Question

A ball having mass $m$ is fastened at the end of a flagpole that is connected to the side of a tall building at point $P$ as shown in figure. The length of the flagpole is $l$, and it makes an angle $\theta$ with the $x$ axis. The ball becomes loose and starts to fall with acceleration $-g\hat{j}$. The angular momentum of the ball about point $P$ as a function of time is

• A. $mglt\cos \theta \hat{k}$
• B. $-mglt\cos \theta \hat{k}$
• C. $mglt\sin \theta \hat{k}$
• D. $-mglt\sin \theta \hat{k}$

Answer 1

The correct option is B $-mglt\cos \theta \hat{k}$

Position vector of ball from $P$,
$\overrightarrow{r}={\overrightarrow{r}}_1+{\overrightarrow{v}}_{i}t+\frac{1}{2}\overrightarrow{a}{t}^2$

$\overrightarrow{r}=(\cos \theta \hat{i}+\sin \theta \hat{j})+0-\left(\frac{1}{2}g{t}^2\right)\hat{j}$

Velocity of ball, $\overrightarrow{v}={\overrightarrow{v}}_i+\overrightarrow{a}t=0-gt\hat{j}$
Angular momentum,

$\overrightarrow{L}=\overrightarrow{r}\times m\overrightarrow{v}$

$m\left[\right(lcos\theta \hat{i}+l\sin \theta \hat{j})+0-\left(\frac{1}{2}g{t}^2\right)\hat{j}]\times (-gt\hat{j})$

$\overrightarrow{L}=-mglt\cos \theta \hat{k}$
Final answer (b)