TS EAMCET · Physics · Rotational Motion
A fly-wheel of mass \(25 \mathrm{~kg}\) has a radius of \(0.2 \mathrm{~m}\). It is making \(240 \mathrm{rpm}\). What is the torque necessary to bring to rest in \(20 \mathrm{~s}\) ?
- A \(2 \pi \mathrm{Nm}\)
- B \(0.4 \pi \mathrm{Nm}\)
- C \(\frac{2}{\pi} \mathrm{Nm}\)
- D \(4 \pi \mathrm{Nm}\)
Answer & Solution
Correct Answer
(B) \(0.4 \pi \mathrm{Nm}\)
Step-by-step Solution
Detailed explanation
\(\alpha=\frac{2 \pi n}{t}=\frac{2 \pi \times \frac{240}{60}}{20}\) \(=\frac{2 \pi \times 4}{20}\) \(\alpha=\frac{2 \pi}{5}\) Torque, \(\quad \tau=I \alpha\) \(\tau=M R^2 \alpha\) \(=25 \times(0.04) \times \frac{2 \pi}{5}\) \(=0.4 \pi \mathrm{Nm}\)
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