A flywheel of mass $4 \mathrm{~kg}$ and radius $10 \mathrm{~cm}$, rotating with a uniform angular velocity of $5 \mathrm{rad} / \mathrm{s}$, is subjected to a torque of $0.01 \mathrm{~N}$.m for 10 seconds.
If the torque increases the speed of rotation, find
(i) the final angular velocity of the flywheel
(ii) the change in its angular velocity
(iii) the change in its angular momentum
(iv) the change in its kinetic energy.
If the torque increases the speed of rotation, find
(i) the final angular velocity of the flywheel
(ii) the change in its angular velocity
(iii) the change in its angular momentum
(iv) the change in its kinetic energy.
Q 105.5
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$
\begin{aligned}
& \text { Data : } M=4 \mathrm{~kg}, \mathrm{R}=10 \mathrm{~cm}=0.1 \mathrm{~m}, \omega_1=5 \mathrm{rad} / \mathrm{s}, \tau=0.01 \mathrm{~N} \cdot \mathrm{m}, \mathrm{t}=10 \mathrm{~s} \\
& I=\frac{M R^2}{2}=\frac{4 \times 0.01}{2}=0.02 \mathrm{~kg} \cdot \mathrm{m}^2 \\
& \alpha=\frac{\tau}{l}=\frac{0.01}{0.02}=0.5 \mathrm{rad} / \mathrm{s}
\end{aligned}
$
(i) The final angular velocity of the flywheel,
$
\begin{aligned}
& \omega_2=\omega_1+\alpha \mathrm{t} \\
& =5+0.5 \times 10=10 \mathrm{rad} / \mathrm{s}
\end{aligned}
$
(ii) The change in its angular velocity
$
=\omega_2-\omega_1=5 \mathrm{rad} / \mathrm{s}
$
(iii) The change in its angular momentum
$
\begin{aligned}
& =\left|\omega_2-\right| \omega_1=\mid\left(\omega_2-\omega_1\right) \\
& =0.02 \times 5=0.1 \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}
\end{aligned}
$
(iv) The change in its kinetic energy
$
\begin{aligned}
& =\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) \\
& =\frac{1}{2} \times 0.02 \times\left[(10)^2-(5)^2\right]=0.75 \mathrm{~J}
\end{aligned}
$
\begin{aligned}
& \text { Data : } M=4 \mathrm{~kg}, \mathrm{R}=10 \mathrm{~cm}=0.1 \mathrm{~m}, \omega_1=5 \mathrm{rad} / \mathrm{s}, \tau=0.01 \mathrm{~N} \cdot \mathrm{m}, \mathrm{t}=10 \mathrm{~s} \\
& I=\frac{M R^2}{2}=\frac{4 \times 0.01}{2}=0.02 \mathrm{~kg} \cdot \mathrm{m}^2 \\
& \alpha=\frac{\tau}{l}=\frac{0.01}{0.02}=0.5 \mathrm{rad} / \mathrm{s}
\end{aligned}
$
(i) The final angular velocity of the flywheel,
$
\begin{aligned}
& \omega_2=\omega_1+\alpha \mathrm{t} \\
& =5+0.5 \times 10=10 \mathrm{rad} / \mathrm{s}
\end{aligned}
$
(ii) The change in its angular velocity
$
=\omega_2-\omega_1=5 \mathrm{rad} / \mathrm{s}
$
(iii) The change in its angular momentum
$
\begin{aligned}
& =\left|\omega_2-\right| \omega_1=\mid\left(\omega_2-\omega_1\right) \\
& =0.02 \times 5=0.1 \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}
\end{aligned}
$
(iv) The change in its kinetic energy
$
\begin{aligned}
& =\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) \\
& =\frac{1}{2} \times 0.02 \times\left[(10)^2-(5)^2\right]=0.75 \mathrm{~J}
\end{aligned}
$
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