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Centripetal acceleration is the acceleration that is produced when a body moves in a circular path. The velocity of the body continuously changes its direction when the object is in circular motion. Due to this change in velocity, an acceleration is produced which is known as Centripetal acceleration.
- Centripetal acceleration always acts radially toward the center of the circular path.
- The force causing this acceleration is known as Centripetal force.
- Centripetal acceleration is given by the magnitude of the square of the velocity of the body divided by the radius of the circular path.
- The SI unit of centripetal acceleration is meter-per-second square (m/s2).
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Key Terms: Centripetal Force, Centripetal Acceleration, Centripetal Acceleration Formula, Circular motion, Linear velocity, Motion, Angular acceleration.
Centripetal Acceleration
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Centripetal acceleration refers to the acceleration that is produced when the body is in a circular motion. If the body is not in circular motion then no centripetal acceleration is produced.
- Centripetal acceleration is always directed toward the center of the circular in which the body is moving and it is perpendicular to the direction of tangential velocity.
- Since the circular motion of a body is due to the centripetal force which acts radially towards the center of the circular path, therefore the acceleration caused due to this force is known as centripetal acceleration.
Read more: Uniform Circular Motion

Centripetal Acceleration
The video below explains this:
Centripetal Acceleration Detailed Video Explanation:
The formula for centripetal acceleration is given by
ac = v2/r
Where
- ac = centripetal acceleration
- v = tangential velocity or linear velocity
- r = radius of the circular path
Read more: Types of Friction
When an object moves in a circular path it has three different types of acceleration: Angular acceleration, Tangential acceleration, and Centripetal acceleration
- Angular acceleration is an angular quantity, whereas Centripetal and Tangential acceleration are linear quantities.
- The major difference between centripetal acceleration and tangential acceleration is that centripetal acceleration is produced due to the change in the direction of the tangential velocity, while tangential acceleration is produced due to the change in the magnitude of tangential velocity.

Centripetal Force
Read more: Centripetal and Centrifugal force
Derivation of Centripetal Acceleration
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Consider a body of mass m moving in a circular path of radius r with velocity v, as shown in the figure.
Centripetal force Fc helps the body to move in a circular path and is directed toward the center of the circle. This force acting on the body of mass m produced an acceleration known as centripetal acceleration ac and is given by
Fc = mac
Also, centripetal acceleration is given by the rate of change of linear velocity with time.
ac = Δv/Δt ...(i)

