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At first, I thought the value of $g$ ($9.8m/s^2$) could be determined simply by placing a ball at the top of a ramp at a known height. The ball was released with no initial velocity, and the final velocity after exiting the ramp was recorded with video tracking software.

I simply used the equation for potential energy, $E = mgh$ and kinetic energy $\frac12 mv^2$ to find the value of $g$.

However, the ball is rotating, which I was told affects it, and is probably why my value is so low: I get $5 m/s^2$.

Is there a better equation I can use with rotational inertia etc.? I don't really know anything about it.

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For a rotating ball you should use for energy $$E=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2+mgh$$ where $I$ is the moment of inertia which for a sphere is $$I=\frac{2}{5}mr^2$$ where $r$ is the radius of the sphere, and $\omega$ it's his angular velocity which is related to the velocity of the center of mass via the equation $$\vec{r}\times \vec{\omega}=\vec{v}$$ in you case if the sphere moves along a single direction you may write $\omega=vr$ where $r$ is the radius, maybe you also want to check if friction is a relevant factor. If not you will get $$g=\frac{7}{10}\frac{v^2_{final}}{l\sin(\alpha)}$$ where l is the distance the sphere travelled on the plane and $\alpha$ is the angle of inclination of the plane

$UPDATE$ (due to comment)

In the presence of friction you will have a relation of the type: $$E(t)-E(0)=\epsilon(t)\rightarrow \frac{7}{10}mv^2(t)-mgl\sin(\alpha)=\epsilon(t)$$ energy is not conserved and the difference between initial and final energy is an unkown function of time,thus you will have $$v^2(t)=\frac{10}{7}g\sin(\alpha)l(t)+\frac{10\epsilon(t)}{7m}$$ so if you plot $v^2$ for different values of $l(t)$ (basically at different times) you should get a line with coefficient $\frac{10}{7}g\sin(\alpha)$ if the contribution of friction, which i stress is an unkown funtion of time, it's strong than it will be probably something else than a line since $\epsilon=\epsilon(t,l(t),v(t),..)$ you dont really know, if you get a line it might not be crossing the origin since for weak it might be you still have a contribution from friction, but in the case of a linear relation you can calculate the angular coefficent $$\frac{10}{7}g\sin(\alpha)$$ to estimate $g$

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  • $\begingroup$ It might be worth pointing out that a measurement like this is best done by plotting A against B, and looking for the slope. In this case, I think (check this) that if you plot $v(t)^2$ as a function of $x(t)$ (measured along the slope), the slope of the straight line you get is $\frac{10 \sin\alpha}{7}$. Multiple points along the slope greatly improve the confidence in the measurement and method (like - if you don't get a straight line through the origin you know you have a problem; for example it will tell you that friction is indeed a problem). $\endgroup$
    – Floris
    Feb 16, 2015 at 19:51
  • $\begingroup$ I did several trials and am going to graph them all $\endgroup$
    – Skyler 440
    Feb 16, 2015 at 21:25
  • $\begingroup$ what did you find? $\endgroup$
    – Fra
    Feb 16, 2015 at 23:02
  • $\begingroup$ The issue is that I don't have the angle or the length of the plane and I only have the data from the trials including the ball mass, diameter, initial height and final velocity $\endgroup$
    – Skyler 440
    Feb 17, 2015 at 2:04
  • $\begingroup$ Consider that $h(t)=h_0-l(t)sin(\alpha)$ so if you can measure the height at all times you should be able to workout $l(t)$ if you can't do that i think that metod is useless $\endgroup$
    – Fra
    Feb 17, 2015 at 12:32

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