Question

A massless rod $S$ having length $2l$ has point masses attached to its two each as shown in the figure. The rod is rotating about an axis passing through its centre and making an angle $\alpha$ with the axis. The magnitude of the rate of change of momentum of rod i.e. $\left| {\dfrac{{dL}}{{dt}}} \right|$ equals

A) $2m{l^3}{\omega ^2}\sin \alpha .\cos \alpha$
B) $m{l^2}{\omega ^2}\sin 2\alpha$
C) $m{l^2}\sin 2\alpha$
D) ${m^{1/2}}{l^{1/2}}\omega \sin \alpha .\cos \alpha$

Answer 1

Hint : In this solution, we will first determine the radius of the circle traced by the ends of the rod. Then we will use the formula for angular momentum to determine the change of angular momentum.

Formula used: In this solution, we will use the following formula:
Angular momentum of a rotating body: $\vec L = \vec r \times m\vec v$ where $\vec r$ is the radius, $m$ is the mass, and $\vec v$ is the velocity vector.

Complete step by step answer
We’ve been given that a massless rod with two masses attached to its ends rotates about its centre. Let us start with the radius of the circle traced by the masses at the ends of the rod. The radius of the circle will be equal to the perpendicular component of the length of the rod which will be determined as,
$\sin \alpha = \dfrac{r}{l}$
$\Rightarrow r = l\sin \alpha$
Then the angular momentum will be
$\vec L = \vec r \times m\vec v$ . Since the velocity vector and the radius vector will be perpendicular in this case, the angular momentum will simplify to
$\vec L = m\omega {r^2}$ $(\because v = \omega r)$
As $r = l\sin \alpha$ , the angular momentum will be
$L = m\omega {l^2}{\sin ^2}\alpha$
Then the rate of angular momentum with respect to time
$\left| {\dfrac{{dL}}{{dt}}} \right| = m\omega {l^2}(2\sin \alpha )\cos \alpha \dfrac{{d\alpha }}{{dt}}$
Since $(2\sin \alpha )\cos \alpha = \sin 2\alpha$ and $\dfrac{{d\alpha }}{{dt}} = \omega$ , we have
$\left| {\dfrac{{dL}}{{dt}}} \right| = m{l^2}{\omega ^2}\sin 2\alpha$ which corresponds to option (B).

Note
Here since the rod is massless, we can directly treat the point masses at the end of the rods directly without any influence of the rod on the dynamics of the situation. Since we only want to find the magnitude of momentum, we don’t have to worry about the direction of the angular momentum of the rod system.