The Forces of Rolling — Question 2

Question 2

A center pull needs backward friction to establish rolling

A horizontal force $F=12\,\mathrm N$ acts at the center of a $3\,\mathrm{kg}$ disk on level ground. The disk rolls without slip and has $I=MR^2/2$.

1. Find center acceleration and static friction.

2. Explain why friction points opposite the applied force.

Question 2 - Solution

A center pull needs backward friction to establish rolling

(a) Derivation and calculation

Choose rightward translation and counterclockwise rotation positive. Then $$F+f=Ma,\qquad Rf=I\alpha,\qquad\alpha=-a/R.$$ Elimination gives $$\boxed{a=\frac{F}{M+I/R^2}=\frac{12}{3+1.5}=\frac83\ \mathrm{m/s^2}},$$ $$\boxed{f=Ma-F=3(8/3)-12=-4\ \mathrm N}.$$

(b) Result and interpretation

The central applied force has zero moment about $C$. Friction supplies $$\tau_C=Rf=-4R,\qquad\alpha=\frac{-4R}{MR^2/2}=-\frac8{3R}.$$ Without friction, $a=F/M$ but $\alpha=0$, so the bottom would accelerate right relative to the ground. Leftward friction opposes that incipient slip.

Physical check

The resulting clockwise angular acceleration satisfies $a+R\alpha=0$.