Centripetal acceleration
From the ΔPQR, we have
\(\vec {PQ}+\vec {QS} = \vec {SP}\)
-v1 + v2 = Δv
Δv = v2 – v1
Now triangle PQR and AOB are similar, therefore
\(\frac {\Delta v}{AB}=\frac {v}{r}\)
Also, we have
AB = arc AB = vΔt
\(\Rightarrow \frac {\Delta v}{v\Delta t}=\frac {v}{r}\)
\(\Rightarrow \frac {\Delta v}{\Delta t}=\frac {v^2}{r}\)
Substituting the above equation in equation (i), we get
\(a_c=\frac {v^2}{r}\)
This is the derivation for the formula of centripetal acceleration.
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Solved Examples
Ques. A rock tied to a string is moving at a fixed speed of 20.0 m/s in a circle having a radius of 6.0 m. Evaluate the approximate magnitude of the centripetal acceleration of the rock.
Ans. Given,
- the velocity of the rock tied to the string, v = 20 m/s
- the radius of the circular path, r = 6 m
From the formula of centripetal acceleration,
ac = v2 / r
ac = (20)2/ 6 = 66.67 m/s2
Therefore, centripetal acceleration is 66.67 m/s2
Ques. A 23 kg girl is riding a merry-go-round with a radius of 4 m. What is the centripetal force on the girl if her velocity is 5 m/s?
Ans. Given,
- mass of the girl, m = 23 kg
- the radius of the circular path, r = 4 m
- the tangential velocity of the girl, v = 5 m/s
The centripetal force acting on the girl is given by
Fc = mv2/r
⇒ Fc = (23 × 52) / 4 = 143.75 N
Things to Remember
- Acceleration refers to a change in speed or velocity, both in magnitude and direction or in any of them.
- In the case of uniform circular motion, the direction of velocity changes continuously. As a result, even though the speed is constant, a related acceleration is always there.
- Centripetal acceleration is the property of motion of an object moving through a circular path. It refers to an object which moves in a circle with an acceleration vector pointing toward the circle's center.
- Centripetal force is the net force that causes an object's centripetal acceleration in a circular motion. The centripetal force is directed toward the center, which is perpendicular to the body's motion.
- When an object moves in a circular motion, the following equation can be used to calculate its speed: ac = v2/r
Sample Questions
Ques. A 25 kg boy is riding a merry-go-round with a radius of 5m. What is the centripetal force on the boy if his velocity is 6 m/s? (1 mark)
Ans. Given,
- m = 25 kg
- r = 5 m
- v = 6 m/s
We have
Centripetal force, Fc = mv2/r = 25 × (6 × 6)/5 = 180N
Ques. A rock tied to a string is moving at a fixed speed of 10.0 m/s in a circle having a radius of 8.0 m. Evaluate the approximate magnitude of the centripetal acceleration of the rock. (2 marks)
Ans. Given,
- v = 10 m/s
- r = 8 m
According to the formula,
ac = v2 / r
ac = (10)2/ 8 = 12.5 m/s2
Therefore, centripetal acceleration is 12.5 m/s2
Ques. Calculate the dimensional formula of centripetal acceleration. (2 marks)
Ans. Dimensional formula of velocity = M0L1T-1
Dimensional formula of radius = M0L1T0
Substituting the above in the equation of the centripetal acceleration formula:
Dimensional formula of centripetal acceleration = M0L1T-2
Ques. In the case of a slot-car set, its maximum centripetal acceleration without being ejected from its track is noted to be 3.8 meters per second squared. It is noticed that these slot cars fly off their track when they exceed 1.1 meters per second. What is the radius of the curve in the track? Answer in meters. (2 marks)
Ans. Here, Maximum possible centripetal acceleration a = 3.8 m/s2
And, the maximum speed which can be attained by these particular cars without flying off its track v is 1.1 m/s.
Using the centripetal acceleration formula:
ac = v2 / r
r = v2 / ac = (1.1 m/s)2 / 3.8 m/s2 = 0.32m
Therefore, the Radius of the curve is 0.32m
Ques. Where is centripetal acceleration the highest? (1 mark)
Ans. The centripetal acceleration is comparatively higher when a body is moving at high speed through sharp curves having very low radii. It is often observed when driving cars.
Ques. The documentation for your slot-car set states that the maximum centripetal acceleration the cars can withstand without being ejected from the track is 4.8 meters/second squared. You notice that the slot cars fly off the track if they go more than 1.8 meters/second. What will be the radius in meters of the curve in the track? (2 marks)
Ans. The maximum centripetal acceleration a = 4.8 m/s2
And the maximum speed at which these slot cars can run without flying off the track v = 1.8 m/s
Using centripetal acceleration formula,
ac = v2 / r
r = v2 / ac = (1.8)2/4.8 = 0.67 m.
Thus, the radius would be 0.67 m
Ques. A ball is attached to a string that is 1.5 m long. It is spun so that it completes two full rotations every second. What is the centripetal acceleration felt by the ball? (2 marks)
Ans. The radius would be equal to the length of the string.
Since it travels in a circle with a radius of 1.5 m and completes two rotations per second. The length of each rotation is the circumference.
C = 2πr = 2π(1.5) = 3π
v = C x f = 3π×2 = 6π m/s
ac = v2 / r = (6 π)2 / 1.5 = 237 m/s2
Ques. There is an amusement ride that is used to teach students about centripetal force. The ride is a circular wall that you place your back on. The wall and floor then begin to spin. Once it reaches a certain rotational velocity, the floor drops, and the students are pinned to the wall as a result of centripetal force. A student of mass 50kg decides to go on the ride. The coefficient of static friction between the student and the wall is 0.8. If the diameter of the ride is 10m, what is the maximum period of the ride's rotation that will keep the student pinned to the wall once the floor drops? (5 marks)
Ans. Given,
- g = 10 m/s2
- m = 50 kg
- µ = 0.8
- Fs = mg = 50×10 = 500N
As we know, Fs = µN = 500
N = Fs/µ = 500/0.8 = 625N
Now, N = Fc = mac
ac = 625/50 = 12.5 m/s2
ac = v2 / r
v = (acr)1/2
v = (12.5 × 5)1/2
v = 7.9 m/s
now, Period = circumference / velocity
P = πD/v = π10/7.9 = 3.97 s
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