Work-energy principle for 3d rigid bodies - translational energy
I begin by considering the kinetic energy of a rigid body. A small mass element \mathrm dm within the body possesses kinetic energy:
T = \frac{1}{2} \mathrm dm \, v^2
where v is the velocity of the element. Summing over all mass elements gives the total kinetic energy:
T = \frac{1}{2} \int (\mathbf{v} \cdot \mathbf{v}) \, \mathrm dm
I express the velocity of any point P in the body, \mathbf{v}_P, using the relative velocity equation:
\mathbf{v}_P = \mathbf{v}_C + \boldsymbol{\omega} \times \mathbf{r}
Here, \mathbf{v}_C is the velocity of the center of mass, \boldsymbol{\omega} is the angular velocity, and \mathbf{r} is the position vector from the center of mass to point P. Substituting this into the kinetic energy equation and performing some vector algebra, I arrive at:
T = \frac{1}{2} m \left(\mathbf{v}_C \cdot \mathbf{v}_C \right) + \frac{1}{2} \boldsymbol{\omega} \cdot \mathbf L_C
where m is the total mass and \mathbf L_C is the angular momentum about the center of mass. This expression separates the kinetic energy into two parts:
- Translational kinetic energy: T_v = \frac{1}{2} m \left(\mathbf{v}_C \cdot \mathbf{v}_C \right)
- Rotational kinetic energy: T_\omega = \frac{1}{2} \boldsymbol{\omega} \cdot \mathbf L_C
If a point in the body is fixed, the translational kinetic energy vanishes, leaving only the rotational component.
I now consider the work done by external forces on the rigid body. Starting with Euler’s first law:
\sum \mathbf{F} = m \mathbf{a}_C
I take the dot product of both sides with \mathbf{v}_C and integrate with respect to time, after few steps I arrive to:
\Delta W = W_{1 \to 2} = \frac{1}{2} m v_C^2 \Big|_{t_1}^{t_2} = T_{v_2} - T_{v_1} = \Delta T_v
This shows that the work done by external forces is equal to the change in translational kinetic energy. The next time I will add rotational kinetic energy and generalize.
For more insights into this topic, you can find the details here.