MHT CET · Physics · Motion In One Dimension
A particle moves along a circle of radius \(r\) with constant tangential acceleration. If the velocity of the particle is \(v\) at the end of second revolution after the revolution has started then the tangential acceleration is
- A \(\frac{v^2}{8 \pi r}\)
- B \(\frac{v^2}{6 \pi r}\)
- C \(\frac{v^2}{4 \pi r}\)
- D \(\frac{v^2}{10 \pi r}\)
Answer & Solution
Correct Answer
(A) \(\frac{v^2}{8 \pi r}\)
Step-by-step Solution
Detailed explanation
Using \(v^2-u^2=2 a s\) and \(u=u_1\) and \(s=4 \pi r\)
\(\therefore 2 a s=v^2 \Rightarrow \frac{v^2}{2 s}=a=\frac{v^2}{2 \times 4 \pi r}=\frac{v^2}{8 \pi r}\)
\(\therefore 2 a s=v^2 \Rightarrow \frac{v^2}{2 s}=a=\frac{v^2}{2 \times 4 \pi r}=\frac{v^2}{8 \pi r}\)
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