When a particle moves in a circular path with uniform angular velocity, it is said to be in uniform circular motion. The particle moves with a constant speed in this type of motion. In circular motion, velocity of the particle at any point is along the tangent to the circle at that point i.e. the direction of the velocity of the particle changes continuously. Therefore, the particle experiences acceleration.

According to Newton’s Second Law of Motion, a force must act on the particle. This force is called centripetal force and the produced acceleration is known as centripetal acceleration. The force always acts towards the centre of the circular path and perpendicular to the velocity at that point.

Expression For Centripetal Force And Acceleration

Consider a particle moving in a circular path of radius $r$ and centred at $O$. Let $P$ be the position of the particle at any instant $t$ and $Q$ be the position of the particle at $t+∆t$. Let $v_P$ and $v_Q$ be the velocities of the particle at position $P$ and $Q$ respectively. The direction of the velocity of the particle at any point on the circular path is tangent drawn on the path from that point. $v_P$ and $v_Q$ are equal in magnitude but different in direction. Let $∆s$ be the displacement from $P$ to $Q$ in time $∆t$. \[\text{Let }|\vec{v}_P|=|\vec{v}_Q|=v\]

The vector change in velocity $∆v$ can be obtained from figure (2) given below. For this, draw $AP’$ and $AQ’$ representing $v_P$ and $v_Q$ respectively and $P’Q’$ represents $∆v$.

Here, triangle $OPQ$ from fig.(1) and triangle $AP’Q’$ from fig.(2) are similar. \[∆OPQ\sim∆CPQ\] \[\frac{PQ}{P’Q’}=\frac{OP}{AP’}\] \[\frac{∆\vec{s}}{∆\vec{v}}=\frac{r}{v}\] \[[AP’=V \text{(Taken in Magnitude)}]\] \[∆\vec{v}=∆\vec{s}\frac{v}{r}\]\[\text{Dividing by ∆t,}\] \[\frac{∆\vec{v}}{∆t}=\frac{∆\vec{s}}{∆t}\frac{v}{r}\] \[\text{Taking }\lim_{∆t\to 0},\] \[\lim_{∆t\to 0}\frac{∆\vec{v}}{∆t}=\lim_{∆t\to 0}\frac{∆\vec{s}}{∆t}\frac{v}{r}\] \[a=v\frac{v}{r}\] \[a=\frac{v^2}{r}\] This is centripetal acceleration. and, \[F = ma\] \[F = \frac{mv^2}{r}\] This is centripetal force.