The Forces of Rolling — Question 3

Question 3

A top pull reverses the required friction direction

A horizontal force $12\,\mathrm N$ acts rightward at the top of a $3\,\mathrm{kg}$ disk. It rolls without slip on level ground; $I=MR^2/2$.

1. Find acceleration and signed friction.

2. Explain how forward friction can coexist with zero frictional work.

Question 3 - Solution

A top pull reverses the required friction direction

(a) Derivation and calculation

Force, moment and rolling equations are $$F+f=Ma,\qquad-FR+fR=I\alpha=-Ia/R.$$ Combining them, $$\boxed{a=\frac{2F}{M+I/R^2}=\frac{24}{4.5}=\frac{16}{3}\ \mathrm{m/s^2}},$$ $$\boxed{f=Ma-F=16-12=+4\ \mathrm N}.$$

(b) Result and interpretation

The frictional power splits into translation and rotation: $$\mathcal P_{f,\mathrm{trans}}=fV,\qquad\mathcal P_{f,\mathrm{rot}}=(fR)\omega=-fV.$$ Hence $$\boxed{\mathcal P_f=fV-fV=0}.$$ Forward friction prevents the top pull from creating excessive clockwise spin and backward contact slip. Its direction is set by relative contact slip, not center velocity.

Physical check

The top applied force delivers power $F(2V)$, while ideal stationary-ground friction delivers none.