Angular quantities in tractor tires

Data
R_front = 40 cm = 0.4 m
R_rear = 1.2 m

φ_front = 10 vueltas
φ_rear = ? vueltas

Resolution

Since the tires have different radii, while the bigger rear tires complete one revolution the smaller front tires will complete more than one revolution, but regardless of the number of revolutions each one does, they both always travel the same distance. Therefore:

$$\begin{gathered} S_{front}=S_{rear} \Rightarrow \\ {\phi }_{front} · R_{front} = {\phi }_{rear} · R_{rear} \end{gathered}$$

If the front tires have completed 10 revolutions, the angular position of these tires will be 10 times the angle of a full circumference, or what is the same 10 times 2π:

$${\phi }_{front}=10 \cancel{revolution} · 2\pi rad/\cancel{revolution}=\underline{20\pi rad}$$

So, the angular position of the rear tires is:

$$\begin{gathered} {\phi }_{front} · R_{front} = {\phi }_{rear} · R_{rear} \Rightarrow \\ {\phi }_{front} =\frac{{\phi }_{front} · R_{front}}{R_{rear}}\Rightarrow \\ {\phi }_{rear}=\frac{20\pi ·0.4}{1.2}\Rightarrow \\ \underline{{\phi }_{rear}=6.67\pi rad} \end{gathered}$$

If each revolution is 2π rad, then the rear tires revolution will be:

$$No. revolutions=\frac{6.67\pi }{2\pi }=\boxed{3.33 }$$