JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Two small balls with masses m and 2 m are attached to both ends of a rigid rod of length d and negligible mass. If angular momentum of this system is \(L\) about an axis (A) passing through its centre of mass and perpendicular to the rod then angular velocity of the system about A is :
- A \( \frac{3}{2}\frac{L}{md^{2}} \)
- B \( \frac{2L}{md^{2}} \)
- C \( \frac{4}{3}\frac{L}{md^{2}} \)
- D \( \frac{2L}{5md^{2}} \)
Answer & Solution
Correct Answer
(A) \( \frac{3}{2}\frac{L}{md^{2}} \)
Step-by-step Solution
Detailed explanation
\(L = I \omega\) and \(\omega=\frac{ L }{ I }\) \(\omega=\frac{L}{m\left(\frac{2 d}{3}\right)^2+2 m\left(\frac{d}{3}\right)^2}=\frac{L}{\frac{4}{9} m d^2+\frac{2}{9} m d^2}=\frac{L}{\frac{6 m d^2}{9}}\) \(\omega=\frac{3 L}{2 md ^2}\)
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