The Forces of Rolling — Question 4

Question 4

A particular pull height eliminates friction entirely

A wheel has inertia $I=\beta MR^2$. A horizontal rightward force $F$ acts at height $y$ above its center, with $-R\le y\le R$. Assume rolling without slip on a level surface.

1. Derive acceleration and signed friction.

2. At what height does friction vanish, and when is that height accessible?

Question 4 - Solution

A particular pull height eliminates friction entirely

(a) Derivation and calculation

With rightward $f$ positive, $$F+f=Ma,\qquad-Fy+fR=-\beta MRa.$$ Insert $f=Ma-F$ into the moment equation: $$M(1+\beta)a=F(1+y/R).$$ Thus $$\boxed{a=\frac{F(1+y/R)}{M(1+\beta)}},\qquad \boxed{f=\frac{F(y/R-\beta)}{1+\beta}}.$$

(b) Result and interpretation

The zero-friction height follows from $$f=0\Rightarrow\boxed{y=\beta R}.$$ This lies in the allowed interval $[-R,R]$ when $0\le\beta\le1$, as for mass confined within $R$. For $F>0$, $f>0$ above this height and $f<0$ below it.

Physical check

For a disk, $y=0$ gives backward friction and $y=R$ gives forward friction, matching the preceding two cases.