Uniform Circular Motion puzzle

[Ludwig]

The position function for a particle moving on a circle (with constant speed) is:
$\vec{r}(t)=\left \langle r\,sin(t), \, r\,cos(t) \right \rangle$
Taking the first and second derivatives,
$\vec{v}(t)=\left \langle -r\,cos(t), \, r\,sin(t) \right \rangle$
$\vec{a}(t)=\left \langle -r\,sin(t), \, -r\,cos(t) \right \rangle$
suggesting that the magnitudes of each are equal (i.e., r = a = v).

However, I frequently see this equation:
$\sum F = ma_{c}=m \frac{v^2}{r}$

...but wouldn't $a_{c}=\frac{v^2}{r}$ simply be equal to r?

If so, why express it in the above form?

 

[Orodruin]

Writing down your expressions, you have implicitly assumed that the angular frequency is 1. This is not always the case and is dimensionally inconsistent as the angular frequency should have units s^-1. Instead, use a general angular frequency $\omega$ such that the argument of the sines and cosines becomes $\omega t$.

 

[Philip Wood]

sin (t) and cos (t) don't make sense. You can't take the sin or cos of something with units. Try looking up sin (10 second) !
t must be multiplied by a constant, $\omega$, with units of s^-1, or rad s^-1.

$\omega$ has an easy physical interpretation. Agreed?

 

[Ludwig]

Writing down your expressions, you have implicitly assumed that the angular frequency is 1. This is not always the case and is dimensionally inconsistent as the angular frequency should have units s^-1. Instead, use a general angular frequency $\omega$ such that the argument of the sines and cosines becomes $\omega t$.

Even with the correct arguments, won't the magnitudes of the vectors still be equal?
I.e., $\sqrt{(\pm rCos(\omega t))^{2}+(\pm rSin(\omega t))^{2}}=\sqrt{(- rSin(\omega t))^{2}+(\pm rCos(\omega t))^{2}} = r$, right?

I'm sure that there is something I'm missing, but I don't know what.

 

[jbriggs444]

what is the first derivative of $cos({\omega}t)$ with respect to t?

 

[Ludwig]

what is the first derivative of $cos({\omega}t)$ with respect to t?

Yes.

 

[Philip Wood]

Jbriggs444 has identified another reason why you're confused. You need to think about the answer to this poster's question. It is not $-sin(\omega t)$.

 

[Ludwig]

Chain rule. Right. Well that clarifies a lot!
So the magnitudes end up being:
$v=\omega r$
$a=\omega^{2}r,$ right?

Excellent. This is a useful result! Thanks for indicating the way out of my confusion.

 

[Philip Wood]

Good!

It's worth remembering that any function (such as sin, cos, ln, exp) that can be expanded as a power series can only have a pure number as argument, otherwise we'd be adding together terms with different units.