A wheel of moment of inertia $1 \mathrm{~kg} \cdot \mathrm{m}^2$ is rotating at a speed of $40 \mathrm{rad} / \mathrm{s}$. Due to the friction on the axis, the wheel comes to rest in 10 minutes. Calculate the angular momentum of the wheel, two minutes before it comes to rest.
Q 101.2
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$
\begin{aligned}
& \text { Data }: \mid=1 \mathrm{~kg} \cdot \mathrm{m}^2, \omega_1=40 \mathrm{rad} / \mathrm{s}, \omega_2=0 \text { at } \\
& \mathrm{t}=10 \text { minutes }=60 \times 10 \mathrm{~s}=600 \mathrm{~s}, \\
& \mathrm{t}^{\prime}=8 \text { minutes }=60 \times 8 \mathrm{~s}=480 \mathrm{~s} \\
& \omega_2=\omega_1+\alpha t \\
& \begin{aligned}
& \therefore=\frac{\omega_2-\omega_1}{t} \\ \\
&=\frac{0-40}{600}=-\frac{1}{15} \mathrm{rad} / \mathrm{s}^2
\end{aligned}
\end{aligned}
$
At time $t^{\prime}$,
$
\begin{aligned}
\omega_3 & =\omega_1+\alpha t^{\prime} \\
& =40-\frac{1}{15} \times 480=40-32=8 \mathrm{rad} / \mathrm{s} \\
\therefore L & =I \omega_3=1 \times 8=8 \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}
\end{aligned}
$
This is the required anqular momentum of the wheel.
\begin{aligned}
& \text { Data }: \mid=1 \mathrm{~kg} \cdot \mathrm{m}^2, \omega_1=40 \mathrm{rad} / \mathrm{s}, \omega_2=0 \text { at } \\
& \mathrm{t}=10 \text { minutes }=60 \times 10 \mathrm{~s}=600 \mathrm{~s}, \\
& \mathrm{t}^{\prime}=8 \text { minutes }=60 \times 8 \mathrm{~s}=480 \mathrm{~s} \\
& \omega_2=\omega_1+\alpha t \\
& \begin{aligned}
& \therefore=\frac{\omega_2-\omega_1}{t} \\ \\
&=\frac{0-40}{600}=-\frac{1}{15} \mathrm{rad} / \mathrm{s}^2
\end{aligned}
\end{aligned}
$
At time $t^{\prime}$,
$
\begin{aligned}
\omega_3 & =\omega_1+\alpha t^{\prime} \\
& =40-\frac{1}{15} \times 480=40-32=8 \mathrm{rad} / \mathrm{s} \\
\therefore L & =I \omega_3=1 \times 8=8 \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}
\end{aligned}
$
This is the required anqular momentum of the wheel.
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