__What is called Centripetal Acceleration?__

Uniform circular motion is an accelerated motion. When a particle is in uniform circular motion, its *speed* remains constant but its *velocity* changes continuously due to the change in direction of motion. Hence, a circular motion is accelerated.

*A body undergoing uniform circular motion is acted upon by an acceleration which is directed along the radius towards the centre of the circular path. This acceleration is called centripetal acceleration.*

### Magnitude of Centripetal Acceleration

Consider about a particle moving with a constant speed ( v ) in a circular path of radius ( r ) as shown in figure.

Suppose at time ( t ) the particle is at P and at time ( t + \delta t ) * *it is at Q . Let, ( v_1 ) and ( v_2 ) are the tangential velocities at points P and Q respectively.

Now, a velocity vector triangle is drawn as shown in figure. In vector triangle –

- Side AB represents velocity vector ( v_2 )
- Side BC represents velocity vector ( - v_1 ) .

Vector AC is now drawn to close the triangle. Then AC will represent the change in velocity [ \Delta v = ( v_2 - v_1 ) ] .

If ( \delta t ) * *is too small then, ( \text {chord} \ PQ \ \simeq \ \text {Arc} \ PQ . Then, sector OPQ can be considered as a triangle.

Now, in triangles \triangle {OPQ} and \triangle {ABC} , we have –

- \angle POQ = \angle ABC = \delta \theta
- \left ( \frac {OP}{OQ} \right ) = 1 ( each is equal to radius of circle r )
- \left ( \frac {AB}{BC} \right ) = 1 ( each is equal to linear velocity \omega r

Hence, ( \triangle {OPQ} ) \ \text {and} \ ( \triangle {ABC} ) are congruent ( By SAS ).

Therefore, \quad \left ( \frac {\Delta v}{\delta s} \right ) = \left ( \frac {v}{r} \right )

Or, \quad \Delta v = \left ( \frac {v}{r} \right ) \delta s

Dividing both sides by ( \Delta t ) ,* *we will get –

\left ( \frac {\delta v}{\delta t} \right ) = \left ( \frac {v}{r} \right ) \left ( \frac {\delta s}{\delta t} \right )

In limiting conditions \quad \left ( \frac {dv}{dt} \right ) = \left ( \frac {v}{r} \right ) \left ( \frac {ds}{dt} \right )

But, \quad \left ( \frac {dv}{dt} \right ) = a \quad and \quad \left ( \frac {ds}{dt} \right ) = v

Therefore, \quad a = \left ( \frac {v}{r} \right ) v

Or, \quad a = \left ( \frac {v^2}{r} \right ) = \omega^2 r \quad

Because, \quad ( v = \omega r )

Therefore, magnitude of centripetal acceleration is given by –

a = \left ( \frac {v^2}{r} \right ) \quad Or, \quad a = \omega^2 r

### Direction of centripetal acceleration

Centripetal acceleration comes to action when a particle moves in curve or circular path. *Uniform circular motion* is an accelerated motion. It is the acceleration which keeps necessary force on the body to move in a circular path.

*Hence, centripetal acceleration always acts radially to the path of motion and its direction is towards the centre of curve or circular path.*

**EXAMPLE –**

Consider about the figure shown above. When the particle is at point ( P ) , the centripetal acceleration is acting in a direction from point ( P \rightarrow O ) and when the particle is at point ( Q ) , the centripetal acceleration is acting from point ( Q \rightarrow O ) .

__Centripetal Force__

*Centripetal force is the force required to make a body to move along a circular path with a uniform speed. It always acts along the radius and towards the centre of circular path.*

According to *Newton’s second law of motion*, the centripetal force required to move a body of *mass* ( m ) along a circular path of radius ( r ) is given by –

\text {Force} = \text {Mass} \ \times \ \text {Acceleration}

So, \quad F = m \times \left ( \frac {v^2}{r} \right ) = \left ( \frac {mv^2}{r} \right )

= \left [ \frac {m \left (r \omega \right )^2}{r} \right ] = m r \omega^2

Examples of centripetal forces are –

- When a stone is rotated in a circle, the
*tension*in the string provides the centripetal force. - For motion of planets around the sun, centripetal force for such motion is provided by
*gravitational force*exerted by sun on the planet. - When a car is taking a circular turn on a horizontal road, the centripetal force is provided by the force of
*friction*between the tyres and the road. *Electrostatic force*between electron and nucleus provide necessary centripetal force for electrons to move around the nucleus.

See the numerical problems based on this article –