# How is torque equal to moment of inertia times angular acceleration divided by g?

How is the following relation true

$$\tau = \large\frac{I}{g} \times \alpha$$

where $\tau$ is torque,

$I$ is moment of inertia,

$g= 9.8ms^{-2}$,

and $\alpha=$ angular acceleration.

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In older texts---especially older engineering texts---you find some strange constructions related to use of "pounds force" and "pounds mass" and the tabulation of results in units that would be strange to eyes raised in a nice clean SI tradition. –  dmckee May 13 '13 at 19:16

## 1 Answer

This is only true for engineering units which have $I$ in ${\rm lbf\,in^2}$. In the metric system the units of $I$ are ${\rm kg\, m^2}$. So to convert force ${\rm lbf}$ to mass you divide by $g$.

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Bloody "pounds mass" again. –  dmckee May 13 '13 at 19:14
I don't doubt that this is the answer, but I have to confess to being a bit bewildered by it. How do you use a moment of inertia defined in lbf $in^2$? What does it even mean? –  Dave May 13 '13 at 19:18
You use it like you see in the original posting. –  ja72 May 13 '13 at 19:28
So what merit is there in definition of moment of inertia that has to be divided by the same constant every time it's used? Was it part of an attempt to avoid ambiguity by not using lbm anywhere? –  Dave May 13 '13 at 20:47
I don't disagree, it is idiotic, but it is a result expressing angular momentum in engineering units. –  ja72 May 14 '13 at 12:06