<< Chapter < Page | Chapter >> Page > |
$$\begin{array}{l}x=f\left(t\right)\end{array}$$
Problem : The angular position (in radian) of a particle under circular motion about a perpendicular axis with respect to reference direction is given by the function in time (seconds) as :
$$\begin{array}{l}\theta ={t}^{2}-0.2t+1\end{array}$$
Find (i) angular position when angular velocity is zero and (ii) determine whether rotation is clock-wise or anti-clockwise.
Solution : The angular velocity is equal to first derivative of angular position,
$$\begin{array}{l}\omega =\frac{\u0111\theta}{\u0111t}=\frac{\u0111}{\u0111t}({t}^{2}-0.2t+1)=2t-0.2\end{array}$$
For ω= 0, we have :
$$\begin{array}{l}2t-0.2=0\\ \Rightarrow t=0.1\phantom{\rule{2pt}{0ex}}s\end{array}$$
The angular position at t = 0.1 s,
$$\begin{array}{l}\theta ={\left(0.1\right)}^{2}-0.2x0.1+1=0.99\phantom{\rule{2pt}{0ex}}\mathrm{rad}\\ \Rightarrow \theta =\frac{0.99x360}{2\pi}=\frac{0.99x360x7}{2x22}={56.7}^{0}\end{array}$$
As the particle makes a positive angle with respect to reference direction, we conclude that the particle is moving in anti-clockwise direction (We shall discuss the convention regarding direction of angular quantities in detail subsequently).
In order to understand the relation, let us consider two uniform circular motions with equal time period (T) along two circular trajectories of radii ${r}_{1}$ and ${r}_{2}$ ( ${r}_{2}>{r}_{2}$ ). It is evident that particle along the outer circle is moving at a greater speed as it has to cover greater perimeter or distance. On the other hand angular speeds of the two particles are equal as they transverse equal angles in a given time.
This observation is key to understand the relation between linear and angular speed. Now, we know that :
$$\begin{array}{l}s=r\theta \end{array}$$
Differentiating with respect to time, we have :
$$\begin{array}{l}\frac{\u0111s}{\u0111t}=\frac{\u0111r}{\u0111t}\theta +r\frac{\u0111\theta}{\u0111t}\end{array}$$
Since, “r” is constant for a given circular motion, $\frac{\u0111r}{\u0111t}=0$ .
$$\begin{array}{l}\frac{\u0111s}{\u0111t}=\frac{\u0111\theta}{\u0111t}r=\omega r\end{array}$$
Now, $\frac{\u0111s}{\u0111t}$ is equal to linear speed, v. Hence,
This is the relation between angular and linear speeds. Though it is apparent, but it is emphasized here for clarity that angular and linear speeds do not represent two separate individual speeds. Remember that a particle can have only one speed at a particular point of time. They are, as a matter of fact, equivalent representation of the same change of position with respect to time. They represent same speed – but in different language or notation.
Problem : The angular position (in radian) of a particle with respect to reference direction, along a circle of radius 0.5 m is given by the function in time (seconds) as :
$$\begin{array}{l}\theta ={t}^{2}-0.2t\end{array}$$
Find linear velocity of the particle at t = 0 second.
Solution : The angular velocity is given by :
$$\begin{array}{l}\omega =\frac{\u0111\theta}{\u0111t}=\frac{\u0111}{\u0111t}({t}^{2}-0.2t)=2t-0.2\end{array}$$
For t = 0, the angular velocity is :
$$\begin{array}{l}\Rightarrow \omega =2x0-0.2=0.2\phantom{\rule{2pt}{0ex}}\mathrm{rad/s}\end{array}$$
The linear velocity at this instant is :
$$\begin{array}{l}\Rightarrow v=\omega r=0.2x0.5=0.1\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
The angular quantities (displacement, velocity and acceleration) are also vector quantities like their linear counter parts and follow vector rules of addition and multiplication, with the notable exception of angular displacement. Angular displacement does not follow the rule of vector addition strictly. In particular, it can be shown that addition of angular displacement depends on the order in which they are added. This is contrary to the property of vector addition. Order of addition should not affect the result. We intend here to skip the details of this exception to focus on the subject matter in hand. Besides, we should know that we may completely ignore this exception if the angles involved have small values.
Notification Switch
Would you like to follow the 'Physics for k-12' conversation and receive update notifications?