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.
• B.
• C.
• D.

Answer 1

The correct option is B

Let us take the origin at P, X-axis along the horizontal and Y-axis along the vertically upwards direction as shown in figure. For the 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,

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

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

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

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

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

=$m(\overrightarrow{ix}+\overrightarrow{jy})\times (\overrightarrow{i{v}_x}+\overrightarrow{j{v}_y})$

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

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

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

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.