Centripetal acceleration of a point

Problem

A point moves in uniform circular motion with speed $v = 4{,}0\;\text{m/s}$ on a circle of radius $r = 2{,}0\;\text{m}$. Calculate the magnitude of the centripetal acceleration.

Solution

Data. $v = 4{,}0\;\text{m/s}$, $r = 2{,}0\;\text{m}$.

Formula. In uniform circular motion the acceleration is purely centripetal: $a_c = \frac{v^2}{r}$

Numerical substitution: $a_c = \frac{(4{,}0)^2}{2{,}0} = \frac{16}{2{,}0} = 8{,}0\;\text{m/s}^2$

Result. $\ev{a_c = 8{,}0\;\text{m/s}^2}$

Links

Topics: Dynamics Concepts: Uniform circular motion · Centripetal acceleration