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This is actually from my physics class, but it's the algebra that's the problem. It's been driving me insane. So we're doing torque problems.

$\Sigma\tau = I \alpha $

We're routinely replacing the angular acceleration (alpha) with its equivalency, $a/R$ so that $ \alpha = a/R$, where $a$ is the tangential acceleration and $R$ is the radius.

The moment of inertia, $I$, includes an $R^2$ term: some fraction times mass times radius squared.

My professor always cancels out all the R terms here, leaving just a fractional $ma$, mass times acceleration. This is the part I don't understand. Why isn't there an R left over in the numerator?

GraphicsMuncher
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user115991
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  • What do you mean with a/R? a being what? Check the units, though. – Lessa121 Dec 15 '13 at 19:45
  • Could you provide a little more detail regarding the equations? It'd probably help with pinpointing what's happening for us who are not familiar with those physics formulae. Thanks. – Sujaan Kunalan Dec 15 '13 at 19:47
  • I've checked my torque problems, and in most of them, I always have an R left in the expresion. Be careful though, that´s the formula for a fixed axis, maybe you are missing another term... I don't want to confuse you. – Lessa121 Dec 15 '13 at 19:54
  • I edited with more info. It's not a units thing, it's a factors thing. How can he eliminate all the R terms when there are two in the numerator (R squared) and one in the denominator (R)? – user115991 Dec 15 '13 at 19:55
  • That's what I´m telling you, there is no way and you can see it with the units. Torque: $m.N$ (meters per newton), Inertia: $kg.m^2$ and then angular acceleration: $rad/s^2$ So you SHOULD have another factor in meters – Lessa121 Dec 15 '13 at 19:59
  • I think I figured it out... his handwriting is utterly illegible. He's setting the net torque equal not only to the inertia times angular acceleration (Ia) but ALSO to force times radius. That extra R is what cancels the second one out. – user115991 Dec 15 '13 at 20:12

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