# Rigid Body Equations in terms of Body Coordinates by Hamilton's Principle

I sought-for the equations of motion of an unrestrained rigid body. The equations of motion are readily available in the literature, but my concern is to derive them by Hamilton's principle.

Expressing the position of an infinitesimal particle within the body as:

$$\vec{R} = \vec{R}_0 + \vec{r}$$

where $$\vec{R}_0$$ and $$\vec{r}$$ are expressed in terms of body coordinates, origin of which is located at the center of mass. Additionally, $$\vec{R}_0$$ is the position of the center of mass measured from inertial frame, $$\vec{r}$$ is the position of the point measured from body frame.

The velocity of this point with respect to the inertial frame can be found as:

$$\vec{V} = \dot{\vec{R}}_0 + \vec{\omega} \times (\vec{R}_0+\vec{r})$$

where $$(\dot{})$$ represents the time derivative with respect to the body frame. To find the acceleration, differentiate with respect to inertial frame once more yields:

$$\vec{a} = \ddot{\vec{R}}_0 + \dot{\vec{\omega}}\times(\vec{R}_0+\vec{r}) + \vec{\omega} \times \dot{\vec{R}}_0 + \vec{\omega} \times \vec{V}$$

One can find the variation of velocity by replacing the time derivatives by variational $$\delta$$ operator and using $$\vec{\delta\theta}$$ infinitesimal rotation vector:

$$\delta \vec{V} = \delta \dot{\vec{R}}_0 + \delta \vec{\omega} \times(\vec{R}_0+\vec{r}) + \vec{\omega} \times \delta \vec{R}_0 + \vec{\delta\theta} \times \vec{V}$$

Now, I can use variation of kinetic energy. For the simplicity, I do not consider potential energy and work done by external forces:

$$\int_{t_1}^{t_2} \delta T dt = 0$$

where

$$\delta T = \int_D \rho \vec{V} \cdot \delta\vec{V} dD$$

By first, calculating the $$\delta \vec{R}_0$$ , I obtain the following:

$$\int_{t_1}^{t_2} \int_D \rho \left[ \left( \dot{\vec{R}}_0 + \vec{\omega} \times (\vec{R}_0+\vec{r}) \right) \cdot \delta \dot{\vec{R}}_0 + \left( \dot{\vec{R}}_0 + \vec{\omega} \times (\vec{R}_0+\vec{r}) \right) \cdot \left( \vec{\omega} \times \delta \vec{R}_0 \right) \right] dD dt$$

The first part of this integral can be integrated by parts, and hence one can obtain the translational equation of motion.

The rotational equations of motion can be obtained from the second part:

$$\int_{t_1}^{t_2} \int_D \rho \left[ \left( \dot{\vec{R}}_0 + \vec{\omega} \times (\vec{R}_0+\vec{r}) \right) \cdot \left( \delta \vec{\omega} \times(\vec{R}_0+\vec{r}) \right) + \left( \dot{\vec{R}}_0 + \vec{\omega} \times (\vec{R}_0+\vec{r}) \right) \cdot \left( \vec{\delta\theta} \times \vec{V} \right) \right] dD dt$$

The second part of the integral is zero, hence we end up with the following form:

$$\int_{t_1}^{t_2} \int_D \rho \left[ \left( \vec{R}_0+\vec{r} \right) \times \left( \dot{\vec{R}}_0 + \vec{\omega} \times (\vec{R}_0+\vec{r}) \right) \right] \cdot \delta \vec{\omega} dD dt$$

My question is here. Since angular velocity is non-holonomic, it cannot be expressed as a derivative of a vector, i.e., one cannot obtain an expression for rotation. I need to evaluate this integral by parts to obtain the rotational equations of motion.

In other words, how can I find the relation between $$\delta \vec{\omega}$$ and $$\vec{\delta \theta}$$?

Please note that the usage of the following expresion does not yield the correct result:

$$\delta \vec{\omega} = \frac{d (\vec{\delta \theta}) }{dt}$$

• you should use this equations $\overrightarrow {\omega }=A\left( \overrightarrow {\theta }\right) \dfrac {d\overrightarrow {\theta }}{dt}$ – Eli Feb 8 at 8:24
• Comment to the post (v2): Consider to include the explicit formulas of the sought-for equations of motion to ensure that everybody is on the same page. – Qmechanic Feb 8 at 8:56
• Related: physics.stackexchange.com/q/74742/2451 , physics.stackexchange.com/q/321748/2451 and links therein. – Qmechanic Feb 8 at 9:13
• @Eli what do you mean by $A(\vec{\theta})$ ? – nicomedian Feb 8 at 11:43
• With this equations $\dfrac {dR\left( \theta \right) }{dt}=R\left( \overrightarrow {\theta }\right) \tilde \omega$ for the rotation matrix R,you get the A matrix with $\theta$ the Euler angels – Eli Feb 8 at 11:52