Angular momentum of a single particle moving with constant speed along circular path:

$A$ time varying force $F = 2t$ is applied on a spool rolling as shown in figure. The angular momentum of the spool at time $t$ about bottommost point is:

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 solid sphere of mass $500\,g$ and radius $5\,cm$ is rotated about one of its diameter with angular speed of $10\,rad \, s ^{-1}$. If the moment of inertia of the sphere about its tangent is $x \times 10^{-2}$ times its angular momentum about the diameter. Then the value of $x$ will be ..............

The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point $O$ is given by $\mathrm{V}(\mathrm{r})=\mathrm{kr}^2 / 2$, where $\mathrm{k}$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $\mathrm{R}$ about the point $\mathrm{O}$. If $\mathrm{v}$ is the speed of the particle and $\mathrm{L}$ is the magnitude of its angular momentum about $\mathrm{O}$, which of the following statements is (are) true?

$(A)$ $v=\sqrt{\frac{k}{2 m}} R$

$(B)$ $v=\sqrt{\frac{k}{m}} R$

$(C)$ $\mathrm{L}=\sqrt{\mathrm{mk}} \mathrm{R}^2$

$(D)$ $\mathrm{L}=\sqrt{\frac{\mathrm{mk}}{2}} \mathrm{R}^2$

A small mass $m$ is attached to a massless string whose other end is fixed at $P$ as shown in the figure. The mass is undergoing circular motion is the $x-y$ plane with centre at $O$ and constant angular speed $\omega$. If the angular momentum of the system, calculated about $O$ and $P$ are denoted by $\vec{L}_O$ and $\vec{L}_P$ respectively, then