# Why does $\vec{F(t)} \cdot \vec{v(t)} = 0$ lead to a circular motion?

Here is a mathematical proof that any force $$F(t)$$, which affects a body, so that $$\vec{F(t)} \cdot \vec{v(t)} = 0$$, where $$v(t)$$ is its velocity cannot change the amount of this velocity.

Further, there is stated that $$\vec{v(t)}$$ itself cannot change, what I think is nonsense. Since:

$$\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \times t = \begin{pmatrix} 1 \\ t \\ 0 \end{pmatrix}$$

But maybe I am just wrong. Now, I am further wondering in how far $$\vec{F(t)} \cdot \vec{v(t)} = 0$$ leads to a circular motion and how to proof this using Netwons laws and calculus? Can you explain why a circle is created, if you let the time tick?

Let the velocity vector be defined as $$v(t) = (s \cos( \theta(t)), s \sin (\theta(t))$$, where $$\theta(t)$$ is a time varying quantity. Note that $$s$$ is a constant, since we have already proved that the speed of the velocity vector does not change.

Let the force vector be defined as $$F(t) = (r(t) \cos(\phi(t)), r(t) \sin(\phi(t))$$ where $$r(t)$$ and $$\phi(t)$$ are time varying quantities.

Now, we know that $$F . v = 0$$. Hence:

\begin{align*} &r(t) s \left[\cos(\theta(t))\cos(\phi(t)) + \sin(\theta(t))\sin(\phi(t)) \right] = 0 \\ &r(t)s[\cos(\theta(t) - \phi(t))] = 0 \quad \text{(since \cos(a - b) = \cos a \cos b + \sin a \sin b)} \end{align*}

Let us assume that $$r(t), s \neq 0$$ for the moment. If they are, then the problem reduces to trivial cases. So, this now means that $$cos(\theta(t) - \phi(t)) = 0$$, or $$\theta(t) - \phi(t) = \pi /2$$. Hence, $$\theta(t) = \pi/2 + \phi(t)$$.

Next, let us assume the mass of the body is 1 (otherwise, we will need to carry a factor of $$m$$ everywhere which is annoying and adds no real insight), and hence $$F = \frac{dv}{dt}$$.

\begin{align*} &F = \frac{dv}{dt} \\ &(r(t) \cos(\phi(t)), r(t) \sin(\phi(t)) = \frac{d \left(s \cos (\theta(t)), s \sin (\theta(t)) \right)}{dt} \\ % &(r(t) \cos(\phi(t)), r(t) \sin(\phi(t)) = \left(-s \sin (\theta(t)) \theta'(t), s \cos (\theta(t)) \theta'(t) \right) \quad \text{(Differentiating with respect to t)} \\ % &(r(t) \cos(\phi(t)), r(t) \sin(\phi(t)) = \left(-s \sin (\pi/2 + \phi(t)) \theta'(t), s \cos (\pi/2 + \phi(t)) \theta'(t) \right)\quad \text{(\theta(t) = \pi/2 + \phi(t))} \\ % &(r(t) \cos(\phi(t)), r(t) \sin(\phi(t)) = \left(-s\theta'(t) \cos (\phi(t)) , -s\theta'(t) \sin (\phi(t)) \right)\quad \\ % \end{align*} Comparing the LHS and the RHs, we conclude that $$r(t) = -s \theta'(t)$$.

If we are interested in uniform circular motion, then we would set $$\theta'(t)$$ to a constant and proceed to solve the system as described on wikipedia

For non-uniform circular motion, I don't know off-hand how to solve the system of equations, but I presume it is possible. I'll update the answer once I go look it up

Since $$\vec{F} = m{d\vec{v}\over dt} =m\cdot \vec{v}'$$

so $$\vec{v}'\cdot \vec{v} =0\implies (\vec{v}^2)' = 0 \implies \vec{v}^2 = constant$$

So $$|\vec{v}|^2 = constant\implies |\vec{v}| = constant_2$$

So the magnitude of velocity is constant, but not the velocity it self.

• Sure, but this does not explain the circle which is created by $\vec{s}(t)$, does it? – TVSuchty Mar 30 at 10:44
• In the other prove s is defined to be the velocity squared. In physics generally the position is meant... – TVSuchty Mar 30 at 11:10
• Well, I'm not sure, but I think that what there was said s^2=vv is not true. It sholud be v =s ′, no? – Maria Mazur Mar 30 at 11:14
• I think I answered this already in my last comment. I assume a merge-conflict here... – TVSuchty Mar 30 at 11:16
• Any curve can be walked through by constant speed. – Berci Mar 30 at 11:34

Try to think like this: the component of the force parallel to the velocity changes the module of the velocity, while component perpendicular to the velocity changes its direction. Since you have only a perpendicular component, the velocity will stay constant in module while changing its direction. Now, if your force rotationally symmetric and is parallel to $$\vec{s}$$, you also have that $$\vec{s}$$ and $$\vec{v}$$ are perpendicular. Therefore by the same euristic argument as before we can say that $$\vec{s}$$ will only change its direction and not its module. This is exactly like saying that the trajectory is a circle