Question

A small ball 'A' of mass 'm' is attached rigidly to the end of a light rod of length 'd'. The structure rotates about an axis perpendicular to the rod and passing through its other end with an angular velocity '$\omega$'. The angular momentum of the system about this axis is.

• A. $\frac{1}{2}m\omega d^2$
• B. $\frac{1}{4}m\omega d^2$
• C. $m\omega d^2$
• D. $2m\omega d^2$

Answer 1

The correct option is C $m\omega d^2$

Angular momentum of A about the axis of rotation,

$\overrightarrow{L}=\overrightarrow{r}X\overrightarrow{p}\{here\overrightarrow{r}\perp \overrightarrow{p}\}$
$\left|\overrightarrow{L}\right|\Rightarrow rpsin{90}^{\circ }=rp$

Now, the magnitude of linear momentum of A, which is d units apart from axis of rotation = mv

By Rotational Kinematics, we know clearly,

$V=\omega r$

$\Rightarrow$ $V=\omega d$.
$\Rightarrow \left|\overrightarrow{L}\right|=m\omega d^2$ $\{\left|\overrightarrow{L}\right|=mvr\}$

Directed along the Axis of Rotation.