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## Derivation of centripetal acceleration using polar coordinates

In this blog, I introduced the idea of polar coordinates as a way to derive the area of a circle. I mentioned at the end of the blog that polar coordinates can also be used to derive the centripetal acceleration, the acceleration an object feels when it moves in a circle. There are a few ways to do this, but I will choose what I think is the easiest way.

## The position of a point on a circle in polar coordinates

Remember, as I mentioned in my blog on deriving the formula for the area of a circle, we can express any point on a circle, which is usually given in terms of its x and y coordinates, instead in terms of the radius vector $\vec{r}$ and the angle the radius vector makes with the x-axis $\theta$. The point $(x,y)$ on the circle can be written in terms of the radius $r$ and the angle $\theta$. We can write that $x = r \cos \theta$ and $y = r \sin \theta$

When we do this, we can write the x-direction vector to this point as $\vec{x}=\vec{r} \cos \theta$ and the y-direction vector as $\vec{y}=\vec{r} \sin \theta$.

## The velocity of circular motion in polar coordinates

The velocity of an object undergoing circular motion is given by $\vec{v}$, the direction of which is tangential to the the radius vector $\vec{r}$, as I described in my blog on the direction of the angular velocity vector here, and shown in the figure below. The velocity of an object moving in a circle is in a direction which is at right angles to the radius. We can split that velocity up into its x and y-components

We can write the x and y-coordinates of the point in terms of their polar coordinates $r \text{ and } \theta$; that is $\vec{x} = \vec{r} \cos \theta \text{ and } \vec{y} = \vec{r} \sin \theta$

But, as I mentioned in this blog here, we can write $\theta$ in terms of the angular velocity $\omega$. From the definition of the angular velocity, which is the angle moved per unit time, we have $\omega = \theta / t$ and so rearranging we can write $\theta = \omega t$. We can then re-write the x and y-coordinates in terms of the polar coordinates, but using $\omega t$ instead of $\theta$. $\vec{x} = \vec{r} \cos (\omega t) \text{ and } \vec{y} = \vec{r} \sin (\omega t)$

We can then use differentiation to find the x and y-components of the velocity $\vec{ v_{x} } \text{ and } \vec{ v_{y}}$. $\vec{ v_{x} } = \frac{ d \vec{x} }{ dt } = \frac{ d }{ dt } \vec{r} \cos (\omega t) = -\omega \vec{r} \sin (\omega t)$ $\vec{ v_{y} } = \frac{ d\vec{y} }{ dt } = \frac{ d }{ dt } \vec{r} \sin (\omega t) = \omega \vec{r} \cos (\omega t)$

To find the components of the acceleration, $\vec{ a_{x} } \text{ and } \vec{ a_{y} }$ we differentiate again, so we have $\vec{ a_{x} } = \frac{ d \vec{ v_{x} } }{ dt } = \frac{ d }{ dt } -\omega \vec{r} \sin (\omega t) = -\omega^{2} \vec{r} \cos (\omega t) = -\omega^{2}\vec{x}$

and $\vec{ a_{y} } = \frac{ d \vec{ v_{y} } }{ dt } = \frac{ d }{ dt } -\omega \vec{r} \cos (\omega t) = -\omega^{2} \vec{r} \sin (\omega t) = -\omega^{2}\vec{y}$

To find the magnitude (size) of the acceleration $| \vec{a} |$ we remember that it is given by $| \vec{a} | = \sqrt{ \vec{ a_{x}^{2} } + \vec{ a_{y}^{2} } }$

so $| \vec{a} | = \sqrt{ (-\omega^{2} \vec{x} )^{2} + (-\omega^{2} \vec{y} )^{2} } = \sqrt{ \omega^{4}x^{2} + \omega^{4}y^{2} } = \omega^{2} \sqrt{ x^{2} + y^{2} } = \omega^{2}r$

(because $\sqrt{ x^{2} + y^{2} } = r$).
To find the direction of the acceleration, we can use a diagram of the two components $\vec{ a_{x} } \text{ and } \vec{ a_{y} }$. The direction of the acceleration can be found from this vector diagram, and is found to be in the negative radial direction, which means towards the centre of the circle

As can be seen from the diagram above, the direction of the centripetal acceleration is in the $- \vec{r}$ direction, which means towards the centre of the circle.

So, in terms of the angular velocity $\omega$ we have $\boxed{ \vec{a} = - \omega^{2} \vec{r} }$

To get the expression we had before in this blog, we need to remember the relationship between the angular velocity and the linear velocity $\vec{v}$. In terms of vectors, $\vec{ v} = \vec{ \omega } \times \vec{ r }$ (where the cross means we are talking about the vector product), but as we have $\omega^{2}$ in our expression for the centripetal acceleration, we don’t need to worry about its direction, only its magnitude, so we can simply write $v = \omega r \rightarrow \omega = \frac{v}{r} \rightarrow \omega^{2} = \left( \frac{v}{r} \right)^{2}$

and so we can write $\vec{a} = - \left( \frac{ v }{ r } \right)^{2} \vec{r}$

and remembering that the unit vector $\hat{r}$ can be written as $\hat{r} = \frac{ \vec{r} }{ | \vec{r} | } \text{ so } \vec{r} = \hat{r} | \vec{r} |$

we can write $\vec{a} = - \left( \frac{ v }{ r } \right)^{2} \cdot \hat{r} | \vec{r} | = - \frac{ v^{2} }{ | \vec{r} | } \hat{r}$