# 6.6 Non-uniform circular motion  (Page 2/4)

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--------------------------- t v ω(s) (m/s) (rad/s) ---------------------------0 0 0.0 1 1 0.12 2 0.2 3 3 0.3---------------------------

The above data set describes just a simplified situation for the purpose of highlighting variation in angular speed. We can visualize this change in terms of angular velocity vector with increasing magnitudes as shown in the figure here :

Note that magnitude of angular velocity i.e. speed changes, but not its direction in the illustrated case. However, it has been pointed out earlier that there are actually two directional possibilities i.e. clockwise and anti-clockwise rotation. Thus, a circular motion may also involve change of direction besides a change in its magnitude.

## Tangential acceleration

The non-uniform circular motion involves a change in speed. This change is accounted by the tangential acceleration, which results due to a tangential force and which acts along the direction of velocity circumferentially as shown in the figure. It is easy to realize that tangential velocity and acceleration are tangential to the path of motion and keeps changing their direction as motion progresses.

We note that velocity, tangential acceleration and tangential force all act along the same direction. It must, however, be recognized that force (and hence acceleration) may also act in the opposite direction to the velocity. In that case, the speed of the particle will decrease with time.

The magnitude of tangential acceleration is equal to the time rate of change in the speed of the particle.

$\begin{array}{l}{a}_{T}=\frac{đv}{đt}\end{array}$

Problem : A particle, starting from the position (5 m,0 m), is moving along a circular path about the origin in xy – plane. The angular position of the particle is a function of time as given here,

$\begin{array}{l}\theta ={t}^{2}+0.2t+1\end{array}$

Find (i) centripetal acceleration and (ii) tangential acceleration and (iii)direction of motion at t =0 .

Solution : From the data on initial position of the particle, it is clear that the radius of the circle is 5 m.

(i) For determining centripetal acceleration, we need to know the linear speed or angular speed at a given time. Here, we differentiate angular position function to obtain angular speed as :

$\begin{array}{l}\omega =\frac{đ\theta }{đt}=2t+0.2\end{array}$

Angular speed is varying as it is a function of time. For t = 0,

$\begin{array}{l}\omega =0.2\phantom{\rule{2pt}{0ex}}\mathrm{rad}/s\end{array}$

Now, the centripetal acceleration is :

$\begin{array}{l}{a}_{R}={\omega }^{2}r={\left(0.2\right)}^{2}x5=0.04x5=0.2\phantom{\rule{2pt}{0ex}}m/{s}^{2}\end{array}$

(ii) For determining tangential acceleration, we need to have expression of linear speed in time.

$\begin{array}{l}v=\omega r=\left(2t+0.2\right)x5=10t+1\end{array}$

We obtain tangential acceleration by differentiating the above function :

$\begin{array}{l}{a}_{T}=\frac{đv}{đt}=10\phantom{\rule{2pt}{0ex}}m/{s}^{2}\end{array}$

Evidently, the tangential acceleration is constant and is independent of time.

(iii) Since, the angular speed is evaluated to be positive at t = 0, it means that angular velocity is positive. This, in turn, means that the particle is rotating anti-clockwise at t = 0.

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