Kinetic energy of Rotating Body (in 3D)

[Curl]

Ok here is what I did:

[tex]
T=\int \vec{\tau} \cdot d\vec{\theta } = \int\frac{\mathrm{d} \vec{L}}{\mathrm{d} t}\cdot d\vec{\theta } = \int \vec{L}\cdot d\vec{\omega} = \int I_{ij}\vec{\omega}\cdot d\vec{\omega}[/tex]
Where I is the inertia matrix.

when I carry that out, I get:
T= 0.5*(I₁₁w_x²+I₂₂w_y²+I₃₃w_z²) + 2*(I₁₂w_xw_y+I₁₃w_xw_z+I₂₃w_zw_y)

but the real answer has no factor of 2 in the second term. What's wrong?

 

[LightHero]

The factor of two is likely due to the fact that you are not taking into account the cross-terms of the inertia matrix. The inertia matrix is composed of terms such as I11wx2, I12wxwy, I13wxwz, I22wy2, I23wzwy, and I33wz2. Because the matrix is symmetric, the cross-terms are equal. That means that I12 = I21, I13 = I31, and I23 = I32. Therefore, when you calculate the second term in your equation, you should take into account the fact that each of these cross-terms appears twice in the inertia matrix, and therefore should be multiplied by two in the equation.