MHT CET · Physics · Motion In Two Dimensions
A particle of mass 'm' is performing U.C.M. along a circle of radius 'r'. The relation between centripetal acceleration 'a' and kinetic energy 'E' is given by
- A \(a=\frac{2 \mathrm{E}}{m r}\)
- B \(a=\left(\frac{2 E}{m r}\right)^{2}\)
- C \(a=\frac{E}{m r}\)
- D \(a=2 \mathrm{Em}\)
Answer & Solution
Correct Answer
(A) \(a=\frac{2 \mathrm{E}}{m r}\)
Step-by-step Solution
Detailed explanation
(C)
\(E=\frac{1}{2} m \omega^{2} r^{2}\)
\(\therefore 2 E=m \omega^{2} r^{2}\)
\(\therefore r \omega^{2}=\frac{2 E}{m r}\)
\(a=\frac{v^{2}}{r}=r \omega^{2}\)
\(a=\frac{2 E}{m r}\)
\(E=\frac{1}{2} m \omega^{2} r^{2}\)
\(\therefore 2 E=m \omega^{2} r^{2}\)
\(\therefore r \omega^{2}=\frac{2 E}{m r}\)
\(a=\frac{v^{2}}{r}=r \omega^{2}\)
\(a=\frac{2 E}{m r}\)
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