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

Concept: For a rotating rigid system: \[ L = I\omega \] where $I$ is the moment of inertia about the given axis. The centre of mass must be located first to calculate distances of individual masses from the axis.
Step 1: Let the distance of mass $m$ from the centre of mass be $x$. Then distance of mass $2m$ from the centre of mass is $d-x$. Using centre of mass condition: \[ m x = 2m(d-x) \] \[ x = \frac{2d}{3} \] Thus, \[ r_1 = \frac{2d}{3}, \quad r_2 = \frac{d}{3} \]
Step 2: Moment of inertia about centre of mass: \[ I = m r_1^2 + 2m r_2^2 \] \[ I = m\left(\frac{2d}{3}\right)^2 + 2m\left(\frac{d}{3}\right)^2 \] \[ I = \frac{4md^2}{9} + \frac{2md^2}{9} = \frac{6md^2}{9} = \frac{2md^2}{3} \]
Step 3: Using angular momentum relation: \[ L = I\omega \Rightarrow \omega = \frac{L}{I} \] \[ \omega = \frac{L}{\frac{2md^2}{3}} = \frac{3L}{2md^2} \] Step 4: Hence, \[ \boxed{\omega = \frac{3L}{2md^2}} \]