Rolling as Translation and Rotation Combined — Question 10

Question 10

A rotating wheel has a whole distribution of rim speeds

A radius-$R$ wheel rolls right at center speed $V$. Label rim position by $\theta$ measured counterclockwise from the rightward horizontal radius.

1. Derive the rim speed as a function of $\theta$.

2. Find the fraction of its circumference whose speed exceeds $V$.

Question 10 - Solution

A rotating wheel has a whole distribution of rim speeds

(a) Derivation and calculation

With $\vec r=R(\cos\theta\hat x+\sin\theta\hat y)$ and $\omega=-V/R$, $$\vec v=V(1+\sin\theta)\hat x-V\cos\theta\hat y.$$ Therefore $$v^2=V^2[(1+\sin\theta)^2+\cos^2\theta] =\boxed{2V^2(1+\sin\theta)}.$$

(b) Result and interpretation

For $V>0$, $$v>V\Longleftrightarrow\sin\theta>-\tfrac12.$$ Over $0\le\theta<2\pi$, the excluded interval is $[7\pi/6,11\pi/6]$, of width $2\pi/3$. Thus $$\boxed{f=\frac{2\pi-2\pi/3}{2\pi}=\frac23}.$$ The equality endpoints have zero arc length and do not affect the fraction.

Physical check

The formula gives zero at the bottom and $2V$ at the top, as required by rolling kinematics.