When a particle moves on a circular path with uniform speed, then its velocity remains changing at each point of the path due to changing the direction of velocity. In this situation an acceleration becomes active on the particle and the direction of this acceleration always remains towards the centre of the path. This acceleration is called centripetal acceleration or normal acceleration or radial acceleration.
Suppose a particle is moving on circular path of radius r with uniform speed v. In time interval ∆t, it reaches from point P to P'. Velocity of the particle at these points is \(\vec{v_1}\) and \(\vec{v_2}\) and the distance along circular part i.e. arc is ∆S. Since the speed is constant;
\(\therefore \ |\vec{v_1}| = |\vec {v_2}|\)= v (speed)
or v1 = v2 = v .........(1)
∵ \(\therefore Angle \ = \frac{arc}{radius}\)
∴ From fig. angle
\(\Delta \theta = \frac{\Delta S}{r}\ ..........(2)\)
If we show vectors \(\vec{v_1}\) and \(\vec{v_2}\) on same vector diagram, then we get the diagram shown in
From this diagram,
On comparing equation (2) and (3), we have
\(\frac{\Delta v}{v} = \frac{\Delta S}{r}\)
With the help of diagram, it can be proved \(\overrightarrow{\Delta v}\) will be towards the centre of the circular path. Therefore this acceleration is called 'Centripetal acceleration' and is denoted by ac.
\(\therefore a_c = \frac{v^2}{r}\ ...(5)\)
Since this acceleration, normal to velocity, therefore it is also called 'normal acceleration' and it is along radius of the path, hence it is called 'radial acceleration.'
when the frequency is given in rotations
Since the magnitude of the particle velocity i.e., speed u is constant, therefore magnitude of \(a_c = \left(\frac{v^2}{r}\right)\) also remains constant but its direction remains changing due to changing the direction of particle velocity. However, its direction remains always perpendicular to velocity and towards centre of the path.
