An Angular Simple Harmonic Oscillator — Question 10

Question 10

Rolling couples translational inertia to angular oscillation

A uniform solid cylinder of mass $M$ and radius $R$ rolls without slipping on a horizontal floor. An ideal horizontal spring of constant $k$ acts through its center. Let center displacement from equilibrium be $x$. Static friction is sufficient for rolling; $I=MR^2/2$.

1. Derive the oscillation frequency using the translation and rotation equations, or the equivalent kinetic energy.

2. Find the static-friction magnitude as a function of $x$ and the amplitude bound imposed by coefficient $\mu_s$.

Question 10 - Solution

Rolling couples translational inertia to angular oscillation

(a) Derivation and calculation

Rolling without slipping gives $v=R\dot\varphi$:$$K=\tfrac12Mv^2+\tfrac12I(v/R)^2=\tfrac12(M+I/R^2)v^2.$$$$(M+I/R^2)\ddot x=-kx,\qquad I=\tfrac12MR^2.$$$$\boxed{\Omega=\sqrt{\frac{2k}{3M}}}.$$

(b) Result and interpretation

With rightward friction $f$ positive,$$Ma=-kx+f,\quad f=-\frac{I}{R^2}a,\quad a=-\frac{2k}{3M}x.$$$$\boxed{f=kx/3},\qquad |f|_{\max}=kA/3\le\mu_sMg.$$$$\boxed{A\le\frac{3\mu_sMg}{k}}.$$At positive $x$, friction points right while the net acceleration points left.

Physical check

Static friction enforces rolling and supplies torque; on a fixed ideal surface its contact point does no work.