# 3.6: Angular Momentum and Torque

- Page ID
- 655

The angular momentum of body, \(dm\), is defined as \[L = r \times U dm\] The angular momentum of the entire system is calculated by integration (summation) of all the particles in the system as \[L_{s} = \int_{m} r \times U dm\] The change with time of angular momentum is called torque, in analogous to the momentum change of time which is the force. \[T_{\tau} = \frac{DL}{Dt} = \frac{D}{Dt}\left(r \times U dm\right) \] where \(T_{\tau}\) is the torque. The torque of entire system is \[T_{\tau s} = \int_{m} \frac{DL}{Dt} = \frac{D}{Dt} \int_{m} \left(r \times U dm \right) \] It can be noticed (well, it can be proved utilizing vector mechanics) that \[T_{\tau} = \frac{D}{Dt}\left(r \times U \right) = \frac{D}{Dt}\left(r \times \frac{Dr}{Dt} \right) = \frac{D^{2}r}{Dt^{2}}\] To understand these equations a bit better, consider a particle moving in x–y plane. A force is acting on the particle in the same plane (x–y) plane. The velocity can be written as \(U = u \hat{i} + v\hat{j}\) and the location from the origin can be written as \(r = x \hat{i} + y \hat{j}\). The force can be written, in the same fashion, as \(F = F_{x} \hat{i} + F_{y} \hat{j}\). Utilizing equation 61 provides \[matrix\] Utilizing equation 63 to calculate the torque as \[matrix\] Since the torque is a derivative with respect to the time of the angular momentum it is also can be written as \[xF_{x} - yF_{y} = \frac{D}{Dt}\left[\left(xv - yu\right)dm\right]\] The torque is a vector and the various components can be represented as \[T_{\tau x} = \hat{i} \cdot \frac{D}{Dt} \int_{m} r \times U dm \] In the same way the component in \(y\) and \(z\) can be obtained.

## Contributors and Attributions

Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.