Prove Total Kinetic Energy of a Rotating System about 3 Axes

[pardesi]

how does one prove this
if a system is rotating about three different perpendicular axes then
total kinetic energy of the system is
$$T=\frac{I_{x}\omega_{x}^{2}}{2} + \frac{I_{y}\omega_{y}^{2}}{2}+\frac{I_{z}\omega_{z}^{2}}{2}$$

 

[AlephZero]

Consider the body as a set of particles.

$$T = \frac{1}{2}\iiint \rho (\dot x^2 + \dot y^2 + \dot z^2) dx\,dy\,dz$$

Then write the translational velocity at (x,y,z) in terms of the rotation velocities and the position (the 3-D version of $$v = r\omega$$).

Rearrange the result and the integrals that define $$I_x$$ etc will appear.

 

[olgranpappy]

well, in general you'll find:
$$T=\frac{1}{2}\sum_{i,j}I_{ij}\omega_i\omega_j$$
then you can change coordinates to diagonalize the I tensor... or, if you made a good choice to begin with the tensor will already be diagonal giving you the expression you want.

 

[olgranpappy]

...(the 3-D version of $$v = r\omega$$).

I.e.
$$\vec v = \vec \omega \times \vec r$$
or
$$v_i = \epsilon_{ijk}\omega_j r_k$$

And use the definition
$$I_{ij}=\int\rho\left( r^2\delta_{ij}-r_ir_j \right)$$

 

[Andrew Mason]

how does one prove this
if a system is rotating about three different perpendicular axes then
total kinetic energy of the system is
$$T=\frac{I_{x}\omega_{x}^{2}}{2} + \frac{I_{y}\omega_{y}^{2}}{2}+\frac{I_{z}\omega_{z}^{2}}{2}$$

I think you can start with the given that:

$$KE = \frac{1}{2}I|\vec\omega|^2$$

What is the square of the length of the vector $\omega$ in terms of the x, y and z components?

AM

 

[olgranpappy]

I think you can start with the given that:

$$KE = \frac{1}{2}I|\vec\omega|^2$$

What is the square of the length of the vector $\omega$ in terms of the x, y and z components?

AM

no. he can't start with that. Only in the special case of
$$I_x=I_y=I_z=I$$