Physics Formulas

Centripetal Acceleration Formula

In this article, we will get to learn about the Centripetal Acceleration Formula. Can an object accelerate if it’s moving with a speed that is constant? Yup! So many people find this counter-intuitive at first. This is just because they forget those changes that are in the direction of motion of an object—even if the object maintains a constant speed—still it gets the count as acceleration.

Centripetal Acceleration Formula and Derivation

A body that is moving in a circular motion (with radius r) at a constant speed (v) is always being accelerated continuously. Thus, the acceleration is at the right angles to the direction of the motion. It is towards the center of the sphere and of magnitude \(v^{2}\)/r.

The direction of the acceleration is deduced through symmetry arguments. If it points out the acceleration out of the plane of the sphere, then the body would leave the plane of the circle. However, it does not, so it isn’t.  If the acceleration that is pointed in any direction other than perpendicular either left or right then the body would increase its speed or also might slow down. It does not.

Get the huge list of Physics Formulas here

Now for the magnitude. Let us consider the distance traveled by the body over a small time increment \(\Delta t\):

We can easily calculate the arc length i.e. s as both the distances traveled (distance = rate \(\times\) time = v \(\Delta\)t) and using the definition of a radian (arc = radius \(\times\) angle in radians = r \(\Delta\theta\)).

s = v \(\Delta t\)

s = r \(\Delta \theta\)

r \(\Delta \theta\) = v \(\Delta t\)

\(\frac{\Delta \theta}{\Delta t}\) = \(\frac{v}{r}\)

\(\frac{d\theta}{dt}\) = \(\frac{v}{r}\)

Therefore, the angular velocity of the object is thus v/r (in radians per unit of time).

The right half portion of the diagram is made by putting the tails of the two v vectors together. Notably, \(\Delta \theta\) is the same in both of the diagrams.

sin\(\frac{\Delta \theta}{2}\) = \(\frac{^{\Delta v}/2}{v}\)

\(\Delta v\) = 2v sin\(\frac{\Delta \theta}{2}\)

\(\frac{\Delta v}{\Delta \theta}\) = 2v\(\frac{sin\frac{\Delta \theta}{2}}{\Delta \theta}\)

\(\frac{\Delta v}{\Delta \theta}\) = v\(\frac{sin\frac{\Delta \theta}{2}}{\frac{\Delta \theta}{2}}\)

Notably, the passing from sin to cos is occurring through the L’Hôpital’s rule.

\(\frac{dv}{d\theta}\) = \(\lim_{\Delta \theta \rightarrow 0}\) v\(\frac{sin\frac{\Delta \theta }{2}}{\frac{\Delta \theta}{2}}\) = v\(\lim_{\Delta \theta \rightarrow 0}\) \(\frac{\left ( cos\frac{\Delta \theta }{2} \right )\frac{1}{2}}{\frac{1}{2}}\) = v

\(\frac{dv}{dt}\) = \(\frac{dv}{d\theta}\) \(\frac{d\theta}{dt}\) = v\(\frac{v}{r}\) = \(\frac{v^{2}}{r}\)

Solved Examples


We twirl a lasso over our head at a constant angular speed of 3.8 radians/second. What will be the centripetal acceleration in meters/second squared of the lasso’s tip? 1.4 meters away from our hand?


20 m/\(s^{2}\)

The tangential speed of the lasso’s tip will be:

v = \(\omega\) r


\(\omega\) = 3.8 radians/second

Is the angular speed and r = 1.4 meters is the radial distance from the middle of the sphere (i.e. our hand) to the lasso’s tip. The centripetal acceleration is:

a = \(\frac{v^{2}}{r}\)

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

= \(\omega^{2}\)r

Plug in the quantities we have to find:

a = \(\omega ^{2}\)r

= \(\left ( 3.8rad/s \right )^{2}\) \(\times\) 1.4m

= 20m/\(s^{2}\)


The documentation for your slot-car set states that the maximum centripetal acceleration the cars can withstand without being ejected from the track is 3.8 meters/second squared. You notice that the slot cars fly off the track if they go more than 1.1 meters/second. What will be the radius in meters of the curve in the track?


0.32 m

The maximum centripetal acceleration we have is a = 3.8 meters/second squared, and the maximum speed at which these slot cars can run without flying off the track is:

V = 1.1 meters/second.

Solve the equation for the centripetal acceleration for the radius and then insert these quantities in it. Then the result will be:

a = \(\frac{v^{2}}{r}\)

r = \(\frac{v^{2}}{a}\)

= \(\frac{\left ( 1.1m/s^{2} \right )}{3.8m/s^{2}}\)

= 0.32m.

Share with friends

Customize your course in 30 seconds

Which class are you in?
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
Ashhar Firdausi
IIT Roorkee
Dr. Nazma Shaik
Gaurav Tiwari
Get Started

5 responses to “Spring Potential Energy Formula”

  1. Typo Error>
    Speed of Light, C = 299,792,458 m/s in vacuum
    So U s/b C = 3 x 10^8 m/s
    Not that C = 3 x 108 m/s
    to imply C = 324 m/s
    A bullet is faster than 324m/s

  2. Malek safrin says:

    I have realy intrested to to this topic

  3. umer says:

    m=f/a correct this

  4. Kwame David says:

    Interesting studies

  5. Yashdeep tiwari says:

    It is already correct f= ma by second newton formula…

Leave a Reply

Your email address will not be published. Required fields are marked *

Download the App

Watch lectures, practise questions and take tests on the go.

Customize your course in 30 seconds

No thanks.