Why does circular motion have a resultant force acting towards the centre?

Question

Why does the resultant force in centripetal force act towards the center of the circle in uniform circular motion and not at any other point, why only the center, always?

Answer 1

Well, in circular motion, there is a particle moving straight and there is a centripetal force $F_{cp} = \frac{mv^2}{r}$.

This centripetal force changes the direction of motion such that it is always tangential to a circle with radius $r$.

As for why it acts towards the center, it is just convention just like we have defined normal force to always act perpendicular to the surface.

Answer 2

The force $\mathbf F\propto \mathbf v'$ can't have a component that is along the velocity vector because that means the magnitude of the velocity (aka speed) changes.

Since the velocity (or motion) is tangential to the circular path, it's perpendicular to the radius and that's where the acceleration and force point; along the radius, also known as towards the center.

You can also look at $$\mathbf v\cdot \mathbf v=C\\2\mathbf v\cdot\frac{d\mathbf v}{dt}=0\\\left|\mathbf v\right|\left|\frac{d\mathbf v}{dt}\right|\cos\Psi=0\\\Psi =\frac\pi 2$$ where $\Psi$ is the angle between the vectors $\mathbf v$ and $\frac{d\mathbf v}{dt}$ and the first equation assumes $\mathbf v\cdot\mathbf v=\|\mathbf v\|^2,$ which is the square of the speed, is a constant. We conclude that the vectors $\mathbf v$ and $\frac{d\mathbf v}{dt}$ are perpendicular.

You can, if you wish, use a similar arguement to show that $\mathbf v=\frac{d\mathbf x}{dt}$ is tangential to the circle in the first place, i.e is perpendicular to the radial position vector $\mathbf x$ since you know $\mathbf x$ has a constant magnitude which is the length of the radius of the circular path.