A torque of $20 \mathrm{~N}$.m sets a stationary circular disc into rotation about a transverse axis through its centre and acts for $2 \pi$ seconds. If the disc has a mass $10 \mathrm{~kg}$ and radius $0.2 \mathrm{~m}$, what is its frequency of rotation after $2 \pi$ seconds ?
Q 105.6
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Data : $\tau=20 \mathrm{~N} . \mathrm{m}, \mathrm{t}=2 \pi \mathrm{s}, \mathrm{M}=10 \mathrm{~kg}, \mathrm{R}=0.1 \mathrm{~m}$ Let $\mathrm{f}_1$ and $\mathrm{f}_2$ be the initial and final frequencies of rotation of the discr and $\omega_1$ and $\omega_2$ be its initial and final angular speeds.
Since the disc was initially stationary, $f_1=\omega_1,=0$ and $\omega_2=2 \pi f_2$.
The $\mathrm{Ml}$ of the disc about the given axis is
$
I=\frac{M R^2}{2}=\frac{10 \times(0.2)^2}{2}=0.2 \mathrm{~kg} \cdot \mathrm{m}^2
$
Torque, $\tau=I \alpha$
Angular acceleration, $\alpha=\frac{\tau}{I}=\frac{20}{0.2}=100 \mathrm{rad} / \mathrm{s}^2$
Now, $\omega_2=\omega_1+\alpha \mathrm{t}=0+\alpha \mathrm{t}$
$
\begin{aligned}
& \therefore 2 \pi f_2=\alpha t \\
& \therefore \mathrm{f}_2=\frac{\alpha t}{2 \pi}=\frac{100(2 \pi)}{2 \pi}=100 \mathrm{~Hz}
\end{aligned}
$
Since the disc was initially stationary, $f_1=\omega_1,=0$ and $\omega_2=2 \pi f_2$.
The $\mathrm{Ml}$ of the disc about the given axis is
$
I=\frac{M R^2}{2}=\frac{10 \times(0.2)^2}{2}=0.2 \mathrm{~kg} \cdot \mathrm{m}^2
$
Torque, $\tau=I \alpha$
Angular acceleration, $\alpha=\frac{\tau}{I}=\frac{20}{0.2}=100 \mathrm{rad} / \mathrm{s}^2$
Now, $\omega_2=\omega_1+\alpha \mathrm{t}=0+\alpha \mathrm{t}$
$
\begin{aligned}
& \therefore 2 \pi f_2=\alpha t \\
& \therefore \mathrm{f}_2=\frac{\alpha t}{2 \pi}=\frac{100(2 \pi)}{2 \pi}=100 \mathrm{~Hz}
\end{aligned}
$
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