The object accelerating in the circular motion experiences the centripetal force as well as the centrifugal force.

**The centripetal acceleration is the variation in the velocity of the object in the circular motion with respect to time and the direction of centripetal acceleration is tangent to the curve of the circular path.**

**Direction of Centripetal Acceleration in Circular Motion**

**The direction of the object moving in a circular motion varies at every discrete change in a distance ‘dx’ traveled by the object.**

Consider an object of mass ‘m’ accelerating in the circular path of radius ‘r’ such that the centripetal force on the object is F=mv^{2}/r

**In the above diagram, you can clearly depict that the direction of the object continuously changes as the object accelerates in a circular path. The direction of the motion of the object in a circular trajectory can be considered as tangential to the path, and so the velocity of the object varies at every small distance.**

The centripetal acceleration is pointing towards the center of the circular path and makes a 90-degree angle with the direction of the velocity of the object in a motion. The acceleration of the object is inward due to the application of the centripetal force imposed on the object in a circular trajectory. **Accordingly, we get the direction of the motion of the object in centripetal acceleration either clockwise or anti-clockwise.**

**Read more on How to find centripetal acceleration: different use cases and problems with facts.**

**What is the direction of the centripetal acceleration of a girl sitting on a ferries wheel of a radius of 10 meters and completing one revolution per minute?**

**Given:** r=10 meters

Time period T= 1rev/min

The circumference of a ferries wheel is

2πr=2 x 3.14 x 10=62.8 meters

The speed of a ferries when is

1 rev/min=62.8 m/60 s =1.05 m/s

Hence, a girl cover 1.05 meter every second on a ferries wheel.

**The direction of the velocity of a girl constantly changes.** It can be clearly understood that while moving from lower height to the highest point above the ground from this ferries wheel, the direction of the velocity of a girl is upward and then as the girl accelerates from that highest point to back to the lowest point near ground, the direction of the velocity of a girl is downward. **Well, the direction of centripetal acceleration is always towards the center of a wheel, thus remaining perpendicular to the velocity of a girl.**

**Read more on Centripetal Force Examples, Critical FAQs.**

**How to Find Direction of Centripetal Acceleration?**

The acceleration of the object in a centripetal motion moving with velocity ‘v’ along the radius ‘r’ of the circular path is a=v^{2}/r.

**The velocity of the object in a circular path is always tangential to the circle, while the centripetal acceleration remains parallel and in the direction equivalent to the centripetal force and perpendicular to the direction of velocity.**

The same is shown in the below figure:-

‘v’ is the velocity of the object traveling in a circular path while covering a distance ‘dx’ in time ‘t’. The force on the object in centripetal motion is

F=mv^{2}/r

According to Newton’s Second Law,

ma=mv^{2}/r

Hence, the centripetal acceleration of the object in circular motion is,

a=v^{2}/r

The direction of the centripetal acceleration is pointing towards the center of the circle as shown in the above figure. The tangential velocity is in a straight path directing outward of the circle every elapses and thus found to be perpendicular to the centripetal acceleration which pulls the body in keeping it in a circular track.

**Read more on Centripetal Acceleration Vs Acceleration: Various Types Acceleration Comparative Analysis.**

**What is a centripetal acceleration of an object moving on a circular path having a radius of 76 meters moving at a speed of 10 m/s? What will be the centripetal acceleration in terms of acceleration due to gravity?**

**Given:** v=10 m/s

r=76 meters

We have,

a=v^{2}/r

=(10)^{2}/76= 100/76 = 1.32 m/s^{2}

Hence, the centripetal acceleration of the object is 1.32 m/s^{2}.

The centripetal acceleration in relation to the term ‘g’ is

a/g=1.32 m/s^{2}/9.8 m/s^{2}=0.132 g

**So, in terms of acceleration due to gravity, centripetal acceleration of an object in a circular trajectory moving with a speed of 10 m/s is 0.132g.**

**Read more on Centripetal force.**

**Why is the Direction of Centripetal Acceleration always Perpendicular to the Velocity?**

The direction of the velocity of an object keeps on varying radially in a 360-degree path.

**The centripetal force so as the centripetal acceleration is always acting towards the center, the object remains in a circular track.**

The object accelerating in a circular path exert a force equal to mv^{2}/r. At the same time, the centrifugal force is also acting on the object that keeps the object away from falling towards the center. This force might have pushed the object at the center. But the force equal in magnitude and opposite in the direction is acting on the object in a circular path, which is a centrifugal force that balances the force, and keeps the motion of the object in a circular path.

