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

To find the angular velocity of the system about an axis passing through its center of mass and perpendicular to the rod, we must first understand the given system: two small balls with masses \(m\) and \(2m\) are attached to the ends of a rigid rod of length \(d\). The axis of rotation passes through the center of mass of the system, and we are provided with the angular momentum \(L\) for this system.

First, let's identify the center of mass of this system:

- The center of mass \(x_{\text{cm}}\) of the rod is given by: \(x_{\text{cm}} = \frac{m \cdot 0 + 2m \cdot d}{m + 2m} = \frac{2md}{3m} = \frac{2d}{3}\)

Now, let's determine the moments of inertia for both masses about the center of mass.

• For mass \(m\) (located at \(\frac{2d}{3}\) from the center of mass):
• For mass \(2m\) (located at \(\frac{d}{3}\) from the center of mass):

The total moment of inertia about the center of mass is:

\(I_{\text{total}} = I_1 + I_2 = \frac{4md^2}{9} + \frac{2md^2}{9} = \frac{6md^2}{9} = \frac{2md^2}{3}\)

Given the angular momentum \(L\) and using the relation:

\(L = I_{\text{total}} \cdot \omega\)

We can solve for the angular velocity \(\omega\):

\(\omega = \frac{L}{I_{\text{total}}} = \frac{L}{\frac{2md^2}{3}} = \frac{3L}{2md^2}\)

Therefore, the angular velocity of the system about the axis through its center of mass is:
The correct answer is: \((3/2) \frac{L}{md^2}\), which corresponds to the option given as the correct answer.