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A disc of mass $10 \mathrm{~kg}$ and radius $0 \cdot 1 \mathrm{~m}$ is rotating at 120 r.p.m. A retarding torque brings it to rest in $10 \mathrm{~s}$. If the same torque is due to force applied tangentially on
the rim of the disc then magnitude of force is
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the rim of the disc then magnitude of force is
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Verified Answer
The correct answer is:
$0.2 \pi \mathrm{N}$
frequency $\mathrm{f}=120 \mathrm{rpm}=\frac{120}{60}=2 \mathrm{rps}$
$\omega=2 \pi \mathrm{f}=2 \pi \times 2=4 \pi \mathrm{rad} / \mathrm{s}$
Angular acceleration $=\propto=\frac{\omega_{2}-\omega_{1}}{t}=\frac{0-4 \pi}{10}=\frac{-2 \pi}{5} \mathrm{rad} / \mathrm{s}^{2}$
Moment of inertia $\quad$ I $=\frac{\mathrm{MR}^{2}}{2}=\frac{10 \times(0.1)^{2}}{2}$
$=0.05 \mathrm{~kg} \mathrm{~m}^{2}$
Torque $\tau=\mathrm{I} \alpha=0.05 \times \frac{2}{5} \pi=0.02 \pi \mathrm{Nm}$.
$\tau=\mathrm{Fr}$
$\therefore \mathrm{F}=\frac{\tau}{\mathrm{r}}=\frac{0.02 \pi}{0.1}=0.2 \pi \mathrm{N}$
$\omega=2 \pi \mathrm{f}=2 \pi \times 2=4 \pi \mathrm{rad} / \mathrm{s}$
Angular acceleration $=\propto=\frac{\omega_{2}-\omega_{1}}{t}=\frac{0-4 \pi}{10}=\frac{-2 \pi}{5} \mathrm{rad} / \mathrm{s}^{2}$
Moment of inertia $\quad$ I $=\frac{\mathrm{MR}^{2}}{2}=\frac{10 \times(0.1)^{2}}{2}$
$=0.05 \mathrm{~kg} \mathrm{~m}^{2}$
Torque $\tau=\mathrm{I} \alpha=0.05 \times \frac{2}{5} \pi=0.02 \pi \mathrm{Nm}$.
$\tau=\mathrm{Fr}$
$\therefore \mathrm{F}=\frac{\tau}{\mathrm{r}}=\frac{0.02 \pi}{0.1}=0.2 \pi \mathrm{N}$
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