When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion. The word “uniform” refers to the speed, which is uniform (constant) throughout the motion. Suppose an object is moving with uniform speed v in a circle of radius R as shown in Fig. 4.19. Since the velocity of the object is changing continuously in direction, the object undergoes acceleration. Let us find the magnitude and the direction of this acceleration.
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| Fig.1: Velocity and acceleration of an object in uniform circular motion. The time interval ∆t decreases from (a) to (c) where it is zero. The acceleration is directed, at each point of the path, towards the centre of the circle. |
Let r and r′ be the position vectors and v and v′ the velocities of the object when it is at point P and P′ as shown in Fig. 4.19(a). By definition, velocity at a point is along the tangent at that point in the direction of motion. The velocity vectors v and v′ are as shown in Fig. 4.19(a1). ∆v is obtained in Fig. 4.19 (a2) using the triangle law of vector addition.
Since the path is circular, v is perpendicular to r and so is v′ to r′. Therefore, ∆v is perpendicular to ∆r. Since average acceleration is along
$\Delta \mathbf{v} \left(\mathbf{\bar{a}}=\frac{\Delta \mathbf{v}}{\Delta t} \right)$
the average acceleration $\mathbf{a}$ is perpendicular to ∆r. If we place ∆v on the line that bisects the angle between r and r′, we see that it is directed towards the centre of the circle. Figure 4.19(b) shows the same quantities for smaller time interval. ∆v and hence a is again directed towards the centre. In Fig. 4.19(c), ∆t → 0 and the average acceleration becomes the instantaneous acceleration. It is directed towards the centre (In the limit ∆t→0, ∆r becomes perpendicular to r. In this limit ∆v → 0 and is consequently also perpendicular to V. Therefore, the acceleration is directed towards the centre, at each point of the circular path.).
Thus, we find that the acceleration of an object in uniform circular motion is always directed towards the centre of the circle. Let us now find the magnitude of the acceleration.
The magnitude of a is, by definition, given by
$|\mathbf{a}|=\lim_{\Delta t\rightarrow 0}\frac{|\Delta \mathbf{v}|}{\Delta t}$
Let the angle between position vectors r and r′ be ∆θ. Since the velocity vectors v and v′ are always perpendicular to the position vectors, the angle between them is also ∆θ . Therefore, the triangle CPP′ formed by the position vectors and the triangle GHI formed by the velocity vectors v, v′ and ∆v are similar (Fig. 4.19a). Therefore, the ratio of the base-length to side-length for one of the triangles is equal to that of the other triangle. That is:
$\frac{|\Delta \mathbf{v}|}{v}=\frac{|\Delta \mathbf{r}|}{R}$
Or, $|\Delta \mathbf{v}|=v\frac{|\Delta \mathbf{r}|}{R}$
Therefore,
$|\mathbf{a}|=\lim_{\Delta t\rightarrow 0}\frac{|\Delta \mathbf{v}|}{\Delta t}=\lim_{\Delta t\rightarrow 0}\frac{v|\Delta \mathbf{v}|}{R\Delta t}=\frac{v}{R} \lim_{\Delta t\rightarrow 0}\frac{|\Delta \mathbf{r}|}{\Delta t}$
If ∆t is small, ∆θ will also be small and then arc PP′ can be approximately taken to be |∆r|:
$|\Delta \mathbf{r}| \cong v\Delta t$
$\frac{\Delta \mathbf{r}}{\Delta t} \cong v$
Or $\lim_{\Delta t\rightarrow 0}\frac{|\Delta \mathbf{r}|}{\Delta t} = v$
Therefore, the centripetal acceleration a$_c$ is:
$a_c=\left(\frac{v}{R}\right)v$
$a_c=\frac{v^2}{R}$
Thus, the acceleration of an object moving with speed v in a circle of radius R has a magnitude $v^2/R$ and is always directed towards the centre. This is why this acceleration is called centripetal acceleration (a term proposed by Newton). A thorough analysis of centripetal acceleration was first published in 1673 by the Dutch scientist Christiaan Huygens (1629-1695) but it was probably known to Newton also some years earlier. “Centripetal” comes from a Greek term which means ‘centre-seeking’. Since v and R are constant, the magnitude of the centripetal acceleration is also constant. However, the direction changes — pointing always towards the centre. Therefore, a centripetal acceleration is not a constant vector.
We have another way of describing the velocity and the acceleration of an object in uniform circular motion. As the object moves from P to P′ in time ∆t (= t′ – t), the line CP (Fig. 4.19) turns through an angle ∆θ as shown in the figure. ∆θ is called angular distance. We define the angular speed ω (Greek letter omega) as the time rate of change of angular displacement:
$\omega=\frac{\Delta \theta}{\Delta t}$
Now, if the distance travelled by the object during the time ∆t is ∆s, i.e. PP′ is ∆s, then:
$v=\frac{\Delta v}{\Delta t}$
but ∆s = R∆θ. Therefore:
$v=R\frac{\Delta \theta}{\Delta t}$
so, $v=\omega R$
We can express centripetal acceleration ac in terms of angular speed:
$a_c=\frac{v^2}{R}=\frac{\omega^2 R^2}{R}$
$a_c=\omega^2 R$
The time taken by an object to make one revolution is known as its time period T and the number of revolution made in one second is called its frequency ν (=1/T). However, during this time the distance moved by the object is s = 2πR.
Therefore, v = 2πR/T =2πRν (4.48)
In terms of frequency ν, we have
ω = 2πν
v = 2πRν
$a_c$ = 4π$^2$ ν$^2$R
Example 1
An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector ? What is its magnitude?
Answer
This is an example of uniform circular motion. Here R = 12 cm. The angular speed ω is given by
ω = 2π/T = 2π × 7/100 = 0.44 rad/s
The linear speed v is:
v = ωR = 0.44 s$^{-1}$ × 12 cm = 5.3 cm.s$^{-1}$
The direction of velocity v is along the tangent to the circle at every point. The acceleration is directed towards the centre of the circle. Since this direction changes continuously, acceleration here is not a constant vector. However, the magnitude of acceleration is constant:
a = ω$^2$R = (0.44/s)$^2$ (12 cm) = 2.3 cm.s $^{-2}$
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