I'm having trouble with the reasoning for the this problem where two particles of mass m (at point a and b) are rotating in circles with a rigid rod connecting them:


More specifically, the last paragraph. Why does the parallel component of the angular velocity vector not contribute any angular momentum? Why does this have to do with mass?


The two particles are assume to be points instead of spheres. So it takes no energy to spin the particles on the axis between them. If you did consider the particles to be spheres then you'd have to integrate the angular momentum of each section of the sphere over its distance to the axis. The distance between the points wouldn't matter.

EDIT *** More fundamentally...

For a point mass $m$ the moment of inertia, $I$, is defined as:

$I = r^2 \cdot m$

where $r$ is the radius of the point mass from the center of rotation.

(1) If $r = 0$ then:

There is no inertia.

(2) If there is no interia, $I$, then there is no angular momentum, $L$.

The angular momentum, $L$ is equal to moment of inertia $I$ and angular speed ω as given by the following equation.

$L = I \cdot ω$

(3) If there is no inertia, then it takes no energy to spin the mass.

The relationship between a object's moment of inertia and its spin rate is given by the following formula:

$ E_{rotational} = \dfrac{1}{2} \cdot I \cdot ω^2$


ω is the angular velocity
$I$ is the moment of inertia around the axis of rotation
$E$ is the kinetic energy

  • $\begingroup$ We didn't really cover energy yet. Is there a way to see this without considering energy? $\endgroup$ – Ayumu Kasugano Jan 26 '17 at 6:00
  • $\begingroup$ You may not understand the relationship between angular momentum and energy, but I'm sure that energy would have been discussed in any physics course before angular momentum. $\endgroup$ – MaxW Jan 26 '17 at 6:04
  • $\begingroup$ Well yes, we have covered energy. But as you said, we have yet to cover energy and angular momentum. $\endgroup$ – Ayumu Kasugano Jan 26 '17 at 6:07

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