Question

A particle is projected at time $t$ = 0 from a point $P$ with a speed $v_0$ at an angle of ${45}^{\circ }$ to the horizontal. Find the magnitude and the direction of the angular momentum of the particle about the point $P$ at time $t$ = $\frac{v_0}{g}$

• A. $\frac{m{v}_0^2}{2\sqrt{2}g}(-\hat{k})$
• B. $\frac{m{v}_0^3}{2\sqrt{2}g}(-\hat{k})$
• C. $\frac{m{v}_0^2}{2\sqrt{2}g}\left(\hat{k}\right)$
• D. $\frac{m{v}_0^3}{2\sqrt{2}g}\left(\hat{k}\right)$

Answer 1

The correct option is B

$\frac{m{v}_0^3}{2\sqrt{2}g}(-\hat{k})$

Let us take the origin at $P$, $X$-axis along the horizontal and $Y$-axis along the vertically upward direction as shown in figure. For horizontal motion during the time $0$ to $t$.

$v_x$ = $v_{0}cos{45}^{\circ }$ = $\frac{v_0}{\sqrt{2}}$

and $x$ = $v_{x}t$ = $\frac{v_0}{\sqrt{2}}$.$\frac{v_0}{g}$ = $\frac{v_0^2}{\sqrt{2}g}$.

For vertical motion,

$v_y$ = $v_{0}sin{45}^{\circ }-gt$ = $\frac{v_0}{\sqrt{2}}-v_0$ = $\frac{(1-\sqrt{2})}{\sqrt{2}}{v}_0$

and $y$ = $\left(v_{0}sin{45}^{\circ }\right)t-\frac{1}{2}g{t}^2$

= $\frac{v_0^2}{\sqrt{2}g}$ - $\frac{v_0^2}{2g}$ = $\frac{v_0^2}{2g}(\sqrt{2}-1)$

The angular momentum of the particle at time $t$ about the origin is

$L$ = $\overrightarrow{r}x\overrightarrow{p}$ = $m\overrightarrow{r}x\overrightarrow{v}$

= $m(\overrightarrow{i}x+\overrightarrow{j}y)x(\overrightarrow{i}{v}_x+\overrightarrow{j}{v}_y)$

= $m(\overrightarrow{k}x{v}_y-\overrightarrow{k}y{v}_x)$

= $m\overrightarrow{k}\left[\left(\frac{v_0^2}{\sqrt{2}g}\right)\frac{v_0}{\sqrt{2}}\right(1-\sqrt{2})-\frac{v_0^2}{2g}(\sqrt{2}-1\left)\frac{v_0}{\sqrt{2}}\right]$

= $-\overrightarrow{k}\frac{m{v}_0^3}{2\sqrt{2}g}$.

Thus, the angular momentum of the particle is $\frac{m{v}_0^3}{2\sqrt{2}g}$ in the negative Z - direction, i.e., perpendicular to the plane of motion, going into the plane.