Obtain the relation between angular momentum of a particle and torque acting on it.
By definition of angular momentum of a particle, $\vec{l}=\vec{r} \times \vec{p}$
Differentiating this equation w.r.t. time $\mathrm{t}, \frac{d \vec{l}}{d t}=\vec{r} \times \frac{d \vec{p}}{d t}+\frac{d \vec{r}}{d t} \times \vec{p}$
But, $\frac{d \vec{p}}{d t}=$ rate of change of linear momentum $=$ force $\overrightarrow{\mathrm{F}}$ and $\frac{d \vec{r}}{d t}=\vec{v}$
$\therefore \quad \frac{d \vec{l}}{d t}=\vec{r} \times \overrightarrow{\mathrm{F}}+\vec{v} \times \vec{p}$
But as $\vec{v}$ and $\vec{p}$ are in the same direction,
$\vec{v} \times \vec{p}=0$
$\therefore \quad \frac{d \vec{l}}{d t}=\vec{r} \times \overrightarrow{\mathrm{F}}=\vec{\tau}$
Thus, "the time rate of change of angular momentum is equal to the torque".
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
A solid sphere rolls without slipping on a rough surface and the centre of mass has a constant speed $v_0$. If the mass of the sphere is $m$ and its radius is $R$, then find the angular momentum of the sphere about the point of contact
$A$ particle of mass $m$ is rotating in a plane is $a$ circular path of radius $r$, its angular momentum is $L$. The centripital force acting on the particle is :
A disc of mass $M$ and radius $R$ moves in the $x-y$ plane as shown in the figure. The angular momentum of the disc at the instant shown is
In the List-$I$ below, four different paths of a particle are given as functions of time. In these functions, $\alpha$ and $\beta$ are positive constants of appropriate dimensions and $\alpha \neq \beta$. In each case, the force acting on the particle is either zero or conservative. In List-II, five physical quantities of the particle are mentioned: $\overrightarrow{ p }$ is the linear momentum, $\bar{L}$ is the angular momentum about the origin, $K$ is the kinetic energy, $U$ is the potential energy and $E$ is the total energy. Match each path in List-$I$ with those quantities in List-$II$, which are conserved for that path.
List-$I$ | List-$II$ |
$P$ $\dot{r}(t)=\alpha t \hat{t}+\beta t \hat{j}$ | $1$ $\overrightarrow{ p }$ |
$Q$ $\dot{r}(t)=\alpha \cos \omega t \hat{i}+\beta \sin \omega t \hat{j}$ | $2$ $\overrightarrow{ L }$ |
$R$ $\dot{r}(t)=\alpha(\cos \omega t \hat{i}+\sin \omega t \hat{j})$ | $3$ $K$ |
$S$ $\dot{r}(t)=\alpha t \hat{i}+\frac{\beta}{2} t^2 \hat{j}$ | $4$ $U$ |
$5$ $E$ |