Question

Two small balls with masses m and 2m 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. 2L/(5md²)
• B. (4/3) L/(md²)
• C. (3/2) L/(md²)
• D. 2L/(md²)

Hint:

The reduced mass formula $\mu = \frac{m_1 m_2}{m_1 + m_2}$ simplifies the calculation of M.I. for two-body systems to $I = \mu d^2$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Angular momentum \(L\) is defined as \(L = I\omega\). To find \(\omega\), we must first determine the moment of inertia \(I\) about the center of mass (CM).
Step 2: Key Formula or Approach:
1. CM distance \(r_1 = \frac{m_2 d}{m_1+m_2}\).
2. \(I = m_1 r_1^2 + m_2 r_2^2 = \mu d^2\), where \(\mu\) is reduced mass.
Step 3: Detailed Explanation:
The position of the CM from mass \(m\): \[ r_1 = \frac{2m \cdot d}{m+2m} = \frac{2d}{3} \] The position of the CM from mass \(2m\): \[ r_2 = d - \frac{2d}{3} = \frac{d}{3} \] Moment of Inertia \(I\): \[ I = m\left(\frac{2d}{3}\right)^2 + 2m\left(\frac{d}{3}\right)^2 = \frac{4md^2}{9} + \frac{2md^2}{9} = \frac{6md^2}{9} = \frac{2}{3}md^2 \] Since \(L = I\omega\): \[ \omega = \frac{L}{I} = \frac{L}{\frac{2}{3}md^2} = \frac{3L}{2md^2} \]
Step 4: Final Answer:
The angular velocity is (3/2) L/(md²).