The Angular Momentum of a Rigid Body Rotating About a Fixed Axis — Question 2

Question 2

A tilted dumbbell exposes the limits of a scalar inertia

Two equal point masses $m$ are connected rigidly at positions $\pm(a,0,b)$ at one instant. The assembly rotates about fixed $z$ with angular velocity $\omega\hat z$. The connector is massless.

1. Find full angular momentum at that instant.

2. Find the angle between angular momentum and $+z$ for $a,b,\omega>0$.

Question 2 - Solution

A tilted dumbbell exposes the limits of a scalar inertia

(a) Derivation and calculation

At this instant, $$\vec r_\pm=\pm(a,0,b),\qquad\vec v_\pm=\vec\omega\times\vec r_\pm=\pm(0,\omega a,0).$$ Each particle contributes $(-mab\omega,0,ma^2\omega)$, so $$\boxed{\vec L=(-2mab\omega,0,2ma^2\omega)}.$$ The axial inertia is $I_z=2ma^2$, consistent with $L_z=I_z\omega$.

(b) Result and interpretation

For $a,b,\omega>0$, $L_z>0$ and $L_x<0$. The angle from $+z$ obeys $$\boxed{\tan\gamma=\frac{|L_x|}{L_z}=\frac ba},\qquad \boxed{\gamma=\tan^{-1}(b/a)}.$$ For nonzero $a$, alignment occurs only when $b=0$. Rotation about a fixed axis alone does not imply $\vec L\parallel\vec\omega$.

Physical check

Both masses give the same transverse moment because both position and velocity reverse sign together.