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Consider a particle of mass m moving with a velocity $\vec{v}$ at a point P.

The angular momentum of this particle about a point O is given by \begin{align} \vec{L}=m\,\vec{\mathrm{OP}}\times\vec{v}=m\,\vec{r}\times\vec{v}, \nonumber \end{align} where $\vec{r}=\vec{\mathrm{OP}}$ is the position vector from O to P.

The angular momentum $\vec{L}$ is defined *about a specific point*. The phrase ``angular momentum about an axis'' means the component of $\vec{L}$ along that axis. This phrase make sense only if component of $\vec{L}$ along the axis is independent of the point about which $\vec{L}$ is defined.

Consider a particle of mass $m$ moving in a circular orbit of radius $r$ with a constant speed $v$. The radial vector $\vec{r}$ and velocity vector $\vec{v}$ lies in the same plane and they are perpendicular to each other. The angular momentum of the particle about the centre of the circle is perpendicular to the plane of motion. The magnitude of angular momentum is given by \begin{align} \vec{L}=mrv=mr^2\omega. \end{align}

Consider plane motion of a rigid body of mass $m$. The moment of inertia of the body about an axis perpendicular to the plane of motion and passing through its centre of mass C is $I_\text{cm}$. Let $\vec{\omega}$ be angular velocity of the body and $\vec{v}_\text{cm}$ be velocity of its centre of mass. The angular momentum of the body about a point O is given by \begin{align} \vec{L}&=\vec{L}_\text{cm}+\vec{L}_\text{about cm} \nonumber\\ &=m\,\vec{r}_\text{cm}\times\vec{v}_\text{cm}+I_\text{cm}\vec{\omega}, \nonumber \end{align} where $\vec{r}_\text{cm}=\vec{\mathrm{OC}}$ is the position vector from O to C.

If a rigid body is rotating about a fixed axis then its 'angular momentum about the axis of rotation' is given by
\begin{align}
L=I_O \omega, \nonumber
\end{align}
where $I_O$ is moment of inertia about the axis of rotation. If the axis of rotation pass through the centre of mass then it is called **spin angular momentum** of the body.

A mass $m$ is moving with a constant velocity along a line parallel to the $x$-axis, away from the origin. Its angular momentum with respect to the origin,

- is zero
- remains constant
- goes on increasing
- goes on decreasing

**Solution:**
The angular momentum of the particle about the origin O is given by,
\begin{align}
\vec{L}&=m\vec{r}\times\vec{v}\nonumber\\
&=m(r\cos\theta\,\hat\imath+r\sin\theta\,\hat\jmath)\times (v\,\hat\imath)\nonumber\\
&=-mvr\sin\theta\,\hat{k}=-mvh\,\hat{k}.\nonumber
\end{align}

A stone of mass $m$, tied to the end of a string, is whirled around in a horizontal circle. (Neglect the force due to gravity). The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by $T=Ar^n$, where $A$ is a constant, $r$ is the instantaneous radius of the circle. Find $n$.

**Solution: **
Consider motion of the particle of mass $m$ at some time instant $t$. The position vector $\vec{r}$ and the velocity vector $\vec{v}$ of the particle are perpendicular to each other (see figure).

The angular momentum of a particle about the centre *O* is given by $\vec{L}=m\,\vec{r}\times\vec{v}=mvr\,\hat{k}$. The tension in the string $T$ provides centripetal acceleration to the particle i.e.,
\begin{align}
T=\frac{mv^2}{r}=\frac{m}{r}\left(\frac{L}{mr}\right)^2=\left(\frac{L^2}{m}\right)r^{-3}.\nonumber
\end{align}
Comparing with $T=Ar^{n}$, we get $n=-3$.

The readers are encouraged to make a small demo to feel the physics. Take a ball pen and remove the refill so that you can pass a thread through the pen cover. Take a thread and tie one of its end to an eraser, pass the other end through the pen cover, and tie a larger mass (key ring etc) to the other end (see figure).

Hold the pen cover and rotate the eraser in horizontal circle. Increase the rotation speed till mass $M$ just starts lifting up. At this instant $T=Mg$. Increase the speed further till mass $M$ touches the pen. Now, pull the mass $M$ downward and see what happens to the speed of mass $m$. Try various ways to feel the relation between $T$, $r$, and $v$.

A disc of mass $M$ and radius $R$ is rolling with angular speed $\omega$ on a horizontal plane (see figure). The magnitude of angular momentum of the disc about the origin O is

- $\frac{1}{2}MR^2\omega$
- $MR^2\omega$
- $\frac{3}{2}MR^2\omega$
- $2MR^2\omega$

**Solution: **
The angular momentum of the disc about the origin O in combined rotation and translation motion is given by
\begin{align}
\vec{L}_\text{O}=\vec{L}_\text{cm}+M\vec{r}_\text{C}\times\vec{v}_\text{C}.\nonumber
\end{align}
The first term $\vec{L}_\text{cm}$ represents the angular momentum of the disc as seen from the centre of mass frame. Thus,
\begin{align}
\vec{L}_\text{cm}=I\vec{\omega}=-\tfrac{1}{2}MR^2\omega\,\hat{k}, \nonumber
\end{align}
where $I=\frac{1}{2}MR^2$ is the moment of inertia of the disc about a perpendicular axis passing through C. The second term $M\vec{r}_\text{C}\times\vec{v}_\text{C}$ equals the angular momentum of the disc if it is assumed to be concentrated at the centre of mass translating with a velocity $\vec{v}_\text{C}$.

- A small mass m is attached to a mass string (IIT JEE 2012)
- A particle of mass 2 kg is on a smooth horizontal table (JEE Mains 2015)
- A particle of mass m is moving along the side of a square (JEE Mains 2016)
- A particle of mass 20 g is released with an initial velocity 5 m/s along the curve (JEE Mains 2019)
- The time dependence of the position vector of a particle (JEE Mains 2019)
- Angular Momentum of a Projectile

- Kinematics of Rigid Bodies
- Moment of Inertia
- Angular Momentum
- Angular Momentum of a Projectile
- Torque
- Conservation of Angular Momentum
- Fixed Axis of Rotation
- Rolling without slipping of rings cylinders and spheres
- Equilibrium of rigid bodies
- Direction of frictional force on bicycle wheels