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 Sub Topics When a particle undergoes uniform circular motion, magnitude of the velocity of that particle remain constant and the direction will change continuously. In every instant, the velocity is along the direction of tangent. Since the particle continuously changes its direction of velocity, a non zero acceleration is present. According to Newton's second law, the acceleration of a particle of mass m is in the same direction of the force acting on it. This statement suggests that, in the case of uniform circular motion, the direction of acceleration should be towards the center. Acceleration which directed towards center is generally called centripetal acceleration or radial acceleration. From the Newton's law we can conclude that a uniform circular motion requires a radial acceleration of constant magnitude which depends on the radius as well as the speed of a rotating body. The behavior of the radial or centripetal acceleration given in the following section.

The acceleration of an object undergoing uniform circular motion is often called radial acceleration. The radial reminds the direction of this acceleration is towards center. This is also known as the centripetal acceleration because centripetal force causes this acceleration.

One of the major component of acceleration of a uniformly rotating body is radial component. The direction of this acceleration is towards center hence the name radial acceleration. It is also known as centripetal acceleration. The more details about this component is given in the following paragraphs.

The mathematical representation of radial acceleration is given as,

$a_{r}$ = $\frac{v^{2}}{r}$
Where $a_{r}$ is the radial component of acceleration
v is the velocity
r is the radius of the circular path
Also we know that, v = rω
Where ω is the angular velocity
Substitute the value of v in the above equation. We get
$a_{r}$ = $\omega ^{2}r$

The unit of radial acceleration (a_{r}) in SI unit is m/s2 . But the unit of ω must be in rad/s and the unit of r is in m.

1. Radial acceleration of an object on earth

Radius of the earth is given by, r = 6371 Km; T = 24 hr

We know the equation of radial acceleration as

$a_{r}$ = $\omega ^{2}r$

$\omega$ = $\frac{2\pi }{T}$

So, $a_{r}$ = $\frac{4\pi ^{2}r}{T^{2}}$

$a_{r}$ = $\frac{4\times 3.14^{2}\times 6371\times 10^{3}}{(24\times 60\times 60)^{2}}$

$a_{r}$ = $\frac{2.512\times 10^{8}}{7.464\times 10^{9}}$

$a_{r}$ = 0.033 m/s2

2. Radial acceleration of the earth towards sun

Radius of the earth's orbit (assumed to be circular), r = 1.50 ×108 Km; T = 365 days and ω = 30Km/s

Substitute these values in the equation, we get

$a_{r}$ = $\frac{\omega ^{2}}{r}$

$a_{r}$ = $\frac{(30\times 10^{3})^{2}}{1.50\times 10^{11}}$

$a_{r}$ = 6 × 10-3 m/s2

There are two different components in the case of an acceleration caused by a uniform circular motion of a particle. One is tangential acceleration and other one is radial acceleration. The notations of tangential and radial acceleration are at and ar respectively. The direction of tangential acceleration is along the tangent and the direction of radial component is towards center. The radial component of acceleration is also known as centripetal acceleration because the force caused for this acceleration is centripetal force.

This figure illustrates the tangential and radial component of acceleration

Some of the problems related to finding the radial acceleration is given below:

### Solved Examples

Question 1: Determine the radial acceleration of a CD which spins at 210 rpm and the diameter of this CD is given as 12 cm.

Solution:

The given entities are T = 210 rpm, d = 12 cm, So r = 6 cm
We can convert T in terms of frequency f
f = 210 $\frac{rev}{min}$ $\times$ $\frac{1}{60}$ $\frac{min}{s}$= 3.5 $\frac{rev}{s}$
For every rotation CD rotates through an angle 2π radians
So the angular velocity is given by,
$\omega$ = 2 $\pi$ f = 2 $\pi$ $\frac{rad}{rev}$ $\times$ 3.5 $\frac{rev}{s}$ = 7.0 $\pi$ $\frac{rad}{s}$
The radial acceleration is given by,
$a_{r}$ = $\omega ^{2}r$
$a_{r}$ = $\left(7.0 \pi \frac{rad}{s} \right)^{2}$ $\times$ 0.060m= 29 m/s2

Question 2: Calculate the radial acceleration of a point 25.4 cm far from the center which is rotating at 78 rpm on a turntable?

Solution:

From the question it is clear that T=78 rpm, r= 0.254 m
So, ω will be
$\omega$ = 78 $\frac{rev}{min}$ $\times$ $\frac{1}{60}$ $\frac{min}{s}$ $\times$ 2$\pi$ $\frac{rad}{rev}$
$\omega$ = 2.6 $\pi$ $\frac{rad}{s}$
$a_{r}$ = $\omega ^{2}r$
$a_{r}$ = $\left(2.6 \times \frac{rad}{s}\right)^{2}$ $\times$ 0.254
$a_{r}$ = 17 m/s2