MHT CET · Physics · Rotational Motion
A particle of mass ' \(m\) ' is performing uniform circular motion along a circular path of radius ' \(r\) '. Its angular momentum about the axis passing through the centre and perpendicular to the plane is ' \(L\) '. The kinetic energy of the particle is
- A \(\frac{\mathrm{L}^2}{2 \mathrm{mr}^2}\)
- B \(\frac{2 \mathrm{~L}^2}{\mathrm{mr}^2}\)
- C \(\frac{\mathrm{L}^2}{\mathrm{mr}^2}\)
- D \(\frac{2 \mathrm{~L}^2}{3 \mathrm{mr}^2}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{L}^2}{2 \mathrm{mr}^2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & E=\frac{1}{2} I \omega^2 \\ & L=I \omega \Rightarrow L^2=I^2 \omega^2 \\ \therefore \quad & E=\frac{1}{2} \cdot \frac{L^2}{I} \\ & B u t I=M r^2 \\ \therefore \quad & E=\frac{1}{2} \frac{L^2}{M r^2}=\frac{L^2}{2 \mathrm{Mr}^2}\end{aligned}\)
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