A disc of mass $10 \mathrm{~kg}$ and radius $0 \cdot 1 \mathrm{~m}$ is rotating at 120 r.p.m. A retarding…

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
  1. $0.2 \pi \mathrm{N}$
  2. $0 \cdot 4 \pi \mathrm{N}$
  3. $0.8 \pi \mathrm{N}$
  4. $0 \cdot 1 \pi \mathrm{N}$

Solution

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 $\tau=\mathrm{I} \alpha=0.05 \times \frac{2}{5} \pi=0.02 \pi \mathrm{Nm}$. $\tau=\mathrm{Fr}$ $\therefore \mathrm{F}=\frac{\tau}{\mathrm{r}}=\frac{0.02 \pi}{0.1}=0.2 \pi \mathrm{N}$

Asked in: MHT CET 2020 (13 Oct Shift 2)

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