This has been one of my biggest hurdles in VCE Physics, as I was not able to derive the formula. I have been told that
truetears has posted a quite elegant solution involving similar triangles, though I cannot find it. [If someone could link/copy it, that'll be great =) ]
Anyhow, bored in my lecture today, here's a derivation that is slightly beyond the level of Specialist but some of you still may enjoy.
Let
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\tilde{r})
be the radius vector pointing away from the centre with magnitude
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?r)
, and
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\theta)
is the angle it makes with the horizontal. Let
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\tilde{i})
be a unit vector pointing in the direction of the horizontal axis, and
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\tilde{j})
be a unit vector pointing in the direction of the vertical axis.
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\implies \tilde{r} = r (\cos \theta \tilde{i} + \sin \theta \tilde{j} ))
Let
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\hat{r})
be the unit vector of
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\tilde{r})
,
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\hat{r} = \cos \theta \tilde{i} + \sin \theta \tilde{j} )
Let
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\tilde{v})
be the velocity vector with the magnitude
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?v)
. Assuming the motion is observed from a point such that the object is traveling in a counter-clockwise direction. Since
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\tilde{v})
is perpendicular to the radius vector, it makes an angle of
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\theta + \frac{\pi}{2})
with the horizontal axis.
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\implies \tilde{v} = v \left(\cos \left( \theta+\frac{\pi}{2}\right) \tilde{i} + \sin \left( \theta+\frac{\pi}{2}\right) \tilde{j} )\right)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\begin{align*}<br />\implies \frac{d\tilde{v}}{d\theta} &= v \left(-\sin \left( \theta+\frac{\pi}{2}\right) \tilde{i} + \cos \left( \theta+\frac{\pi}{2}\right) \tilde{j}\right) \\<br />&= v (-\cos \theta \tilde{i} - \sin \theta \tilde{j}) \\<br />&= -v \hat{r} \\ \end{align*})
Using the arc-length
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?r \theta = L = v \cdot t)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\implies \theta = \frac{vt}{r})
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\implies \frac{d\theta}{dt} = \frac{v}{r})
Let the acceleration vector be
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\tilde{a})
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\implies \tilde{a} = \frac{d\tilde{v}}{dt} = \frac{d\tilde{v}}{d\theta} \cdot \frac{d\theta}{dt} = -v \hat{r} \cdot \frac{v}{r} = -\frac{v^2}{r}\hat{r})
Magnitude of acceleration:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\| \tilde{a} \| = \frac{v^2}{r})
The direction of acceleration (
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\tilde{a})
) is
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?-\hat{r})
, opposite the radius vector, i.e. in towards the centre.