Torque considering the mass of the lever arm Consider a mass pulling on a lever.
The well known formula for torque is: $\vec{M}=\vec{r} \times \vec{F} \space$
Is there a general formula that also considers the mass and geometry of the lever itself?
When the mass pulling on the lever is much greater than the mass of the lever itself, the error should be minimal, right?
How would you proceed calculating this?
 A: Great question. What you are looking for is called moment of inertia and is defined as $I=\int_0^M r^2dm$. Where r is the distance to the pivot point. Intuitively this formula says we are summing up each small chunk of mass dm in the lever arm weighted by the radius (to the pivot) of said chunk of mass squared. For a simple lever bar with uniform mass which pivots on one end, the moment of inertia is simply $I= {1\over3}ML^2$, where L is the length of the rod. Then the formula for torque is simply $\vec M = I \vec \alpha$. where alpha is the angular acceleration Notice the similarity to F=ma.
Your are mostly correct in your second question, but one needs to consider the geometry of the lever, not just its mass, then one can compare the relative torque due to each object and see if it will be negligible.
A: The answer above seems to assume you are concerned about the movement of the system, and whether the mass of the lever is accelerated (in angular acceleration). If that is the question then +1 to above. The Moment of Inertia of the lever simply adds to that of the rotating load - taken about the same axis. Look up the 'Parallel Axis Theorem' to transfer MOI from the lever's CoM to the body's.
However if you are asking simply about the torque then the answer is different. It depends on the geometry of the system.
For instance if your applied force is a mass hanging on a horizontal lever then the mass of the lever can be considered to add a torque equal to the weight (=mg) of the lever acting through its centre of mass.
If however the lever is vertical (and the applied force is not) then it has no effect on the resultant torque.
A: Your torque formula,
$$\vec M=\vec r\times \vec F,$$
is the general formula for torque.
What you are asking to when focusing on mass and geometry is not the torque, but the effect of the torque. If you apply the same torque (calculated from the above formula) on a bike wheel and on an engine flywheel, then the result - the induced rotational motion - will be different.
And is there a formula for the effect from a torque on a particular object? Yes, there sure is. When you apply a force $\vec F$, then its effect - represented by the acceleration $\vec a$ - is found via Newton's 2nd law:
$$\sum \vec F=m\vec a.$$
When you apply a torque $\vec M$, then its effect - represented by the angular acceleration $\vec\alpha$ - is found via the rotational equivalent of Newton's 2nd law:
$$\sum \vec M=I\vec \alpha.$$
In both cases there is inertia resisting the effect. Against forces, the inertia is mass $m$. Against torques, the inertia is the rotational equivalent of mass, called moment-of-inertia $I$.
And this moment-of-inertia can indeed be calculated from mass and geometry, so your guess was perfectly correct. (In general, the rotational equivalents of "normal" properties are typically calculated by including geometry in some way. Just like the torque formula above, which is simply the force with the lever arm direction and length taken into account.) Basically, the moment-of-inertia sums up all points that the object consists of and takes into again the mass of each point as well as distance $d$ from a chosen rotational centre:
$$I=\int d^2\,\mathrm dm.$$
The tougher flywheel is not just tougher to rotate with an applied torque because of its mass but even more because of how far that mass is located from the rotational midpoint.
A: We do this sort of calculation in statics all the time. Consider a cantilever beam with total mass $m$ and length $\ell$ and try to find the supporting moment needed on one end.
You can ignore any external loading other than its own weight since any such loading can be easily accounted for with $r \times F$ as mentioned in the question.
Problem Definition

Determine the support force $S$ and equipollent moment $M$ of a beam with distributed load $w = \frac{m g}{\ell}$ equal to the weight of the beam.
Analytical Approach
Consider each slice of the beam ${\rm d}x$ located at a distance $x$ from the support and add up all the contributions to the supporting force ${\rm d}S$ and ${\rm d}M$ moment.

$$ \left. \begin{aligned}
 {\rm d}S & = w\,{\rm d}x \\ \\
 {\rm d}M & = x\,(w\,{\rm d}x)
\end{aligned} \right\} \begin{aligned}
 S & = \int_0^\ell w\,{\rm d}x  = w \ell\\
 M & = \int_0^\ell x\,(w\,{\rm d}x) =w \tfrac{\ell^2}{2}
\end{aligned} $$
and since the uniform distributed load is $w = m g/\ell$ the result is
$$ \begin{aligned} S &= m g \\ M & = \tfrac{\ell}{2} m g \end{aligned} $$
Notice that the result is equivalent to placing the entire weight of the beam $m g$ at the center of mass $\ell/2$.
General approach.
In statics, if there is a general load distribution $w(x)$ you find the equipollent moment using the following procedure


*

*Calculate the total loading $W = \int_0^\ell w(x)\,{\rm d}x$

*Calculate the centroid of the loading $c = \frac{\int_0^\ell x\, w(x)\,{\rm d}x}{W}$

*The equipollent moment is the total load going through the centroid $M = c\,W$.

The centroid is like the center of mass of the load distribution curve. In the case of uniform loading $c = \ell/2$ and $W = m g$ so you see how the general approach produces the same result as the first approach.

If you are asking about the dynamic behavior of the massive rod then it is an entirely different question than what I understood it to be.
