A torque of $100 \mathrm{~N} . \mathrm{m}$ is applied to a body capable of rotating about a given axis. If the body starts from rest and acquires kinetic energy of $10000 \mathrm{~J}$ in 10 seconds, find
(i) its moment of inertia about the given axis
(ii) its angular momentum at the end of 10 seconds.
(i) its moment of inertia about the given axis
(ii) its angular momentum at the end of 10 seconds.
Q 114.7
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Data : $\tau=100 \mathrm{~N} \cdot \mathrm{m}, \omega_i=0, \mathrm{E}_{\mathrm{i}}=0, \mathrm{E}_{\mathrm{f}}=10^4 \mathrm{~J}, \mathrm{t}=10 \mathrm{~s}$
$
\tau=\frac{\Delta L}{\Delta t}=\frac{L_{\mathrm{f}}-L_{\mathrm{i}}}{\Delta t}
$
Since the body starts from rest, its initial angular momentum, $L_i=0$.
The final angular momentum,
$
L_f=\tau \Delta t=(100)(10)=10^3 \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}
$
The final rotational kinetic energy, $E_f=\frac{1}{2} L_{\mathrm{f}} \omega_{\mathrm{f}}$
$
\begin{aligned}
\therefore \omega_{\mathrm{f}} & =\frac{2 E_{\mathrm{f}}}{L_{\mathrm{f}}} \\
& =\frac{2 \times 10^4}{10^3}=20 \mathrm{rad} / \mathrm{s}
\end{aligned}
$
The moment of inertia of the body,
$
\begin{aligned}
I & =\frac{L_f}{\omega_f} \\
& =\frac{10^3}{20}=50 \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}
$
$
\tau=\frac{\Delta L}{\Delta t}=\frac{L_{\mathrm{f}}-L_{\mathrm{i}}}{\Delta t}
$
Since the body starts from rest, its initial angular momentum, $L_i=0$.
The final angular momentum,
$
L_f=\tau \Delta t=(100)(10)=10^3 \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}
$
The final rotational kinetic energy, $E_f=\frac{1}{2} L_{\mathrm{f}} \omega_{\mathrm{f}}$
$
\begin{aligned}
\therefore \omega_{\mathrm{f}} & =\frac{2 E_{\mathrm{f}}}{L_{\mathrm{f}}} \\
& =\frac{2 \times 10^4}{10^3}=20 \mathrm{rad} / \mathrm{s}
\end{aligned}
$
The moment of inertia of the body,
$
\begin{aligned}
I & =\frac{L_f}{\omega_f} \\
& =\frac{10^3}{20}=50 \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}
$
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