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

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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