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NEET · Physics · STD 11 - 3.2 motion in plane
A particle starting from rest, moves in a circle of radius \(r\). It attains a velocity of \(\mathrm{V}_{0} \;\mathrm{m} / \mathrm{s}\) in the \(\mathrm{n}^{\text {th }}\) round. Its angular acceleration will be
- A \(\frac{\mathrm{V}_{0}}{\mathrm{n}}\; \mathrm{rad} / \mathrm{s}^{2}\)
- B \(\frac{\mathrm{V}_{0}^{2}}{2 \pi \mathrm{nr}^{2}} \; \mathrm{rad} / \mathrm{s}^{2}\)
- C \(\frac{\mathrm{V}_{0}^{2}}{4 \pi \mathrm{nr}^{2}}\; \mathrm{rad} / \mathrm{s}^{2}\)
- D \(\frac{V_{0}^{2}}{4 \pi n r}\; \mathrm{rad} / \mathrm{s}^{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{V}_{0}^{2}}{4 \pi \mathrm{nr}^{2}}\; \mathrm{rad} / \mathrm{s}^{2}\)
Step-by-step Solution
Detailed explanation
\(\theta=(2 \pi n), \omega_{0}=0, \omega=V_{0} / r\) \(\alpha=\frac{\omega^{2}-\omega_{0}^{2}}{2 \theta}=\frac{\left(\mathrm{V}_{0} / \mathrm{r}\right)^{2}-0}{2(2 \pi \mathrm{n})}\)\(=\frac{\mathrm{V}_{0}^{2}}{4 \pi \mathrm{nr}^{2}}\)
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