20.2 Second law of motion in angular form  (Page 3/2)

The heading of this module (law of motion in angular form - not law of motion in rotational form) points to the subtle difference about angular quantities as applicable to general motion vis-a-vis rotation. The title also indicates that law of motion in angular form is just yet another form of Newton's law - not specific to any particular type of motion. What it means that we can use angular concepts and angular law (including the one being developed here) to even pure translational motion along a straight line.

In earlier module titled Rotation of rigid body , we wrote Newton's second law for rotation as :

This relation is evidently valid for rotation, which involves circular motion of particles constituting the rigid body or that of a particle in a plane perpendicular to a fixed axis of rotation. This is so because, moment of inertia is defined with rotational motion only. It is evident that we can not use this form of Newton's second law for any other motion.

In this module, we shall develop the form of Newton's law, which is based on the concept of angular momentum of a particle about a fixed point in the coordinate reference. The law so derived, ofcourse, will then be shown to yield the version of Newton's second law for rotation, which is a special case.

Newton's second law for a particle in general motion

It was stated in the previous module that angular momentum is defined such that its first time derivative gives torque on the particle. This condition was specified keeping in mind about Newton's second law for translation. Following the logic, let us consider angular momentum of a moving particle moving with respect to a point and find whether first derivative actually yields torque as expected or not?

Angular momentum of the particle about the origin of the coordinate system is :

The " r " and " v " vectors represent position and velocity vectors respectively as shown in the figure. Taking differentiation of the terms with respect to time, we have :

$$\begin{array}{l}\frac{d\mathbf{\ell }}{dt}=\frac{d}{dt}\left\{m\right(\mathbf{r}\mathbf{x}\mathbf{v}\left)\right\}\end{array}$$

For constant mass,

$$\begin{array}{l}\Rightarrow \frac{d\mathbf{\ell }}{dt}=m\frac{d}{dt}(\mathbf{r}\mathbf{x}\mathbf{v})\end{array}$$

$$\begin{array}{l}\Rightarrow \frac{d\mathbf{\ell }}{dt}=m(\mathbf{r}\mathbf{x}\frac{d\mathbf{v}}{dt}+\frac{d\mathbf{r}}{dt}\mathbf{x}\mathbf{v})\end{array}$$

By definition, the first time derivative of velocity is the acceleration and first time derivative of position vector is the velocity of the particle. Putting the appropriate terms for these quantities,

$$\begin{array}{l}\Rightarrow \frac{d\mathbf{\ell }}{dt}=m(\mathbf{r}\mathbf{x}\mathbf{a}+\mathbf{v}\mathbf{x}\mathbf{v})\end{array}$$

But, the vector product of a vector with itself is equal to zero as sinθ = sin0° = 0. Hence,

$$\begin{array}{l}\Rightarrow \frac{d\mathbf{\ell }}{dt}=m(\mathbf{r}\mathbf{x}\mathbf{a})\end{array}$$

Rearranging, we have :

$$\begin{array}{l}\Rightarrow \frac{d\mathbf{\ell }}{dt}=\mathbf{r}\mathbf{x}m\mathbf{a}=\mathbf{r}\mathbf{x}\mathbf{F}\end{array}$$

Indeed the first derivative of angular momentum equals the toque on the particle as expected. Since force on the particle is external force, we can qualify the above relation that first time derivative of angular momentum equals external torque applied on the particle. Next, we should think about a situation when more than one force acts on the particle. According to Newton's second law in translation, we have :