The Forces of Rolling — Question 5

Question 5

A drive torque produces forward traction

A wheel with $M=4\,\mathrm{kg}$, $R=0.25\,\mathrm m$ and $I=0.125\,\mathrm{kg\,m^2}$ receives a clockwise axle torque of $3\,\mathrm{N\,m}$. There is no other horizontal applied force. It rolls without slip.

1. Find acceleration and friction.

2. Find the required minimum $\mu_s$ on level ground with $g=9.8$.

Question 5 - Solution

A drive torque produces forward traction

(a) Derivation and calculation

Let the drive-torque magnitude be $\tau_d=3$ N m. The equations are $$f=Ma,\qquad-\tau_d+Rf=I\alpha=-Ia/R.$$ Therefore $$\boxed{a=\frac{\tau_d}{R(M+I/R^2)} =\frac3{.25[4+.125/(.25)^2]}=2.00\ \mathrm{m/s^2}},$$ $$\boxed{f=Ma=8.00\ \mathrm N\quad\text{right}}.$$

(b) Result and interpretation

Normal balance gives $N=Mg=4(9.8)=39.2$ N. Thus $$\boxed{\mu_s\ge\frac{|f|}{N}=\frac8{39.2}=\frac{10}{49}=.2041}.$$ The road supplies the forward external force. The applied axle torque supplies energy; static ground contact has zero power.

Physical check

The clockwise drive moment exceeds the opposing friction moment by $3-8(.25)=1$ N m.