Dgdrxrt

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

Further, there is stated that $vecv(t)$ itself cannot change, what I think is nonsense. Since:

$$beginpmatrix 1 \ 0 \ 0 endpmatrix + beginpmatrix 0 \ 1 \ 0 endpmatrix times t = beginpmatrix 1 \ t \ 0 endpmatrix $$

But maybe I am just wrong.
Now, I am further wondering in how far $vecF(t) cdot vecv(t) = 0$ 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:

beginalign*
&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 $cos(a - b) = cos a cos b + sin a sin b)$
endalign*

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 $F = fracdvdt$.

beginalign*
&F = fracdvdt \
&(r(t) cos(phi(t)), r(t) sin(phi(t)) =
fracd 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(-stheta'(t) cos (phi(t)) , -stheta'(t) sin (phi(t)) right)quad \
%
endalign*
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 $$ vecF = mdvecvover dt =mcdot vecv'$$

so $$vecv'cdot vecv =0implies (vecv^2)' = 0 implies vecv^2 = constant$$

So $$|vecv|^2 = constantimplies |vecv| = constant_2$$

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

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 $vecs$, you also have that $vecs$ and $vecv$ are perpendicular. Therefore by the same euristic argument as before we can say that $vecs$ will only change its direction and not its module. This is exactly like saying that the trajectory is a circle