Why is the relationship between centripetal force and velocity straight, in circular motion? I don't understand why the graph of force vs velocity^2 is a straight line in uniform circular motion. The equation for force in UCM is $F=m\frac{v^2}{r}$, I thought the graphs of such equations would be quadratic as the highest degree is 2. What am I missing to fully grasp the relationship between the two?
 A: This is simpler than you think. You are quite correct that:
$$ F = \frac{m}{r}v^2 $$
So:
$$ F \propto v^2 $$
and a graph of $F$ against $v$ is a parabola. So far so good.
But now let's define a new variable $u = v^2$ so our equation becomes:
$$ F = \frac{m}{r}u $$
Now $F \propto u$ and if we draw a graph of $F$ against $u$ it will obviously be a straight line. You say:

I don't understand why the graph of force vs velocity$^2$ is a straight line in uniform circular motion.

and it's because when you plot $v^2$ on the $x$ axis, instead of just $v$, you are drawing the graph of $F$ against $u$ that I described above and it's a straight line.
A: The axis are different.
If you use F, and v, on the x and y axis, you will expectedly get a quadratic curve.
But, because the graph is a Force--Velocity-squared one, the bottom (x) axis uses $v^2$ instead of $v$.
So now, the graph is straight, because the equation being plotted uses linearly transformed values of the axes (the are no longer any exponentials, $ a \cdot v^2$ is a simple multiplication of the x axis, not raised to a power.
If you made the bottom x axis $v^4$, you would get a square-root-looking graph, you can guess why.
A: The graph between F (x-axis) and V (y-axis) shows a a parabolic relationship where F∝V. Therefore, even though an increase in force impact the increase in velocity of the circular motion, the distribution is not linear.
