Question

A particle of mass $m$, is projected on horizontal ground with an initial velocity of $u$, making an angle $\theta$ with horizontal. Find out the angular momentum at any time $t$ of particle $p$, about :
(i)$yaxis$
(ii) $z-axis$

Answer 1

Let the velocity at any time t

$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$

initial velocity

$\overrightarrow{u}=u\cos \theta \hat{i}+u\sin \theta \hat{j}$

Ref. image

At time t

$x=u\cos \theta t$

$y=u\sin \theta t-\frac{1}{2}g{t}^2$

$v_x=u\cos \theta$

$v_y=(4\sin \theta -gt)$

(i) Angular momentum about y-axis

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

$=x\hat{i}\times m(v_x\hat{i}+v_y\hat{j})$

$=mx{v}_y\hat{k}$

$=m\left(u\cos \theta t\right)(u\sin \theta -gt)\hat{k}$

(II) Angular momentum about 2-axis

$l_z=vecr\times m\overrightarrow{v}$

$=(x\hat{i}+y\hat{j})\times m(v_x\hat{i}+v_y\hat{j})$

$=m[x{v}_y\hat{k}+y{v}_x(-\hat{k}\left)\right]$

$=m[x{v}_y-y{v}_x]\hat{k}$

$=m\left[u\cos \theta t\right(u\sin \theta -gt)-(u\sin \theta t-\frac{1}{2}g{t}^2\left)u\cos \theta \right]\hat{k}$

$=mu\cos \theta t[u\sin \theta -gt-u\sin \theta +\frac{1}{2}gt]\hat{k}$

$=-m\frac{u\cos \theta t^{2}g}{2}\hat{k}$