Angular Momentum

Angular momentum

The moment of linear momentum of a particle about an axis is called the angular momentum of the particle. The angular momentum of a particle rotating about a vertical axis is,\begin{align*} \overrightarrow{J}=\overrightarrow{r}\times \overrightarrow{P}=m(\overrightarrow{r}\times\overrightarrow{v})\end{align*} where, \(\overrightarrow{P}\) is the lineaar momentum of the body. It is a vector quantity having having direction perpendicular to the plane containing\(\overrightarrow{r}\) and\(\overrightarrow{v}\) as defined by the right hand thumb rule.

Angular momentum of a system of particles

Consider a system of n number of particles with masses \(m_{1},m_{2},m_{3}...m_{n}\) and respective position vectors\(\overrightarrow{r_{1}},\overrightarrow{r_{2}},\overrightarrow{r_{3}}...\overrightarrow{r_{4}}\) rotating about the vertical axis. The angular momentum of the system is the vector sum of angular momentum of the individual particles ie.\begin{align*} \overrightarrow{J}=\overrightarrow{J_{1}}+\overrightarrow{J_{2}}+\overrightarrow{J_{3}}...+\overrightarrow{J_{n}}\end{align*}\begin{align*} or, \overrightarrow{J}=\overrightarrow{r_{1}}\times m_{1}\overrightarrow{v_{1}}+\overrightarrow{r_{2}}\times m_{2}\overrightarrow{v_{2}}+\overrightarrow{r_{3}}\times m_{3}\overrightarrow{v_{3}}+...\overrightarrow{r_{n}}\times m_{n}\overrightarrow{v_{n}}\end{align*}\begin{align*} or, \overrightarrow{J}=\sum_{k=1}^{n}\overrightarrow{r_{k}}\times m_{k}\overrightarrow{v_{k}}-----1 \end{align*} where, \(\overrightarrow{r_{k}}\) and\(\overrightarrow{v_{k}}\) respectively are the position and velocity of the particles taken in lab frame. Transforamtion equations for\(\overrightarrow{r_{k}}\) and\(\overrightarrow{v_{k}}\) betwwen lab frame and CM frame are,\begin{align*} \overrightarrow{r_{k}}=\overrightarrow{r'_{k}}+\overrightarrow{R}-----2 \end{align*} and\begin{align*} \overrightarrow{v_{k}}=\overrightarrow{v'_{k}}+\overrightarrow{v}-----2 \end{align*} where, \(\overrightarrow{R}\) and\(\overrightarrow{v}\) are the position and velocities of CM taken in lab frame. Using equation 2 in equation 1, we get,\begin{align*} \overrightarrow{J}=\sum_{k=1}^{n}(\overrightarrow{r'_{k}}+\overrightarrow{R})\times m_{k}(\overrightarrow{v'_{k}}+\overrightarrow{v}) \end{align*}\begin{align*} or, \overrightarrow{J}=\sum [\overrightarrow{r'_{k}}\times m_{k}\overrightarrow{v'_{k}}+\overrightarrow{r_{k}}\times m_{k}\overrightarrow{v}+\overrightarrow{R}\times m_{k}\overrightarrow{v'_{k}}+\overrightarrow{R}\times m_{k}\overrightarrow{v}]\end{align*}\begin{align*} or, \overrightarrow{J}=\overrightarrow{R}\times M\overrightarrow{V}+\sum\overrightarrow{r'_{k}}\times m_{k}\overrightarrow{v'_{k}}+\sum\overrightarrow{r'_{k}}m_{k}\times \overrightarrow{v}+\overrightarrow{R}\times \sum m_{k}\overrightarrow{v'_{k}}---3 \end{align*} we have,\begin{align*} \overrightarrow{r'_{k}}=\overrightarrow{r_{k}}-\overrightarrow{R}\end{align*}\begin{align*} or, \sum m_{k}\overrightarrow{r'_{k}}=\sum m_{k}\overrightarrow{r_{k}}-\sum m_{k}\overrightarrow{R}-----4\end{align*} from definition of CM,\begin{align*} \sum m_{k}\overrightarrow{r_{k}}=\sum m_{k}\overrightarrow{R}-----5\end{align*} from equations 4 and 5,\begin{align*} \sum m_{k}\overrightarrow{r'_{k}}=0-----6\end{align*} again,\begin{align*} \sum m_{k}\overrightarrow{v_{k}}=\sum m_{k}\overrightarrow{v}-----6\end{align*} By definition of CM,\begin{align*} \sum m_{k}\overrightarrow{v_{k}}=\sum m_{k}\overrightarrow{v}-----8 \end{align*} then, from equations 7 and 8,\begin{align*} \sum m_{k}\overrightarrow{v'_{k}}=0-----9 \end{align*} with equations 6 and 9, equation 3 becomes,\begin{align*} or, \overrightarrow{J}=\overrightarrow{R}\times M\overrightarrow{V}+\sum\overrightarrow{r'_{k}}\times m_{k}\overrightarrow{v'_{k}} \end{align*}\begin{align*} or, \overrightarrow{J}=\overrightarrow{R}\times \overrightarrow{P}+\overrightarrow{J_{cm}} \end{align*} where, \(\overrightarrow{J_{o}}=\overrightarrow{R}\times \overrightarrow{P}\) is the angular momentum of the CM about the fixed axis, also known as the orbital angular momentum of the rigid body and,\( \overrightarrow{J_{cm}}=\sum\overrightarrow{r'_{k}}\times m_{k}\overrightarrow{v'_{k}} \) is the angular momentum of the system about the CM, also called as the spin angular momentum of the system.

Torque of a particle

The torque acting on a particle is defined as the moment of force about the axis of rotation. mathematically, \begin{align*} \overrightarrow{\tau}=\overrightarrow{r}\times \overrightarrow{F}\end{align*} it is a vector quantity with direction along the perpendicular to the plane containing \(\overrightarrow{r}\) and\(\overrightarrow{F}\) as defined by the right hand thumb rule. It can also be expressed as the rate of change of angular momentum ie. \(\tau=\frac{\mathrm{d\overrightarrow{J}} }{\mathrm{d} t}\).

Torque of a system

Consider a system of n number of particles. Then the torque of the system is defined as the vector sum of the torques acting on individual particles. Mathematically, \(\overrightarrow{\tau}=\overrightarrow{\tau_{1}}+\overrightarrow{\tau_{2}}+\overrightarrow{\tau_{3}}...+\overrightarrow{\tau_{n}}\).

Conservation of Angular Momentum

In the absence of external torque, angular momentum of a body remains conserved. We know,\(\tau=\frac{\mathrm{d\overrightarrow{J}} }{\mathrm{d} t}\). When, \(\tau=0\), then\(\frac{\mathrm{d\overrightarrow{J}} }{\mathrm{d} t}=0\). It implies that the angular momentum is constant and hence conserved.

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.