1.17 Understanding motion  (Page 3/3)

• If “v” is speed, then it can not be negative.
• If “v” is velocity, then it can be either positive or negative.
• If possible follow the convention as under :

$$\begin{array}{l}v=\mathrm{velocity}\\ \left|v\right|=\mathrm{speed}\end{array}$$

Similarity / Difference 10 : In the case of uniform motion (unidirectional motion), there is no distinction between scalar and vector attributes at all. The distance .vs. displacement and speed .vs. velocity differences have no relevance. The paired quantities are treated equal and same. Motion is one dimensional and unidirectional; there being no question of negative value for attributes with direction.

Similarity / Difference 11 : Since velocity is a vector quantity being the time rate of change of position vector (displacement), there can be change in velocity due to the change in position vector (displacement) in any of the following three ways :

• change in magnitude
• change in direction
• change in both magnitude and direction

This realization brings about important subtle differences in defining terms of velocity and their symbolic representation. In general motion, velocity is read as the "time rate of change of position vector" :

The speed i.e. the magnitude of velocity is read as the "absolute value (magnitude) of the time rate of change of position vector" :

But the important thing to realize is that “time rate of change in the magnitude of position vector” is not same as “magnitude of the time rate of change of position vector”. As such the time rate of change of the magnitude of position vector is not equal to speed. This fact can be stated mathematically in different ways :

We shall work out an example of a motion in two dimensions (circular motion) subsequently in this module to illustrate this difference.

However, this difference disappears in the case of one dimensional motion. It is so because we use scalar quantity to represent vector attribute like position vector and velocity. Physically, we can interpret that there is no difference as there is no change of direction in one dimensional motion. It may be argued that there is a change in direction even in one dimensional motion in the form of reversal of motion, but then we should realize that we are interpreting instantaneous terms only – not the average terms which may be affected by reversal of motion. Here, except at the point of reversal of direction, the speed is :

Similarity / Difference 12 : Understanding of the class of motion is important from the point of view of analysis of motion (solving problem). The classification lets us clearly know which tools are available for analysis and which are not? Basically, our success or failure in understanding motion largely depends on our ability to identify motion according to a certain scheme of classification and then apply appropriate tool (formula/ defining equations etc) to analyze or solve the problem. It is, therefore, always advisable to write down the characteristics of motion for analyzing a situation involving motion in the correct context.

A simple classification of translational motion types, based on the study up to this point is suggested as given in the figure below. This classification is based on two considerations (i) dimensions of motion and (ii) nature of velocity.

Problem : The position vector of a particle in motion is :

$$\begin{array}{l}\mathbf{r}=a\cos \omega t\mathbf{i}+a\sin \omega t\mathbf{j}\end{array}$$

where “a” is a constant. Find the time rate of change in the magnitude of position vector.

Solution : We need to know the magnitude of position vector to find its time rate of change. The magnitude of position vector is :

$$\begin{array}{l}r=\left|\mathbf{r}\right|=\surd \{{\left(a\cos \omega t\right)}^2+{\left(a\sin \omega t\right)}^2\}\\ \Rightarrow r=a\surd ({\cos }^2\omega t+{\sin }^2\omega t)\\ \Rightarrow r=a\end{array}$$

But “a” is a constant. Hence, the time rate of change of the magnitude of position vector is zero :

$$\begin{array}{l}\frac{dr}{dt}=0\end{array}$$

This result is an important result. This highlights that time rate of change of the magnitude of position vector is not equal to magnitude of time rate of change of the position vector (speed).

The velocity of the particle is obtained by differentiating the position vector with respect to time as :

$$\begin{array}{l}\mathbf{v}=\frac{d\mathbf{r}}{dt}=-a\omega \sin \omega t\mathbf{i}+a\omega \cos \omega t\mathbf{j}\end{array}$$

The speed, which is magnitude of velocity, is :

$$\begin{array}{l}v=\left|\frac{d\mathbf{r}}{dt}\right|=\surd \{{(-a\omega \sin \omega t)}^2+{\left(a\omega \cos \omega t\right)}^2\}\\ \Rightarrow v=a\omega \surd ({\sin }^2\omega t+{\cos }^2\omega t)\\ \Rightarrow v=a\omega \end{array}$$

Clearly, speed of the particle is not zero. This illustrates that even if there is no change in the magnitude of position vector, the particle can have instantaneous velocity owing to the change in the direction.

As a matter of fact, the motion given by the position vector in the question actually represents uniform circular motion, where particle is always at constant distance (position) from the center, but has velocity of constant speed and varying directions. We can verify this by finding the equation of path for the particle in motion, which is nothing but a relation between coordinates. An inspection of the expression for “x” and “y” coordinates suggest that following trigonometric identity would give the desired equation of path,

$$\begin{array}{l}{\sin }^2\theta +{\cos }^2\theta =1\end{array}$$

For the given case,

$$\begin{array}{l}\mathbf{r}=a\cos \omega t\mathbf{i}+a\sin \omega t\mathbf{j}\\ \Rightarrow x=a\cos \omega t\\ \Rightarrow y=a\sin \omega t\mathbf{j}\end{array}$$

Rearranging, we have :

$$\begin{array}{l}\cos \omega t=\frac{x}{a}\end{array}$$

and

$$\begin{array}{l}\sin \omega t=\frac{y}{a}\end{array}$$

Now, using trigonometric identity :

$$\begin{array}{l}\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{a^2}=1\\ \Rightarrow x^2+y^2=a^2\end{array}$$

This is an equation of a circle of radius “a”.

In the nutshell, for motion in general (for two/three dimensions),

$$\begin{array}{l}\frac{dr}{dt}=\frac{d\left|\mathbf{r}\right|}{dt}\ne v\\ \frac{dr}{dt}=\frac{d\left|\mathbf{r}\right|}{dt}\ne |\frac{d\mathbf{r}}{dt}\end{array}$$

It is only in one dimensional motion that this distinction disappears as there is no change of direction as far as instantaneous velocity is concerned.