Computing rotational kinetic energy from first principles

Question

I want to calculate the expression of the rotational kinetic energy of a rectangular plate when the axis of rotation does not go through the center of gravity from first principles. However, I am not getting the right expression. Preamble to define the notation and setup. Consider two points attached to a rigid body whose coordinates are expressed with respect to the world coordinate system. Then, their speed is related by the expression: $$ \dot{\mathbf{x}}_W=\dot{\mathbf{y}}_W+\Omega\left[\mathbf{x}_W-\mathbf{y}_W\right] $$ If we choose $\dot{\mathbf{y}}_W=\dot{\mathbf{g}}_W$ to be the center of gravity, $\rho$ to represent the mass density, then the kinetic energy can be written as: $$ T = \frac{1}{2}\int_V\rho(\mathbf{x})\left[\dot{\mathbf{g}}+\Omega(\mathbf{x}-\mathbf{g})\right]^T\left[\dot{\mathbf{g}}+\Omega(\mathbf{x}-\mathbf{g})\right]d\mathbf{x} $$ which can be decomposed into the term responsible for the linear kinetic energy: $$ T_l = \frac{1}{2}M\dot{\mathbf{g}}^T\dot{\mathbf{g}} $$ and in the term responsible for the rotational kinetic energy: $$ T_r = \frac{1}{2}\int_V\rho(\mathbf{x})(\mathbf{x}-\mathbf{g})^T\Omega^T\Omega(\mathbf{x}-\mathbf{g})d\mathbf{x} $$ Using the notation above, I want to compute the rotational kinetic energy of a rectangular plate whose axis of rotation is parallel to the y-axis and goes through the bottom (x,y) face of the plate itself: Thus: $$ \Omega^T\Omega = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \omega_y^2 $$ The plate dimensions are $a$, $b$, and $c$, and I assume the density is constant, hence $\rho(\mathbf{x}) = \frac{M}{abc}$. The center of gravity is $\mathbf{g}=\left[\begin{array}{ccc}0 & 0 & \frac{c}{2}\end{array}\right]^T$. Then the integral becomes: $$ T_r = \frac{1}{2}\frac{M}{abc}\left[\int_{-\frac{a}{2}}^{\frac{a}{2}} \int_{-\frac{b}{2}}^{\frac{b}{2}} \int_{0}^{c} \left[x^2 + \left(z-\frac{c}{2}\right)^2\right]dzdydx\right]\omega_y^2 $$ which is equal to $\frac{1}{2}\frac{M(a^2+c^2)}{12}\omega_y^2$. Unfortunately, the previous expression is the same as if the axis of rotation was going through the center of gravity. Can you help identify where is the error in the steps I described above?

Answer 1

Can you help identify where is the error in the steps I described above?

Although I was unable to follow your derivation, it appears there is no error in the result of your derivation.

The result (correctly) shows that the rotational kinetic energy (KE) of the rectangular prism about an axis on the base parallel to the axis passing through the center of mass (COM) is the rotational KE about the axis through the COM. The total kinetic energy of your object is then the rotational KE plus the kinetic energy of the motion of the COM about the axis of rotation. See the figure below.

This follows Konig's second theorem discussed in the Wikipedia link below, which Wikipedia summarizes as follows:

"Specifically, it states that the kinetic energy of a system of particles is the sum of the kinetic energy associated to the movement of the center of mass and the kinetic energy associated to the movement of the particles relative to the center of mass".

Note it refers to the "movement" of the COM. That movement could be due to the translational motion of the rigid body, or due to the circular movement of the COM about the axis of rotation, the latter applying in your example. Note that the circular motion of the COM about the axis is not "rotational" KE since rotational KE applies to an extended body, not a point mass.

See discussion of Second part of Konig's Theorem: https://en.wikipedia.org/wiki/K%C3%B6nig%27s_theorem_(kinetics)#:~:text=Specifically%2C%20it%20states%20that%20the,to%20the%20center%20of%20mass.

Hope this helps.