Subtopic: Design and selection of machine elements – gears, pulleys, chains and sprockets and belts
Question Statement:
A flat belt drive connects two pulleys. The driving pulley of diameter $250\text{ mm}$ rotates at $1440\text{ rpm}$. If belt slip on each pulley is $2\%$, the rotational speed of the driven pulley of diameter $500\text{ mm}$ in $\text{rpm}$ is ________ (round off to 1 decimal place).
Answer Key:
691.5
Step-by-Step Solution & Explanation
Speed ratio of belt drive considering total slip $s$:
$\frac{N_2}{N_1} = \frac{d_1}{d_2} \left(1 - \frac{s}{100}\right)$
Total slip $s = s_1 + s_2 = 2\% + 2\% = 4\%$.
$N_2 = 1440 \times \frac{250}{500} \times \left(1 - \frac{4}{100}\right) = 1440 \times 0.5 \times 0.96 = 691.2\text{ rpm}$
Acceptable range: 689.0 to 693.0 rpm.