Lets take a string with the stone knotted to the other end. Whirl it from one end you could see the string fly off in a straight line moving towards the center.This is nothing but the existence of centripetal force.
Centripetal acceleration also known as center seeking force plays a essential part of a body moving in a circular path. |

If an object moves in a circle it acceleration varies at every point even though it is moving with uniform speed. It has two motions

- One is accelerating towards the center
- The other is a constant velocity perpendicular to that acceleration.

The one which is accelerating towards the center is what we call the centripetal acceleration denoted as a

_{c}._{}If a body moves in a circular path it moves with constant speed but its velocity is changing due to the direction change. The acceleration is towards the center of the circular path and is called centripetal acceleration.

It is given by

It is given by

a

_{c}= $\frac{v_t^2}{r}$Here a

The centripetal acceleration of moon = 0.00258 m/s

The centripetal acceleration of earth = 0.005954 m/s

_{c}is the centripetal acceleration, v is the velocity of the body and r is the radius of circle.It tells that if the speed is doubled the acceleration is increased by 4.The centripetal acceleration of moon = 0.00258 m/s

^{2}The centripetal acceleration of earth = 0.005954 m/s

^{2}The centripetal acceleration is the acceleration that points towards the center of the circular path.Let's take a body moves along a curve with radius r and tangential velocity v

_{t}the centripetal acceleration is given asa

Here a_{c}= lim_{$\Delta$t->0}$\frac{dv_t}{dt}$ = $\frac{v_t}{t}$_{c}= centripetal acceleration

v

_{t}= tangential acceleration

If you consider that body moves in the curve with radius r and tangential velocity v

_{t }the centripetal acceleration is

a

_{c}= $\frac{v_t^2}{r}$The total angular acceleration is the sum of tangential acceleration a

_{t}and centripetal acceleration a_{c}given asa

_{total}= $\sqrt{a_c^2 + a_t^2}$Lets take a body moving with constant speed v in a circle of radius r. If it moves from point A to B in the time interval $\Delta$ t. Since distance = speed $\times$ time the arc AB = v $\delta$ t. The definition of angle in radians is arc AB = r $\delta$ $\theta$ thus r $\delta$ $\theta$ = v $\delta$ t that is same as $\delta$ $\theta$ = v $\frac{\delta t}{r}$.

The vectors v

Acceleration is the change in velocity over time = $\frac{X}{\delta t}$ therefore a = $\frac{v^2}{r}$.

_{a}and v_{b}drawn tangentially at A and B represent the velocities at these points. The change in velocity between A and B is obtained by subtracting V_{a}and V_{b}i.e., V_{b}- V_{a}.The angle is small then the distance between the two velocity vector lines is v $\delta$ $\theta$.x = v $\delta$ $\theta$ however from first equation $\delta$ $\theta$ = v $\frac{\delta t}{r}$x = $\frac{v^2}{r}$ $\times$ $\delta$ tAcceleration is the change in velocity over time = $\frac{X}{\delta t}$ therefore a = $\frac{v^2}{r}$.

In centripetal acceleration the velocity of the body varies at every point in the circular motion.Since acceleration is nothing but the rate of change of velocity the centripetal acceleration cannot be a constant as shown below

Centripetal acceleration a_{c}= $\frac{dv_c}{dt}$ $\neq$ Constant if v

_{c}$\neq$ 0.

So centripetal acceleration can be constant only when body is at rest.

Here are given some centripetal acceleration graphs - Fig 1 shows the centripetal acceleration increasing linearly with time
- Fig 2 tells about constant acceleration what ever might be the time
- Fig 3 shows that the centripetal acceleration gets decreased with time
- Fig 4 shows that the centripetal acceleration increases exponentially with time.

**Here are few examples on centripetal acceleration you can go through it:**

- A man jogs many rounds in circular path garden
- A race car riding in a curved track
- A skater riding in circles round and round.

**Here are a few problems on calculating centripetal acceleration you can go through it:**

### Solved Examples

**Question 1:**A cat sits at 1.50 m from the center of a merry-go-round and revolves at a tangential speed of 1.80 m/s. Find its centripetal acceleration?

**Solution:**

Using the formula of centripetal acceleration, we have

$a_{c}$ = $\frac{v_{t}^{2}}{r}$

= $\frac{1.8^{2}}{1.5}$

= $\frac{3.24}{1.5}$

The acceleration is $a_{c}$ = 2.16 m/s

^{2}

Hence the centripetal acceleration experienced by cat is 2.16 m/s

^{2}.

**Question 2:**Calculate the magnitude of centripetal acceleration of a truck traveling at 10 m/s and radius of the circle is 100m.

**Solution:**

Given: Velocity v = 10 m/s and radius r = 100m

The centripetal acceleration is

a

_{c}= $\frac{v^2}{r}$ = $\frac{10^2}{100}$ = 1 m/s

^{2}.

Truck is moving with the centripetal acceleration of 1 m/s

^{2}.