# Rigid body in lagrangian mechanics

Imagine we´re dealing with a rigid body and we want to apply for it lagrangian mechanics, in order to obtain its equations of motion. If we want to compute the generalized force associated to a specific generalized coordinate, I´ve observed that, when gravity is the only force, and we consider gravitational field to be uniform in the region, generalized force can be written as the total force on the body times the displacement of the center of mas with respect to the generalized coordinate.

It would be equivalent to consider the rigid body to be a point mass at which the total force on the body is acting, which is located at the center of mass of the rigid body, and whose mass equals the total mass of the rigid body.

Why is this possible? Would it be extensible to non-uniform forces?

• Why is this possible? Because in analytical mechanics, you will be working in Configuration space. There will be a one-to-one correspondence between the generalised coordinates and the ordinary coordinates. – user36790 Oct 15 '16 at 12:48
• It would not work if the force were not uniform. – Lewis Miller Oct 15 '16 at 13:04

First of all, the product of the generalized coordinate and the generalized force corresponding to that coordinate gives the amount of work done on that object. For example, if the coordinate is y, the displacement along the vertical axis, then the corresponding force is mg and therefore the work done is mg$\Delta$y. If the coordinate was $\theta$, the angle with respect to some axis, then the corresponding generalized force is the torque about that axis $\tau$ and therefore the work done is $\tau\theta$.
Now if you are asking, why the gravitational potential enenrgy of a rigid body is given by $mgy_{c.o.m.}$ and whether this formula is valid even if the gravitational potential was not uniform then that is a different question.