# 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
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 $g$.
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