The Angular Momentum of a Rigid Body Rotating About a Fixed Axis — Question 8

Question 8

A rolling body has more angular momentum about contact than about its center

A disk of mass $2\,\mathrm{kg}$ and radius $0.50\,\mathrm m$ rolls right at $3\,\mathrm{m/s}$. Let $P$ be the fixed ground point coincident with contact at this instant. Positive $z$ is out of the page.

1. Find angular momentum about the center.

2. Find angular momentum about $P$, separating spin and center-motion contributions.

Question 8 - Solution

A rolling body has more angular momentum about contact than about its center

(a) Derivation and calculation

The disk inertia and signed spin are $$I_C=\tfrac12MR^2=\tfrac12(2)(.50)^2=.250\ \mathrm{kg\,m^2},$$ $$\omega_z=-V/R=-3/.50=-6.00\ \mathrm{rad/s}.$$ Thus $$\boxed{L_{C,z}=I_C\omega_z=-1.50\ \mathrm{kg\,m^2/s}}.$$

(b) Result and interpretation

From the specified fixed ground point, $\vec R_C=.50\hat y$ m and $M\vec V=6\hat x$ kg m/s. Hence $$\vec L_{\mathrm{orb}}=\vec R_C\times M\vec V=-3\hat z\ \mathrm{kg\,m^2/s}.$$ Adding spin, $$\boxed{L_{P,z}=-3-1.5=-4.50\ \mathrm{kg\,m^2/s}}.$$ At this instant it also equals $(I_C+MR^2)\omega_z=(.25+.50)(-6)=-4.50$.

Physical check

The orbital and spin contributions are both clockwise about the contact location.