Centripetal Acceleration

Centripetal acceleration refers to how quickly an object’s velocity changes as it moves along a circular path. This acceleration always acts toward the center of the circle, ensuring the object continues following its curved route instead of flying off in a straight line. In physics, centripetal acceleration is key to understanding circular motion — whether it’s planets orbiting the Sun, a car taking a sharp turn, or an object tied to a string being spun around.Grasping this concept helps students, engineers, and science enthusiasts understand how different forces work in rotational systems and how factors like speed, mass, and radius affect motion. 

1.0Definition of Centripetal Acceleration

The velocity of the particle changes while moving on the curved path, this change in velocity is brought by a force known as centripetal force and the acceleration so produced in the body is known as centripetal acceleration.

${\vec{a}}_c=\vec{\omega }\times \vec{v}$

• Centripetal acceleration is always perpendicular to the velocity at each point , therefore it is responsible for change in direction of velocity.
• In terms of magnitude $a_c=\omega v={\omega }^{2}r=\frac{v^2}{r}$

2.0Acceleration of a Particle in Uniform Circular Motion

Uniform circular motion: If a particle is moving in a circle with constant speed then motion is called Uniform circular motion (UCM).    

$$\begin{gathered} \overrightarrow{\Delta v}={\vec{v}}_B-{\vec{v}}_A \\ \text{Magnitude of change} \\ \left|\overrightarrow{\Delta v}\right| \end{gathered}$$$=\left|{\vec{v}}_B-{\vec{v}}_A\right|=\sqrt{v_B^2+v_A^2+2v_A{v}_B\cos (\pi -\theta )}$

$$\begin{gathered} v_A=v_B=v,\text{since speed is same} \\ \therefore \left|\overrightarrow{\Delta v}\right|=2v\sin \frac{\theta }{2} \\ \text{Distance travelled by particle between A and B }=r\theta \\ \text{Hence time taken, }\Delta t=\frac{r\theta }{v} \\ \text{Average Acceleration,} \\ \left|\overrightarrow{\Delta a}\right|=\left|\frac{\overrightarrow{\Delta v}}{\Delta t}\right|=\frac{2v\sin \frac{\theta }{2}}{\frac{r\theta }{v}}=\frac{v^2}{r}\frac{2\sin \frac{\theta }{2}}{\theta } \\ \text{If}\Delta t\to 0,\text{then}\theta \text{is small} \\ \sin \left(\frac{\theta }{2}\right)=\frac{\theta }{2} \\ {\lim }_{\Delta t\to 0}\left|\frac{\Delta \vec{v}}{\Delta t}\right|=\left|\frac{d\vec{v}}{dt}\right|=\frac{v^2}{r} \\ \text{Instantaneous acceleration is }\frac{v^2}{r} \\ \therefore a=a_c=\frac{v^2}{r} \end{gathered}$$

3.0Key points Centripetal Acceleration

1. The above acceleration is due to change in direction of velocity. It is called centripetal acc. or normal acceleration $(a_N)$…….
2. $(a_C)$…..is always directed towards the centre i.e. ⊥ to the velocity. It always acts towards the centre. So, it is centripetal.
3. In U.C.M, magnitude of $a_C$…… is constant but direction is changing.

Illustration-1.Find centripetal acceleration of given points (A and B) as shown in figure.

Solution:

${\left(a_c\right)}_A={\omega }^{2}R=5\times 5\times 1=25{\text{m/sec}}^2$

${\left(a_c\right)}_A={\omega }^{2}R=5\times 5\times 2=250{\text{m/sec}}^2$

Illustration-2.A body of mass 2 kg rests on a smooth horizontal surface and is attached to a string 3 m long. It is whirled in a horizontal circle at a rate of 60 revolutions per minute. Determine the centripetal acceleration of the body.

Solution: m=2 kg, r=3 m

$\omega =60\text{rev./minute}=\frac{60\times 2\pi }{60}\text{rad/sec}=2\pi \text{rad/sec}$

Because the angle described during 1 revolutions is 2 radian

$v=r\omega =2\pi \times 3\text{m/s}=6\pi \text{m/s}$

$\text{Now }{a}_c={\omega }^{2}r=(2\pi {)}^2\times 3=12{\pi }^2=118.4{\text{m/s}}^2$