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?

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    $\begingroup$ a minute point: in order for the ball to roll there needs to be friction. In fact, the condition that there is friction between the ball and the plane such that there is no slipping is often useful in solving problems about balls/disks rolling down slopes. $\endgroup$ – user27182 Sep 4 '12 at 17:34
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    $\begingroup$ Dolev, where did you get that "simplified equation" from? $\endgroup$ – Yrogirg Sep 5 '12 at 6:22
  • $\begingroup$ $I = \frac{2}5 mr^2$ is for a solid sphere. A basketball full of air is closer to a thin spherical shell $I = \frac{2}3 mr^2 –$ $\endgroup$ – user20716 Feb 8 '13 at 2:19

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 $2/3 \space g \space sin\theta$).

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    $\begingroup$ I = 2/5 mr^2 is for a solid sphere. A basketball full of air is closer to a thin spherical shell (I = 2/3 mr^2) $\endgroup$ – Tim Goodman Sep 4 '12 at 16:52
  • $\begingroup$ Oops, yes of course :-) Thanks. Hang on though, wouldn't that make the equation $3/5 \space g \space sin\theta$? $\endgroup$ – John Rennie Sep 4 '12 at 16:57
  • $\begingroup$ Yes, I agree. I don't know where the asker got 2/3. Thinking of a disc, maybe? $\endgroup$ – Tim Goodman Sep 4 '12 at 17:42

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.


protected by Qmechanic Feb 8 '13 at 3:06

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