Angular Momentum!

  • #1
How do you prove that angular momentum is conserved by using Euler's equations, for a rigid body not subject to any torques?

I can show that angular momentum is constant because there's no torque acting on the body and no torque means no change in angular momentum.

But how do I show this with Euler's equations.
 

Answers and Replies

  • #2
60
0
When you say Euler's equations for a rigid body with no external forces I assume you mean:


[tex](I_{2}-I_{3})\omega_{2}\omega_{3}-I_{1}{\dot{\omega_{1}}}=0[/tex]
[tex](I_{3}-I_{1})\omega_{3}\omega_{1}-I_{2}{\dot{\omega_{2}}}=0[/tex]
[tex](I_{1}-I_{2})\omega_{1}\omega_{2}-I_{3}{\dot{\omega_{3}}}=0[/tex]

Try solving for:

[tex]\omega_{1},\omega_{2},\omega_{3}[/tex]

Once you do that, compute

[tex]\omega=\sqrt{\omega_{1}^2+\omega_{2}^2+\omega_{3}^2}[/tex]

and see if the nature of that answer gives you enough insight to finish the problem.
 
  • #3
When you say Euler's equations for a rigid body with no external forces I assume you mean:


[tex](I_{2}-I_{3})\omega_{2}\omega_{3}-I_{1}{\dot{\omega_{1}}}=0[/tex]
[tex](I_{3}-I_{1})\omega_{3}\omega_{1}-I_{2}{\dot{\omega_{2}}}=0[/tex]
[tex](I_{1}-I_{2})\omega_{1}\omega_{2}-I_{3}{\dot{\omega_{3}}}=0[/tex]

Try solving for:

[tex]\omega_{1},\omega_{2},\omega_{3}[/tex]

Once you do that, compute

[tex]\omega=\sqrt{\omega_{1}^2+\omega_{2}^2+\omega_{3}^2}[/tex]

and see if the nature of that answer gives you enough insight to finish the problem.

[tex]\omega=\sqrt{\omega_{1}^2+\omega_{2}^2+\omega_{3}^2}[/tex] that tells me the magnitude of angular velocity, it could be constant, but its direction doesn't have to be. Plus angular momentum is not necessarily parallel to angular velocity, so more help please!!
 
  • #4
60
0
Plus angular momentum is not necessarily parallel to angular velocity, so more help please!!

Right. When are L and w parallel? This one's tricky--they are parallel when the elements of the inertia tensor are diagonalized (all off diagonal elements are zero). This causes the inertia tensor to act as a 'constant' (I don't know the right word.) Check the matrix multiplication to see what I mean. If inertia is a 3x3 matrix and w is a 3x1 column matrix, and inertia is diagonalized, can you see how the elements of inertia 'pick out' their appropriate values in the w column matrix? Then you get something like:

[tex]L=I_{xx}\omega_x+I_{yy}\omega_y+I_{zz}\omega_z[/tex]
and the L and w vectors must be parallel. The simpler name for this is rotation about a principal axis. Regurgitating, when a rigid object is rotated about one of it's principal axes, the inertia tensor is diagonalized, and L and w are parallel.

And here's seemingly trivial part after all this analysis:
Euler's equations are derived under the assumption that a rigid object is being rotated about one of it's principal axes; therefore, it must have parallel angular momentum and angular velocity.

So now, angular velocity and angular momentum are parallel, and angular velocity has a constant magnitude, and I is constant. L is now a product of constants.
 

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