Question

The nominal radius of the rear wheels of a 2WD tractor is 0.5 m. The rolling radius of the tire is 6% less than the nominal radius. While conducting a field test, if the actual distance covered for 20 revolutions of the rear wheel is found to be 56 m, the rear wheel slip is _________% (Rounded off to 2 decimal places).

Hint:

To calculate wheel slip, remember that the slip is the difference between the theoretical and actual distances covered, divided by the theoretical distance, and multiplied by 100 to get the percentage.

Answer 1

Step 1: Calculate the rolling radius.
The nominal radius of the rear wheel is 0.5 m, and the rolling radius is 6% less than the nominal radius. Thus,
\[ {Rolling radius} = 0.5 - (6\% \times 0.5) = 0.5 - 0.03 = 0.47 \, {m}. \]

Step 2: Calculate the theoretical distance for 20 revolutions.
The theoretical distance that the wheel would cover for 20 revolutions, assuming no slip, is:
\[ {Theoretical distance} = 20 \times (2 \pi \times {Rolling radius}) = 20 \times (2 \pi \times 0.47) = 59.04 \, {m}. \]

Step 3: Calculate the slip.
The slip is the difference between the theoretical distance and the actual distance covered, divided by the theoretical distance and multiplied by 100:
\[ {Slip} = \frac{{Theoretical distance} - {Actual distance}}{{Theoretical distance}} \times 100. \] Substitute the values:
\[ {Slip} = \frac{59.04 - 56}{59.04} \times 100 = \frac{3.04}{59.04} \times 100 = 5.00\%. \]

Thus, the rear wheel slip is 5.00%.