# Rigid body in a force vector field

If I let a particle with coordinates $$x(t)$$ move in a force vector field $$F$$ the equation I have to solve is $$x'(t) = F(x(t))$$, right? But if, instead of a particle, I have a rigid body, which equation I have to solve?

• What Law are you expecting the particle to obey? – Gert Apr 21 at 19:00
• @Gert what do you mean? – Marco Nervo Apr 21 at 19:01
• No Marco, the question is what do you mean by $x'(t) = F(x(t))$? – Gert Apr 21 at 19:03
• @Gert I am thinking about $F$ as a water/wind field; what I want to describe is a ship moving in a river – Marco Nervo Apr 21 at 19:08
• $x'(t)'$ is the derivative in time of the position coordinate, that's a velocity. Really what you need is an Equation of Motion, like Newton's third: $F(x)=mx''(t)$. But that's only in one dimension, unlike your river. – Gert Apr 21 at 19:21

The motion of a rigid body is determined by only two quantities:

• The net force: vector sum of all the forces applied

• The net torque: vector sum of all the torques applied.

The first one gives you the motion of the center of mass

$$m\frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2}=\vec{F}=\int_V \mathrm{d}\vec{F}=\int_V \vec{f}(\vec{r})\mathrm{d}V$$

assuming a diferential force on each volume element (for instante, for a charged volume, would be $$\mathrm{d}\vec{F}=\rho(\vec{r})\vec{E}(\vec{r})\mathrm{d}V$$)

The second one gives you the change on the orientation through the Euler equations. It can be expressed in terms of the angular momentum with respect to the center of mass

$$\frac{\mathrm{d}\vec{L}_G}{\mathrm{d}t}=\vec{M}_G=\int_V (\vec{r}-\vec{r}_G)\times \mathrm{d}\vec{F}$$

or with respect to a fixed point O

$$\frac{\mathrm{d}\vec{L}_O}{\mathrm{d}t}=\vec{M}_O=\int_V \vec{r}\times \mathrm{d}\vec{F}$$

A simple example is given by the gravitational force. We usually assume that for a rigid body there is only one weight acting on the center of mass, but in fact there are millions of small weights, one for each particle.

The net force is given by the sum of all weights

$$\vec{F}=\int_M \vec{g}\,\mathrm{d}m=\vec{g}\int_M \mathrm{d}m = m\vec{g}$$

while the net torque is

$$\vec{M}_O=\int_M \vec{r}\times\vec{g}\,\mathrm{d}m=\left(\int_M\vec{r} \mathrm{d}m\right)\times \vec{g} = \vec{r}_G\times(m\vec{g})$$

So, the system of all weights is in fact equivalent to just one weight acting on the CM (or in any point of the vertical of the CM).