It is possible extend the work-energy principle already discussed for particles here and for rigid bodies in planar motion here to rigid bodies in 3D motion.
For a small mass element \mathrm dm, its kinetic energy is:
T = \frac{1}{2} \mathrm dm \, v^2
Where v is the velocity of the differential mass element. To calculate the total kinetic energy of the rigid body, we integrate over all \mathrm dm in the body:
T = \frac{1}{2} \int (\mathbf{v} \cdot \mathbf{v}) \, \mathrm dm
The velocity of any point \mathbf P in the body can be expressed using the relative velocity equation as the sum of the velocity of the center of mass (\mathbf{v}_C) and the rotational velocity component due to angular velocity (\boldsymbol{\omega}) about the center of mass:
\mathbf{v}_P = \mathbf{v}_C + \boldsymbol{\omega} \times \mathbf{r} Substituting:
\begin{aligned} T & = \frac{1}{2} \int (\mathbf{v} \cdot \mathbf{v}) \, \mathrm dm \\ & = \frac{1}{2} \int \left[\left(\mathbf{v}_C + \boldsymbol{\omega} \times \mathbf{r}\right) \left(\mathbf{v}_C + \boldsymbol{\omega} \times \mathbf{r}\right)\right]\mathrm dm \\ & = \frac{1}{2} \mathbf{v}_C \cdot \mathbf{v}_C \int \mathrm dm + \frac{1}{2} \int \left(\boldsymbol{\omega} \times \mathbf{r}\right) \cdot \left(\boldsymbol{\omega} \times \mathbf{r}\right)\mathrm dm + \mathbf{v}_C \cdot \left[\boldsymbol{\omega} \times \int \mathbf{r}\mathrm dm\right] \\ & = \frac{1}{2} \mathbf{v}_C \cdot \mathbf{v}_C \int \mathrm dm + \frac{1}{2} \int \left(\boldsymbol{\omega} \times \mathbf{r}\right) \cdot \left(\boldsymbol{\omega} \times \mathbf{r}\right)\mathrm dm \\ \end{aligned}
since:
\mathbf{v}_C \cdot \left[\boldsymbol{\omega} \times \int \mathbf{r}\mathrm dm\right] = 0
for the definition of mass center.
Using the vector identity:
(A \times B) \cdot (C \times D) = A \cdot [B \times (C \times D)]
The second equation can be written as:
\begin{aligned} T & = \frac{1}{2} m \left(\mathbf{v}_C \cdot \mathbf{v}_C \right) + \frac{1}{2} \boldsymbol{\omega} \cdot \int \left[\mathbf r \times \left(\boldsymbol{\omega} \times \mathbf r \right) \right] \, dm \\ & = \frac{1}{2} m \left(\mathbf{v}_C \cdot \mathbf{v}_C \right) + \frac{1}{2} \boldsymbol{\omega} \cdot \mathbf L_C \end{aligned} This can be consider as sum of two term, a translational kinetic energy:
T_v = \frac{1}{2} m \left(\mathbf{v}_C \cdot \mathbf{v}_C \right)
and a rotational kinetic energy:
T_\omega = \frac{1}{2} \boldsymbol{\omega} \cdot \mathbf L_C
If the point is fixed in space, naturally there is no translational kinetic energy and there is just the rotational energy T_\omega.
For the works term, we consider first the translational kinetic energy. Let’s start from Euler first law :
\sum \mathbf{F} = m \mathbf{a}_C
Dotting each side with \mathbf v_C and integrating from time t_1 to t_2:
\int_{t_1}^{t_2} \sum \mathbf{F} \cdot \mathbf{v}_C \, \mathrm{dt} = \int_{t_1}^{t_2} m \mathbf{a}_C \cdot \mathbf{v}_C \, \mathrm{dt} using the definition of \mathbf v_C and \mathbf a_C:
\int_{t_1}^{t_2} \sum \mathbf{F} \cdot \frac{\mathrm{d}\mathbf{r}}{\mathrm{dt}} \, \mathrm{dt} = \int_{t_1}^{t_2} m \frac{\mathrm{d}\mathbf{v}_C}{\mathrm{dt}} \cdot \mathbf{v}_C \, \mathrm{dt} Canceling \mathrm dt and integrating:
\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
as the left hand side \mathbf F \cdot \mathbf r is the definition of work, and the right hand side is the change in translational kinetic energy.
Similarly, it is possible to compute the work for the rotational kinetic energy. Let’s start from Euler second law:
\sum \mathbf{M}_C = \dot{\mathbf{L}}_C
Dotting each side with \boldsymbol{\omega}:
\boldsymbol{\omega} \cdot \sum \mathbf{M}_C = \boldsymbol{\omega} \cdot \dot{\mathbf{L}}_C = \frac{1}{2} \frac{\mathrm{d}}{\mathrm{dt}} (\boldsymbol{\omega} \cdot \mathbf{L}_C) The right hand side come from the expansion of:
\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dt}} (\boldsymbol{\omega} \cdot \mathbf{L}_C) = \frac{1}{2} (\dot{\boldsymbol{\omega}} \cdot \mathbf{L}_C + \boldsymbol{\omega} \cdot \dot{\mathbf{L}}_C) and from:
\dot{\boldsymbol{\omega}} \cdot \mathbf{L}_C = \boldsymbol{\omega} \cdot \dot{\mathbf{L}}_C
Therefore:
\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dt}} (\boldsymbol{\omega} \cdot \mathbf{L}_C) = \frac{1}{2} (2 \boldsymbol{\omega} \cdot \dot{\mathbf{L}}_C) = \boldsymbol{\omega} \cdot \dot{\mathbf{L}}_C
Integrating from time t_1 to t_2:
\int_{t_1}^{t_2} \sum \mathbf{M}_C \cdot \boldsymbol{\omega} \, \mathrm{dt} = \int_{t_1}^{t_2} \frac{1}{2} \frac{\mathrm{d}}{\mathrm{dt}} (\boldsymbol{\omega} \cdot \mathbf{L}_C)\, \mathrm{dt} using the definition of \boldsymbol{\omega}:
\int_{t_1}^{t_2} \sum \mathbf{M}_C \cdot \frac{\mathrm{d}\theta}{\mathrm{dt}} \, \mathrm{dt} = \frac{1}{2} \int_{t_1}^{t_2} \frac{\mathrm{d}}{\mathrm{dt}} (\boldsymbol{\omega} \cdot \mathbf{L}_C)\, \mathrm{dt}
Canceling \mathrm dt and integrating:
\Delta W = W_{1 \to 2} = \frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{L}_C \Big|_{t_1}^{t_2} = T_{\omega_2} - T_{\omega_1} = \Delta T_\omega
as the left hand side \mathbf M_c \cdot \mathbf \theta is also the definition of work, and the right hand side is the change in kinetic rotational energy.
Combining these two equations and generalizing for forces and applied moments at any point in the body:
\Delta W = W_{1 \to 2} = \int_{t_1}^{t_2} \sum \left(\mathbf{F}_i \cdot \mathbf{v}_i + \mathbf{M}_i \cdot \boldsymbol{\omega} \right) \, \mathrm{dt} = \Delta T