Question

A uniform stick of length $l$ and mass $m$ lies on a smooth table. It rotates with angular velocity $\omega$ about an axis perpendicular to the table and through one end of the stick. The angular momentum of the stick about the end is

• A. $ml^2\omega$
• B. $\frac{ml^2\omega}{3}$
• C. $\frac{ml^2\omega}{12}$
• D. $\frac{ml^2\omega}{6}$

Answer 1

Answer: (B)

$\frac{ml^2\omega}{3}$

Answer 2

The angular momentum $L$ of a rigid body rotating about a fixed axis is given by the product of its moment of inertia $I$ about that axis and its angular velocity $\omega$, i.e., $L = I\omega$.

For a uniform stick of length $l$ and mass $m$, the moment of inertia about an axis passing through its center of mass (CM) and perpendicular to its length is:

$I_{CM} = \frac{ml^2}{12}$

The problem states that the stick rotates about an axis through one end. We use the parallel axis theorem to find the moment of inertia about this new axis. The distance $d$ from the center of mass to one end is $l/2$.

According to the parallel axis theorem:

$I = I_{CM} + md^2$ $I = \frac{ml^2}{12} + m\left(\frac{l}{2}\right)^2$ $I = \frac{ml^2}{12} + m\frac{l^2}{4}$ $I = \frac{ml^2}{12} + \frac{3ml^2}{12}$ $I = \frac{4ml^2}{12}$ $I = \frac{ml^2}{3}$

Now, substitute the moment of inertia $I$ and the given angular velocity $\omega$ into the angular momentum formula:

$L = I\omega$ $L = \left(\frac{ml^2}{3}\right)\omega$ $L = \frac{ml^2\omega}{3}$