Is Angular Momentum Conserved in Euler's Equations?

[Ted123]

Homework Statement

[PLAIN]http://img836.imageshack.us/img836/5302/euler.jpg

Homework Equations

The Attempt at a Solution

Am I doing this right?

\displaystyle \frac{d}{dt} \left( A{\omega_1}^2+B{\omega_2}^2 + C{\omega_3}^2 )= 2A\dot{\omega}_1 + 2B\dot{\omega}_2 + 2C\dot{\omega}_3 = 2(B-C)\omega_2\omega_3 + 2(C-A)\omega_3\omega_1 + 2(A-B)\omega_1\omega_2

But I can't get this to = 0. Anyone help?

 

[micromass]

You made a (hard-to-spot) mistake at the derivative:

\frac{d}{dx}(f^2)=2f\frac{d}{dx}{f}

and not

\frac{d}{dx}(f^2)=2\frac{d}{dx}{f}

If you correct this, then your equation will yield zero.

 

[Ted123]

You made a (hard-to-spot) mistake at the derivative:

\frac{d}{dx}(f^2)=2f\frac{d}{dx}{f}

and not

\frac{d}{dx}(f^2)=2\frac{d}{dx}{f}

If you correct this, then your equation will yield zero.

So is this correct?:

\frac{d}{dt} (A{\omega_1}^2 + B{\omega_2}^2 + C{\omega_3}^2 ) = 2A\omega_1 \dot{\omega}_1 + 2B\omega_2 \dot{\omega}_2 + 2C\omega_3 \dot{\omega}_3

=2\omega_1 (B-C)\omega_2\omega_3 + 2\omega_2 (C-A)\omega_3 \omega_1 + 2\omega_3 (A-B) \omega_1 \omega_2

2B\omega_1 \omega_2 \omega_3 - 2C\omega_1 \omega_2 \omega_3 + 2C\omega_1 \omega_2 \omega_3 - 2A\omega_1 \omega_2 \omega_3 + 2A\omega_1 \omega_2 \omega_3 - 2B\omega_1 \omega_2 \omega_3 = 0 \Rightarrow \text{(i)\;is\;constant}

\frac{d}{dt} (A^2{\omega_1}^2 + B^2{\omega_2}^2 + C^2{\omega_3}^2 ) = 2A^2\omega_1 \dot{\omega}_1 + 2B^2\omega_2 \dot{\omega}_2 + 2C^2\omega_3 \dot{\omega}_3

=2A\omega_1 (B-C)\omega_2\omega_3 + 2B\omega_2 (C-A)\omega_3 \omega_1 + 2C\omega_3 (A-B) \omega_1 \omega_2

2AB\omega_1 \omega_2 \omega_3 - 2AC\omega_1 \omega_2 \omega_3 + 2BC\omega_1 \omega_2 \omega_3 - 2AB\omega_1 \omega_2 \omega_3 + 2AC\omega_1 \omega_2 \omega_3 - 2BC\omega_1 \omega_2 \omega_3 = 0 \Rightarrow \text{(ii)\;is\;constant}

 

[micromass]

This is fine