Formula for a ball rolling down an Inclined Plane Suppose we set up an experiment where we have an inclined ramp, and a spherical basketball. If we were to assume the ball to be perfectly round, and rolls down in a vertical manner and the situation friction less. The simplified equation that would be used would be $\frac{2}{3} G x \sin\theta$. I'm wondering why is the $2/3$ the constant in the equation. Could it be the force of gravity vs the force of the ramp pushing the ball up? or could it be that spherical objects follow a constant k rate?
 A: If you have an object sliding down a frictionless ramp then after it has fallen some vertical distance $h$, the potential energy has turned into kinetic energy:
$$ mgh = \frac{1}{2}mv^2 $$
With some minor manipulation this gives you the acceleration $a = g \space sin\theta$. With a ball rolling down the plane, and assuming there is no slipping between the ball and the plane, the potential energy turns into translational kinetic energy and rotational kinetic energy so:
$$ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $$
So you have the extra term to consider. Use $v = r\omega$ and $I = 2/5 \space mr^2 $ and do the same manipulation as before and you get $a = 5/7 \space g \space sin\theta$ (not $2/3 \space g \space sin\theta$).
A: This is either you mean a solid cylinder rolling down an inclined plane with angle $\theta$ between the plane and the horizontal or linearized approximation on experimental data. I assume you mean the second one.
Theoretically, $a=\frac{2}{3}g\sin\theta$ is suggested for a nonlinear relation between $a$ and $\theta$. A convenient method of testing $a=\frac{2}{3}g\sin\theta$ is using linearized approximation to make an easier analysis possible. Plot a graph of experimental data, $a$ versus $g\sin\theta$. Consider $a$ as $y$-axis and $g\sin\theta$ as $x$-axis. The plotted data will perform a linear equation $y=mx+c$. In your case, it will be $a=m\cdot g\sin\theta + c$, where $m$ is the slope of the graph and $c$ is the $y$-intercept. The expected value of the slope is $\frac{2}{3}$. Unfortunately, I cannot provide you the proof about this value but I have a strong feeling that this can be obtained by using Taylor series approximation and finite difference method for a second-order expansion.
