MHT CET · Physics · Rotational Motion
A body of mass' \(\mathrm{m}\) ' and radius of gyration ' \(\mathrm{K}\) ' has an angular momentum ' \(L\) '. Then its angular velocity is
- A \(\frac{\mathrm{L}}{\mathrm{mK}^2}\)
- B \(\frac{\mathrm{mK}^2}{\mathrm{~L}}\)
- C \(\frac{\mathrm{K}^2}{\mathrm{~mL}}\)
- D \(\mathrm{mK}^2 \mathrm{~L}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{L}}{\mathrm{mK}^2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{I}=\mathrm{mK}^2 \text { and } \mathrm{L}=\mathrm{I} \omega=\mathrm{mK}^2 \omega \\ & \therefore \omega=\frac{\mathrm{L}}{\mathrm{mK}^2}\end{aligned}\)
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