20.3 Law of motion in angular form for a system of particles  (Page 2/4)

$$\begin{array}{l}\frac{d\mathbf{L}}{dt}=\sum {\mathbf{\tau }}_i\end{array}$$

The most critical aspect of this equation is that the sum of torques involves both internal as well as external torques. When we talk of torques on the system of particles, the classification of internal and external depends on the boundary of closed system. This, in turn, depends which of the particles are included and which of the particles are excluded from the system. The forces (constituting torques) applied by particles included in the system constitute internal torques. The remaining ones are external torques. A particle can not have internal torque anyway. We must also understand that when we say a torque is applied on a particle (without any qualification), we implicitly mean that torque is external to the particle.

According to Newton's third law, the internal forces (torques) appear always in the pair of equal and opposite forces (torques). As such, they cancel each other. The expression of angular momentum, then, reduces to :

$$\begin{array}{l}\frac{d\mathbf{L}}{dt}=\sum {\mathbf{\tau }}_i={\mathbf{\tau }}_{\mathrm{net}}\end{array}$$

where " ${\mathbf{\tau }}_{\mathrm{net}}$ " represents net external torue on the system of particles. Finally, the second law of motion in angular form for a system of particles is expressed as :

This relationship is same as that of a single particle. Only difference is that net external torque, now, is equal to the first time derivative of the vector sum of angular momentum (L) of the system of particles as against a single particle. Evidently, this relation holds for measurement of angular momentum and torque about the same reference point.

Newton's second law for a rigid body in rotational motion

If the particles in a system are closely packed and its mass distribution about the axis of rotation is fixed, then the aggregation of particles is known as rigid body. The case of rigid body differs to a system of particles in following aspects :

• The rigid body moves about an axis of rotation instead of a point.
• The angular quantities can be written in scalar forms with appropriate signs, as they have only two directions.

The angular momentum (ℓ) is the product of moment arm ( $r_{\perp }$ ) and linear momentum. The angular momentum of the rigid body (as a system of particles) is given by a scalar sum as given here :

$$\begin{array}{l}L=\sum {\ell }_i=\sum m_i{p}_i=\sum m_i{r}_{i\perp }{v}_i=\sum m_i{r_{i\perp }}^2\omega \end{array}$$

Scalar summation is possible in the case of rotation as angular momentum is always along the axis of rotation - either in the positive or negative direction. We must be careful to remember that $r_{i\perp }$ is moment arm of about the axis of rotation - not about the point as in the general case. Now, each of the particle rotates with same angular velocity. As such, it can be taken out of the summation sign.

$$\begin{array}{l}L=\omega \sum m_i{r_{i\perp }}^2=I\omega \end{array}$$

We note that moment of inertia of the rigid body does not change with time. The only quantity that may change with is angular velocity. Hence, the first time derivative of the total angular momentum is :

$$\begin{array}{l}\frac{dL}{dt}=I\frac{d\omega }{dt}=I\alpha \end{array}$$

However, we have derived the following relation earlier for rotation of a rigid about an axis :