First off, I don't understand the math formatting thing on here, but none of these formulae are too complicated. Take a tire rolling down a hill with angle t and distance d. Modeling it as a hollow shell with no thickness with $I = MR^2$, I got velocity at the bottom $v = \sqrt{gd*sin(t)}$, using energy, which is correct. However, a solid cylinder with $I = \frac{1}{2}MR^2$, I got $\sqrt{2gd*sin(t)}$, which apparently isn't. Why would this be?
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1 Answer
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Your calculations are wrong, Hint:
$\frac{1}{2}I\omega^2 + \frac{1}{2}Mv^2 = Mgd\sin(t)$
$wr=v$
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$\begingroup$ The first one gave me a correct velocity. Also, the potential energy on the right part should have a mass? $\endgroup$– AaronFCommented Nov 13, 2015 at 5:23
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$\begingroup$ Yes, You wont get \sqrt{2gd*sin(t)} if you use the above formula $\endgroup$– CourageCommented Nov 13, 2015 at 5:32
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$\begingroup$ Yeah, you're right. Algebra mistake. Horrible algebra mistake. $\endgroup$– AaronFCommented Nov 13, 2015 at 23:45