The Angular Momentum of a Rigid Body Rotating About a Fixed Axis — Question 1

Question 1

Summing particle moments produces the axial inertia law

A rigid assembly rotates about fixed $z$ with $\omega=4\,\mathrm{rad/s}$. It contains masses $1$, $2$, and $3\,\mathrm{kg}$ at perpendicular radii $0.20$, $0.30$, and $0.40\,\mathrm m$.

1. Derive $L_z=I_z\omega$ from individual particle velocities and evaluate it.

2. Explain why axial height does not enter $L_z$ but can enter transverse components.

Question 1 - Solution

Summing particle moments produces the axial inertia law

(a) Derivation and calculation

Fixed-$z$ rotation gives $\vec v_i=(-\omega y_i,\omega x_i,0)$. Thus $$L_{z,i}=m_i(x_iv_{iy}-y_iv_{ix})=m_i\omega(x_i^2+y_i^2).$$ Summing, $$I_z=1(.20)^2+2(.30)^2+3(.40)^2=.700\ \mathrm{kg\,m^2},$$ $$\boxed{L_z=I_z\omega=.700(4)=2.80\ \mathrm{kg\,m^2/s}}.$$

(b) Result and interpretation

Axial height does not enter $I_z=\sum m_i(x_i^2+y_i^2)$. However the full particle cross product includes $$L_{x,i}=-m_i\omega z_ix_i,\qquad L_{y,i}=-m_i\omega z_iy_i.$$ These contributions need not cancel in an asymmetric assembly. The scalar relation fixes $L_z$ without guaranteeing that the full vector is parallel to the axis.

Physical check

Only perpendicular distance to the specified axis enters its axial moment of inertia.