MHT CET · Physics · Rotational Motion
Two identical particles each of mass ' \(\mathrm{m}\) ' are separated by a distance ' \(d\) '. The axis of rotation passes through the midpoint of ' \(\mathrm{d}\) ' and is perpendicular to the length \(\mathrm{d}\). If ' \(\mathrm{K}\) ' is the average rotational kinetic energy of the system, then the angular frequency is
- A \(2 \mathrm{~d} \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}\)
- B \(\frac{\mathrm{d}}{2} \sqrt{\frac{\mathrm{K}}{\mathrm{m}}}\)
- C \(\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{K}}{\mathrm{m}}}\)
- D \(\frac{\mathrm{d}}{4} \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}\)
Answer & Solution
Correct Answer
(C) \(\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{K}}{\mathrm{m}}}\)
Step-by-step Solution
Detailed explanation
Moment of inertia \(\mathrm{I}=2 \mathrm{~m}\left(\frac{\mathrm{d}}{2}\right)^2=\frac{\mathrm{md}^2}{2}\)
Kinetic energy \(K=\frac{1}{2} I \omega^2\)
\(\therefore \omega^2=\frac{2 \mathrm{~K}}{\mathrm{I}}=2 \mathrm{~K} \cdot \frac{2}{\mathrm{md}^2}=\frac{4 \mathrm{~K}}{\mathrm{md}^2}\)
\(\therefore \omega=\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\)
Kinetic energy \(K=\frac{1}{2} I \omega^2\)
\(\therefore \omega^2=\frac{2 \mathrm{~K}}{\mathrm{I}}=2 \mathrm{~K} \cdot \frac{2}{\mathrm{md}^2}=\frac{4 \mathrm{~K}}{\mathrm{md}^2}\)
\(\therefore \omega=\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\)
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