Question

A ball of mass $m$ is thrown at an angle $\alpha$ to the horizontal with the initial velocity $v_0$. Find the time dependence of the magnitude of the ball's angular momentum vector relative to the point from which the ball is thrown. Find the angular momentum $M$ (in $kg{m}^2/s$) at the highest point of the trajectory if $m=130g$, $\alpha ={45}^{\circ }$, and $v_0=25m/s$. The air drag is to be neglected.

Answer 1

$\overrightarrow{M}\left(t\right)=\overrightarrow{r}\times \overrightarrow{p}=(\overrightarrow{v_0}t+\frac{1}{2}\overrightarrow{g}{t}^2)\times m(\overrightarrow{v_0}+\overrightarrow{g}t)$
$=m{v}_{0}g{t}^2\sin (\frac{\pi }{2}+\alpha )(-\overrightarrow{k})+\frac{1}{2}m{v}_{0}g{t}^2\sin (\frac{\pi }{2}+\alpha )\left(\overrightarrow{k}\right)$
$=\frac{1}{2}m{v}_{0}g{t}^2\cos \alpha (-\overrightarrow{k})$
Thus $M\left(t\right)=\frac{m{v}_{0}g{t}^2\cos \alpha }{2}$
Thus angular momentum at maximum height
i.e., at $t=\frac{\tau }{2}=\frac{v_0\sin \alpha }{g}$,
$M\left(\frac{\tau }{2}\right)=\left(\frac{m{v}_0^3}{2g}\right){\sin }^2\alpha \cos \alpha =37kg{m}^2/s$