# Angular Momentum of a Particle and a Rigid Body

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.

## Angular Momentum of a Rigid Body

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.

## Solved Problems from IIT JEE

### Problem from IIT JEE 1997

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,

1. is zero
2. remains constant
3. goes on increasing
4. 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}

Since $v$ and $h$ are constant, $\vec{L}$ remains constant.

### Problem from IIT JEE 1993

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 $v$.

### Problem from IIT JEE 1999

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

1. $\frac{1}{2}MR^2\omega$
2. $MR^2\omega$
3. $\frac{3}{2}MR^2\omega$
4. $2MR^2\omega$

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 $\vec{v}_\text{C}$.

In pure rolling, velocity of the contact point P is zero. Thus, velocity of the centre of mass and the term $M\vec{r}_\text{C}\times\vec{v}_\text{C}$ are \begin{align} \vec{v}_\text{C} &=\vec{v}_\text{P}+\vec{\omega} \times \vec{r}_\text{CP} \\ &= \omega R\,\hat\imath, \nonumber \end{align} \begin{align} M\vec{r}_\text{C}\times\vec{v}_\text{C}&=M(x_0\,\hat\imath+R\,\hat\jmath)\times (\omega R\,\hat\imath) \\ &=-M\omega R^2\,\hat{k}. \nonumber \end{align} Hence, angular momentum of the disc about the origin O is \begin{align} \vec{L}_\text{O} & =\vec{L}_\text{cm} + M\vec{r}_\text{C}\times\vec{v}_\text{C} \\ &= -\tfrac{3}{2}M\omega R^2\,\hat{k}.\nonumber \end{align}