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Centripetal Acceleration Formula

Definition

For an object undergoing a circular motion or a curvilinear motion experiences a force directed towards the centre of curvature of the path. Here, the velocity of the object may be constant or changing throughout, but the direction of the tangential velocity always changes throughout the motion. Since velocity is a vector quantity, the change in direction changes the velocity. The rate of change of tangential velocity is defined as the centripetal acceleration. Since the centripetal force, responsible for the circular motion of the object acts inwards, the centripetal acceleration also acts inwards, along the radius vector of the circular path.

Formula

As mentioned above, the centripetal acceleration is defined as the rate of change of tangential velocity, we can write it as,

Centripetal Acceleration = rate of change of tangential velocity

We can also write it as,

a_{c}=lim_{\Delta} t->0\frac{\Delta vt}{\Delta t}

The centripetal force acting on an object moving on a circular or curvilinear path is given as,

F=\frac{mV^{2}t}{r}=m\omega^{2}r

So, the acceleration acting on the object can be given as,

a_{c}= V^{2}_{t}=\omega ^{2}r

Also, as we know, the tangential acceleration is given as,

a_{t}=r\alpha

Where,

r = radius of curvature,

α = angular acceleration and

ω =  angular speed.

The total acceleration in a circular path is

a_{total}=\sqrt{a^{2}_{c}+a^{2}_{t}}=\sqrt{r^{2}\omega ^{4}+r^{2}\alpha ^{2}}

Real Life Example

The most common real life application of the centripetal acceleration and force that you can easily relate to is that of the roller coasters. Unlike the previously used circular loops, the roller coasters are today designed with the teardrop-shaped clothoid loops. Centripetal acceleration is experienced in the clothoid loops.

Solved Examples

Example 1: A plane is circulating around a path of radius 4 km. The speed of the jet is a constant 10 km/s. What is the centripetal acceleration?

Solution:

Radius of the circular path = 4 km = 4 × 1000 m = 4000 m

Velocity = 10 km/sec = 10 × 1000 m/sec = 10,000 m/sec

Therefore, centripetal acceleration = \frac{v^{2}}{r}=\frac{(10000)^{2}}{4000}= 25000 m/sec^{2}

Example 2: A ball has a mass of 0.2 kg. It moves around a circular path of radius(r) 80 cm. Calculate the centripetal force given that it completes 1 round in every 3 seconds.

Solution:

Mass of the ball = 0.2 kg

Radius of the path = 80 cm = 0.8 m,

Path velocity = \frac{2\pi r}{3}=\frac{( 2\pi \times 0.8)}{3}=1.67m/s

Now, we can calculate the centripetal acceleration by using the formula:

a_{c}=\frac{v^{2}}{r}=\frac{1.67 \times 1.67}{0.8}= 3.504 m/s^{2}

Centripetal Force is given by:

F_{c}=ma_{c}=0.2 kg \times 3.504 m/s^{2}

= 0.7008 kgm/s2

To learn more about centripetal acceleration and related topics by watching video lectures and to solve more examples on them, download Byju’s The Learning App.

 

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