Question

Masses $m$ and $2m$ are connected by a massless rod of length $d$. If angular momentum about an axis passing through centre of mass and perpendicular to the rod is $L$, then the angular speed $(\omega)$ of the system is:

• A. $\dfrac{3L}{2md^2}$
• B. $\dfrac{L}{2md^2}$
• C. $\dfrac{5L}{3md^2}$
• D. $\dfrac{3L}{md^2}$

Hint:

Always locate the centre of mass first in two-particle rotation problems. Then calculate moment of inertia using $I=\sum mr^2$ before applying $L=I\omega$.

Answer 1

The Correct Option is A

To determine the angular speed \((\omega)\) of the system, we'll follow these steps:

Determine the position of the center of mass (COM).

For two masses \(m\) and \(2m\) connected by a rod of length \(d\), the position of the COM from mass \(m\) is given by:

\(x_{\text{COM}} = \frac{m \times 0 + 2m \times d}{m + 2m} = \frac{2md}{3m} = \frac{2d}{3}\)

Calculate the moments of inertia about the COM.

The moment of inertia \((I)\) of the system about an axis passing through the COM and perpendicular to the rod is:

\(I = m \left(\frac{2d}{3}\right)^2 + 2m \left(\frac{d}{3}\right)^2\)

Simplifying this,

\(I = m \frac{4d^2}{9} + 2m \frac{d^2}{9} = \frac{4md^2}{9} + \frac{2md^2}{9} = \frac{6md^2}{9} = \frac{2md^2}{3}\)

Use the relation between angular momentum and angular speed.

The angular momentum \((L)\) is related to the moment of inertia and angular speed by the formula:

\(L = I \omega\)

Substituting the expression for \(I\), we have:

\(L = \left(\frac{2md^2}{3}\right) \omega\)

Solving for \(\omega\) gives:

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

Therefore, the angular speed of the system is \(\frac{3L}{2md^2}\), which matches the given correct answer.