## Statement

A farmer has a yellow tractor whose front tires have a radius of 40 cm and the rear tires have a radius of 1.20 m. When the front wheels have completed 10 revolutions, how many revolutions will the rear tires have completed?

## Solution

**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:

$${S}_{front}={S}_{rear}\Rightarrow \phantom{\rule{0ex}{0ex}}{\phi}_{front}\xb7{R}_{front}={\phi}_{rear}\xb7{R}_{rear}\phantom{\rule{0ex}{0ex}}$$

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:

$${\phi}_{front}=10\overline{)revolution}\xb72\pi rad/\overline{)revolution}=\overline{)20\pi rad}$$

So, the angular position of the rear tires is:

$${\phi}_{front}\xb7{R}_{front}={\phi}_{rear}\xb7{R}_{rear}\Rightarrow \phantom{\rule{0ex}{0ex}}{\phi}_{front}=\frac{{\phi}_{front}\xb7{R}_{front}}{{R}_{rear}}\Rightarrow \phantom{\rule{0ex}{0ex}}{\phi}_{rear}=\frac{20\pi \xb70.4}{1.2}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{){\phi}_{rear}=6.67\pi rad}$$

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

$$No.revolutions=\frac{6.67\pi}{2\pi}=\overline{)3.33}\phantom{\rule{0ex}{0ex}}$$