Centripetal Force – Definition, Formula and Unit

Centripetal force is the force required to move a body uniformly in a circle. This force acts along the radius and towards the center of the circle.

When a rigid body moves along a circular path with uniform speed, its direction changes continuously. Due to inertia, at every point of the circular path, the body tends to move along the tangent to the circular path at that point. According to Newton’s second law of motion, a change in the direction of motion of the body can take place only if some external force acts on the body. Thus, to move a body along a circular path, an external force is required, which will deflect the body from its straight path to the circular path at every point of the path.

It is this external force which is called centripetal force.

Centripetal Force Formula

As the body is moving in a circular path, there is a continuous change in the direction of motion. The continuous change in the direction of motion leads to a change in the velocity of the body, and hence it undergoes an acceleration, called centripetal acceleration or radial acceleration given by

$\displaystyle a=\frac{{{{v}^{2}}}}{r}=r{{\omega }^{2}}$

where $\displaystyle v$ is linear velocity, $\displaystyle \omega $ is angular velocity of the body and $\displaystyle r$ is the radius of the circular path.

It is directed along the radius of the circular path and towards the center.

Since, according to Newton’s second law of motion $\displaystyle F=ma$

Therefore,

Centripetal force = mass × centripetal acceleration

i.e. $\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle F=\frac{{m{{v}^{2}}}}{r}=mr{{\omega }^{2}}}}$

The centripetal force acts along the radius and is directed towards the center of the circular path.

Unit of Centripetal Force

Since, it is a type of force which acts only when an object is moving in a circular path, it has the same unit as that of force i.e. N (Newton). Dimensional formula is [M¹ L¹ T^-2]

Example

A stone of mass 1.3 kg tied to the end of a string in a horizontal plane is whirled round in a circle of radius 1 m with a frequency of 40 rev min\(^{-1}\). What is the tension in the string? What is the maximum speed with which the stone can be whirled around, if the string can withstand a maximum tension of 200 N?

Solution:

Here, $ \displaystyle M=0.3\text{kg}$, $\displaystyle r=1\text{m}$, $\displaystyle v=40\text{rev mi}{{\text{n}}^{{-1}}}$ $\displaystyle =40\text{rev}\times {{(60\text{s})}^{{-1}}}$ $\displaystyle =\frac{2}{3}\text{r}\text{.p}\text{.s}$

$ \displaystyle \omega =2\pi v=2\pi \times \dfrac{2}{3}$ $\displaystyle =\dfrac{{4\pi }}{3}\text{rad }{{\text{s}}^{{-1}}}$

The tension in the string provides the necessary centripetal force. Therefore,

$\displaystyle T=\dfrac{{M{{v}^{2}}}}{r}=Mr{{\omega }^{2}}$

$\displaystyle =0.3\times 1\times {{\left( {\dfrac{{4\pi }}{3}} \right)}^{2}}=5.264\text{N}$

Let $\displaystyle {{v}_{{\max }}}$ be the maximum speed at which tension would become 200N. Then,

$\displaystyle \begin{array}{l}\dfrac{{Mv_{{\max }}^{2}}}{r}=200\\\Rightarrow {{v}_{{\max }}}=\sqrt{{\dfrac{{200\times r}}{M}}}\\\Rightarrow {{v}_{{\max }}}=\sqrt{{\dfrac{{200\times 1}}{{0.3}}}}\\=25.82\text{ m}{{\text{s}}^{{-1}}}\end{array}$