac{1}{2} I \omega_2^2-\frac{1}{2} I \omega_1^2=\frac{1}{2} I\left(\omega_2^2-\omega_1^2\right)^{\prime} \\
& =\frac{1}{2} \times 4 \times(1600-400)=\mathbf{2 4 0 0} \mathrm{J}
\end{aligned}
$
If $\mathrm{I}$ is the $\mathrm{MI}$ of the body, its initial angular momentum is $\mathrm{I} \omega_1$, and final angular momentum is $\mathrm{I}_2$.
Change in angular momentum
$
\begin{aligned}
& =1 \omega_2-1 \omega_1\left(\omega_2-\omega_1\right) \\
& \therefore 80=1(40-20) \\
& \therefore I=4 \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}
$
$
\begin{aligned}
\text { Change in KE } & =\frac{1}{2} I \omega_2^2-\frac{1}{2} I \omega_1^2=\frac{1}{2} I\left(\omega_2^2-\omega_1^2\right)^{\prime} \\
& =\frac{1}{2} \times 4 \times(1600-400)=\mathbf{2 4 0 0} \mathrm{J}
\end{aligned}
$
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