**If there was no such force balancing the centripetal force then the electron revolving around the nucleus with high kinetic energy would have collapsed into the nucleus vanishing the charge. There would have not been any kind of charge and thus the existence of energy.**

So, both forces are equally vital for centripetal acceleration to occur. The direction of the object in a circular path radially shifts as the position of the object changes in a circular path. But, as though, the centripetal acceleration is always acting towards the center, the direction of centripetal acceleration remains perpendicular to the velocity of the object.

**Read more on Centripetal Force vs Centripetal Acceleration: Comparative Analysis.**

**Why Centripetal Acceleration Changes the Direction of Velocity?**

The centripetal acceleration is radially acting inward irrespective of the clockwise or anti-clockwise motion of the object in a circular path.

**The centripetal acceleration keeps the object in a circular motion and according to the path of motion of the object varies constantly after every displacement.**

Look at figure 1, it clearly portrays the variation in the direction of the velocity of the object in a centripetal motion. After traveling every discrete length of the distance, the direction of the velocity which is tangent to the circular path changes according to the centripetal acceleration.

If there was no centripetal acceleration, then the object would have traveled in a straight line until some external force was applied on the object to change its velocity and direction and the path would have not been a circular path.

**Read more on How to Find Tangential Velocity: Several Insights and Problem Examples.**

**A car taking a turn on a curved path takes 3 seconds to reach a green line from the red line and covers a distance of 12 meters in seconds. Calculate the velocity of a car. What is the direction of the centripetal acceleration? How the direction of the velocity does vary?**

We have, t=3sec, d=12m

Hence, the velocity of a car while crossing a curved path is

v= d/t

v=12/3=4 m/s

**The velocity of the car is 4m/s. This is the average velocity of the car as there is a variation in the velocity of a car between the red and green lines because the direction and acceleration of the car change constantly.**

**A person jogging in a park on a circular track of radius 38 meters at a speed of 2 m/s. What is the centripetal acceleration of a person while jogging and the direction of his centripetal acceleration?**

**Given:-** v=2m/s

r= 38 meters

We have,

a=v^{2}/r

=(2)^{2}/38=4/38=0.105 m/s^2

The centripetal acceleration that a person has to maintain is 0.105 m/s2 while jogging in a circular path of a radius of 38 meters. The centripetal acceleration is to keep a person on a circular track that is acting inward. Though the direction of the velocity of the object is always changing which is tangential to the circular path, the velocity of the person is a tangential velocity that is perpendicular to the direction of the acceleration of the object.

**Read more on How to Find Centripetal Force: Problem and Examples.**

**Frequently Asked Questions**

**What is centripetal force exerting on the object having a mass of 5 kg completing one revolution around a circular track of diameter 28 meters in a minute?**

**Given:** m= 5kg

d= 28 meters

r=14 meters

Hence, the circumference of a circular track is

C=2πr

C=2 x 3.14 x 14=87.92 m

The time required to cover 87.92 meters is 1 min. Therefore, the speed of an object is

v = d/t

v = 87.92m/60s = 1.46m/s

The velocity of an object in a circular motion is 1.46 m/s.

Now, the centripetal acceleration of an object will be

a=v^{2}/r

a = (1.46)^{2}/14 = 2.13/14=0.15m/s^{2}

The centripetal acceleration of an object is found to be 0.15 m/s2.

Hence, the centripetal force on the object is

F=ma

F=5 x 0.15=0.75N

**The force exerted on the object having a mass of 5kg moving on a circular path is 0.75N.**

**How is a centrifugal force related to the centripetal force?**

The centrifugal force on the object is also equal to mv^{2}/r but exerted in a straight opposite direction.

**If the centripetal force is acting inward of a circular path, the centrifugal force is force-directed outwards against the direction of centripetal force. As both the force are equal and opposite, the object can accelerate in a circular path.**

**Also Read:**

- How to find acceleration projectile motion
- Ways to find acceleration
- How to find acceleration of gravity
- Law of acceleration
- Centripetal acceleration in moon
- Average acceleration formula
- Negative acceleration example
- How acceleration differs from speed
- How to find acceleration down a ramp
- How to find acceleration with gravity

Hi, I’m Akshita Mapari. I have done M.Sc. in Physics. I have worked on projects like Numerical modeling of winds and waves during cyclone, Physics of toys and mechanized thrill machines in amusement park based on Classical Mechanics. I have pursued a course on Arduino and have accomplished some mini projects on Arduino UNO. I always like to explore new zones in the field of science. I personally believe that learning is more enthusiastic when learnt with creativity. Apart from this, I like to read, travel, strumming on guitar, identifying rocks and strata, photography and playing chess.