maline
- 436
- 69
Well, sure enough! Here is an example of a helix that can rotate freely around its axis:
The helix has ##n## complete turns, parametrized as above by ##\vec r (\theta) = ( \cos \theta , \sin \theta, p \theta)## where ## -n\pi \leq \theta \leq n\pi##. Start with a uniform mass distribution of ##\lambda## units per radian. As calculated there, if this rotates with ##\vec \omega = \omega \hat z##, the angular momentum has a non-parallel component ##J_y =\lambda \int_{-n\pi}^{n\pi} \, d\theta\, y(\theta) \omega z(\theta) =(-1)^n 2\pi n \omega p \lambda##. To fix this, we can simply add two point masses to the helix, at points ##\theta = \pm \left( n- \frac 3 2 \right) \pi,## with masses of ##\frac n {n-\frac 3 2} \lambda.## Their contribution to the integral is equal and opposite to the above value, so ##J_y =0.## They lie in the ## yz## plane, so they have no effect on ##J_x=0##, nor on the ##x## component of the center of mass. Their ##y## coordinates are equal and opposite ##(\pm 1)##, so the center of mass remains on the ##z## axis, and free rotation is possible.
So the paradox remains! In the frame boosted by ##\vec v = - \frac 1 {\omega p} \hat z##, (which exists provided ##\omega > \frac 1 p##, in turn requiring ##p>1##), we have a straight rod, including the point masses, revolving around an axis parallel to itself, seemingly a violation of linear momentum conservation!
The helix has ##n## complete turns, parametrized as above by ##\vec r (\theta) = ( \cos \theta , \sin \theta, p \theta)## where ## -n\pi \leq \theta \leq n\pi##. Start with a uniform mass distribution of ##\lambda## units per radian. As calculated there, if this rotates with ##\vec \omega = \omega \hat z##, the angular momentum has a non-parallel component ##J_y =\lambda \int_{-n\pi}^{n\pi} \, d\theta\, y(\theta) \omega z(\theta) =(-1)^n 2\pi n \omega p \lambda##. To fix this, we can simply add two point masses to the helix, at points ##\theta = \pm \left( n- \frac 3 2 \right) \pi,## with masses of ##\frac n {n-\frac 3 2} \lambda.## Their contribution to the integral is equal and opposite to the above value, so ##J_y =0.## They lie in the ## yz## plane, so they have no effect on ##J_x=0##, nor on the ##x## component of the center of mass. Their ##y## coordinates are equal and opposite ##(\pm 1)##, so the center of mass remains on the ##z## axis, and free rotation is possible.
So the paradox remains! In the frame boosted by ##\vec v = - \frac 1 {\omega p} \hat z##, (which exists provided ##\omega > \frac 1 p##, in turn requiring ##p>1##), we have a straight rod, including the point masses, revolving around an axis parallel to itself, seemingly a violation of linear momentum conservation!