MHT CET · Physics · Rotational Motion
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
- A \(0.2 \pi \mathrm{N}\)
- B \(0 \cdot 4 \pi \mathrm{N}\)
- C \(0.8 \pi \mathrm{N}\)
- D \(0 \cdot 1 \pi \mathrm{N}\)
Answer & Solution
Correct Answer
(A) \(0.2 \pi \mathrm{N}\)
Step-by-step Solution
Detailed explanation
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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