A particle of mass $1 kg$ is subjected to a force which depends on the position as $\vec{F}=-k(x \hat{i}+y \hat{j}) kgms ^{-2}$ with $k=1 kgs ^{-2}$. At time $t=0$, the particle's position $\vec{r}=\left(\frac{1}{\sqrt{2}} \hat{i}+\sqrt{2} \hat{j}\right) m$ and its velocity $\vec{v}=\left(-\sqrt{2} \hat{i}+\sqrt{2} \hat{j}+\frac{2}{\pi} \hat{k}\right) m s^{-1}$. Let $v_x$ and $v_y$ denote the $x$ and the $y$ components of the particle's velocity, respectively. Ignore gravity. When $z=0.5 m$, the value of $\left(x v_y-y v_x\right)$ is. . . . . $m^2 s^{-1}$

A particle of mass $2\, kg$ is moving such that at time $t$, its position, in meter, is given by $\overrightarrow r \left( t \right) = 5\hat i – 2{t^2}\hat j$ . The angular momentum of the particle at $t\, = 2\, s$ about the origin in $kg\, m^{-2}\, s^{-1}$ is

A spherical shell of $1 \,kg$ mass and radius $R$ is rolling with angular speed $\omega$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin $O$ is $\frac{a}{3} R^{2} \omega$. The value of a will be …………..

A particle of mass $m = 5$ is moving with a uniform speed $v = 3\sqrt 2$ in the $XOY$ plane along the line $Y = X + 4$ . The magnitude of the angular momentum of the particle about the origin is …….