Question #d076b

Jul 23, 2016

$F = \frac{m {v}^{2}}{r}$

or

$\vec{F} = - \frac{m {v}^{2}}{r} \setminus \hat{r}$

Explanation:

There's a nice vector way to do this for a particle moving in a circle of constant radius with position $\vec{r} \left(t\right)$ at uniform speed $| \vec{v} \left(t\right) |$

we can say that

${v}^{2} = \vec{v} \cdot \vec{v} = c o n s t$

so differentiating wrt time using product rule for the vectors

$\vec{\dot{v}} \cdot \vec{v} + \vec{v} \cdot \vec{\dot{v}} = 0 = 2 \vec{v} \vec{\dot{v}} q \quad \star$

and vecause
${r}^{2} = \vec{r} \cdot \vec{r} = c o n s t$

we can do the same
$\vec{\dot{r}} \cdot \vec{r} + \vec{r} \cdot \vec{\dot{r}} = 0 = 2 \vec{r} \vec{\dot{r}} q \quad \triangle$

$\triangle$ amounts to $\vec{r} \cdot \vec{v} = 0$
$\star$ amounts to $\vec{v} \cdot \vec{a} = 0$

this means, ignoring obvious trivial conclusions, that $\vec{r}$ and $\vec{v}$ are perpendicular, which we already know from the circular motion, but also that $\vec{v}$ and $\vec{a}$ are perpendicular

which means that $\vec{r}$ and $\vec{a}$ are pointing in the same direction or opposite directions which means we already know that the particle is accelerating radially inward out outward. so there must be a radial force acting to cause the acceleration - Newton"s Second Law.

if we differentiate $\triangle$ again using the product rule, we get

$\vec{\dot{r}} \cdot \vec{v} + \vec{r} \cdot \vec{\dot{v}} = 0$
or
$\vec{v} \cdot \vec{v} + \vec{r} \cdot \vec{a} = 0$
or
$\vec{r} \cdot \vec{a} = - {v}^{2} q \quad = | \vec{r} | | \vec{a} | \cos \psi \circ$

so we now conclude from the definition of the dot product that $\psi = - \pi$, acceleration is inwardly radial, and in scalar terms $a = {v}^{2} / r$

So from Newton's Second Law

$F = \frac{m {v}^{2}}{r}$

or

$\vec{F} = - \frac{m {v}^{2}}{r} \setminus \hat{r}$