Physics - Rotation Question with Solution | TestHub

Using 1st-equation of motion$v=u+at$along horizontal direction,



${\text{v}}_x=\left({\text{v}}_0/\sqrt{2}\right)+0\times \text{t}=\text{0.7v}\left[\text{as}a=0\right]$

and along vertical direction

${\text{v}}_{\text{y}}=\left({\text{v}}_0/\sqrt{2}\right)-\text{g}\left({\text{v}}_0/\text{g}\right)=-{\text{0.3v}}_0$

So, the velocity of particle at time t = (v₀/g) in vector form will be

$\overrightarrow{\text{v}}=\overrightarrow{\text{i}}\left({\text{0.7v}}_0\right)-\overrightarrow{\text{j}}\left({\text{0.3v}}_0\right)\text{....(i)}$

$\mathrm{Now\; from\; 2nd\; equation\; of\; motion }\text{s}=\text{ut}+\left(\frac{1}{2}\right){\text{at}}^2$along horizontal direction,

$\text{x}=\frac{{\text{v}}_0}{\sqrt{2}}\times \frac{{\text{v}}_0}{\text{g}}=\text{0.7}\frac{{\text{v}}_0^2}{\text{g}}$

and along vertical direction,

$\text{y}=\frac{{\text{v}}_0}{\sqrt{2}}\times \frac{{\text{v}}_0}{\text{g}}-\frac{1}{2}\text{g}{\left[\frac{{\text{v}}_0}{\text{g}}\right]}^2=\text{0.2}\frac{{\text{v}}_0^2}{\text{g}}$

So the position vector$\overrightarrow{\text{r}}$at the t ( = v₀/g) will be

$\overrightarrow{\text{r}}=\overrightarrow{\text{i}}\left[\text{0.7}\frac{{\text{v}}_0^2}{\text{g}}\right]+\overrightarrow{\text{j}}\left[\text{0.2}\frac{{\text{v}}_0^2}{\text{g}}\right]\text{....(ii)}$

$\text{Now as by definition}\overrightarrow{\text{L}}=\left(\overrightarrow{\text{r}}\times \overrightarrow{\text{p}}\right)=\text{m}\left(\overrightarrow{\text{r}}\times \overrightarrow{\text{v}}\right)$

$\text{So}\overrightarrow{\text{L}}=\text{m}\left|\begin{array}{ccc}\overrightarrow{\text{i}}& \overrightarrow{\text{j}}& \overrightarrow{\text{k}}\\ \frac{{\text{0.7v}}_0^2}{\text{g}}& \frac{{\text{0.2v}}_0^2}{\text{g}}& 0\\ {\text{0.7v}}_0& -{\text{0.3v}}_0& 0\end{array}\right|$

$=\text{m}\overrightarrow{\text{k}}\left[-\text{0.21}\frac{{\text{v}}_0^3}{\text{g}}-\text{0.14}\frac{{\text{v}}_0^3}{\text{g}}\right]$

$\text{i.e.,}\overrightarrow{\text{L}}=-\left({\text{0.35mv}}_0^3/\text{g}\right)\overrightarrow{\text{k}}$

i.e., angular momentum has magnitude$\text{0.35}{\text{mv}}_0^3/\text{g}$and is directed along negative z-axis.