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## Derivation of the centripetal force

I have mentioned a few times in previous blogs that an object moving in a circle at a constant speed does so because of a force acting towards the centre. We call this force the centripetal force. The force is given by the equation

$F = \frac{ mv^{2} }{ r }$

where $m$ is the mass of the object moving in the circle, $v$ is its speed, and $r$ is the radius of the circle. More correctly, remembering that force is a vector, it should be written

$\boxed{ \vec{F} = - \frac{ mv^{2} }{ |\,\vec{r}\,| } \hat{r} }$

where $|\,\vec{r}\,|$ is the magnitude (size) of the radial vector $\vec{r}$, and $\hat{r}$ is the unit vector in the direction of $\vec{r}$.

But, from where does this formula come?

## The acceleration of an object moving in a circle

The acceleration of any object is defined at the change in its velocity divided by the change in time. Both acceleration and velocity are vectors, so mathematically we can write this as

$\vec{a} = \frac{ \Delta \vec{v} }{ \Delta t }$

We are going to consider an object moving at a constant speed in a circle, as illustrated in the diagram below. In a time $\Delta t$ the object has moved through an angle $\theta$, and its initial velocity $v_{1}$ has changed to $v_{2}$, where the only change in the velocity is its direction. Remember, the velocity is tangential to the radius, so makes a right angle with the radius vector $\vec{r}$. The direction of the radius vector is, by definition, radially outwards.

We will consider an object moving in a circle with a constant speed. It moves through an angle $\theta$ in time $t$.

In my blog about vectors I mentioned that, to combine vectors which have different directions, we need to split the vectors into components, add the components and then recombine the resultant components. The components need to be at right-angles to each other, and usually (but not always) we choose the x and y-directions when the vectors are in two dimensions.

To find the acceleration of our object in this example, we want to find the change or difference in the velocity, that is $\Delta \vec{ v } = \vec{ v_{2} } - \vec{ v_{1} }$. We start by splitting the two vectors into their x and y-components.

$\vec{ v_{1} } = v_{1}\hat{x} + v_{1}\hat{y} \; \; \text{ and } \; \; \vec{ v_{2} } = v_{2}\hat{x} + v_{2}\hat{y}$

Looking at our diagram, we can write

$\vec{ v_{1} } = 0\hat{x} + v\hat{y} \; \; \text{ and } \; \; \vec{ v_{2} } = v\sin(\theta)\hat{x} + v\cos(\theta)\hat{y}$

where $v$ is the speed, the size of the vectors $\vec{ v_{1} } \text{ and } \vec{ v_{2} }$.

The change in the velocity in the x-direction, which we will call $(\Delta v)\hat{x}$ is then just

$(\Delta v)\hat{x} = v\sin(\theta) - 0 = v\sin(\theta)$

Similarly, the change in the velocity in the y-direction, which we will call $(\Delta v)\hat{y}$ is given by

$(\Delta v)\hat{y} = v\cos(\theta) - v = v(\cos(\theta) - 1)$

To correctly calculate the acceleration, we need to find the change in velocity with time as the time interval tends to zero. This means the angle $\theta$ tends towards zero also, and when $\theta$ is very small (and expressed in radians) we can write

$\cos(\theta) \rightarrow 1 \; \; \text{ and } \sin(\theta) \rightarrow \theta$

So we can then write

$(\Delta v)\hat{x} \rightarrow v\theta \; \; \text{ and } (\Delta v)\hat{y} \rightarrow v(1 - 1) =0$

The overall change in velocity, $\Delta \vec{v}$ is then just the change in velocity in the x-direction, $\Delta \vec{v} = v\theta$. The direction of the change in velocity is in the positive x-direction, which as $\theta \rightarrow 0$ is along the radial vector towards the centre of the circle (that is, in the $- \vec{r}$ direction).

However, we can do a substation for the angle $\theta$. Remember, the arc-length, which we shall call $l$ is related to the angle $\theta$ via our definition of the radian (see this blog here), $\theta = l/r$, so we can write

$\Delta \vec{v} = \frac{ vl }{ |\,\vec{r}\,| }\hat{r} \; \text{ (Equation 1)}$

The acceleration $\vec{a} = \Delta \vec{v} / \Delta t$. But the speed $v$ and time $t$ are related via $v = l/t$, so we can write that $l = vt$. Substituting this into equation 1 above gives

$\frac{ vl }{ r } = \frac{ v^{2}t }{ r } \; \text{ (Equation 2)}$

and so the acceleration $\vec{a}$ is given by

$\vec{a} = \frac{ \Delta \vec{v} }{ t } = \frac{ v^{2}t }{ |\,\vec{r}\,| t } = \boxed{ - \frac{ v^{2} }{ |\,\vec{r}\,| } \hat{r} }$

where we have added the minus sign to remind us that the change in velocity, and hence the acceleration, is directed towards the centre of the circle.

If you prefer to think of vectors pictorially, then the direction of $\Delta\vec{v} = \vec{v_{2}} - \vec{v_{1}}$ can be seen from this diagram:

This shows the direction of $\Delta \vec{v} = \vec{v_{2}} - \vec{v_{1}}$, and it is along the radius vector towards the centre of the circle, in the opposite direction to the radius vector $\vec{r}$.

The centripetal force is found by remembering that $\vec{F} = m\vec{a}$ (Newton’s 2nd law), so finally we can write that the centripetal force is

$\boxed{ \vec{F} = - \frac{ mv^{2} }{ |\,\vec{r}\,| } \hat{r} }$

### 9 Responses

1. on 12/03/2014 at 06:07 | Reply prashanth

I follow your blog on a daily basis, its been a wonderful journey getting to know about concepts of physics which i always wanted to do.

• Thank you, I am really pleased you are enjoying reading it. It makes my effort worth it 🙂

2. Thanks for sharing, & for interesting posts on this blog.
My only comment is that the proof is significantly shorter by using the polar coordinates (r, theta) instead of (x, y). Will be glad to share if you are interested.

• Thank you. I know that proof but erred away from it as I was targeting this explanation/derivation at people who’ve never heard of polar coordinates and might be a little scared by them 😉

• I may derive it using polar coordinates in a future blog for completeness 🙂

3. […] I showed in this blog, when an object is moving in a circle it must have a force acting on it, because its velocity (in […]

4. […] and Newton was also the first person to produce an equation to describe gravity. For example, in this blog I showed how we can derive the acceleration felt by a body travelling in a circle, which we call […]

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