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?

  • 1
    $\begingroup$ force and velocity are both vectors not scalars, which is why they should be straight, they both constantly change direction at the same rate $\endgroup$ Mar 26, 2017 at 10:10
  • $\begingroup$ then why is the force vs velocity, without velocity being squared result in a curved line? $\endgroup$
    – Son Jerm
    Mar 26, 2017 at 10:15

3 Answers 3


This is simpler than you think. You are quite correct that:

$$ F = \frac{m}{r}v^2 $$


$$ F \propto v^2 $$

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.


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.


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.


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