Solve the problem.
-A pulley with a diameter of 27 inches is driven by a belt which is moving $902 \mathrm { ft } / \mathrm { min }$. To the nearest unit, how many revolutions per minute are made by the pulley?
A) $128 \mathrm { rpm }$
B) $233 \mathrm { rpm }$
C) $226 \mathrm { rpm }$
D) $140 \mathrm { rpm }$

Question

Quizplus · Accepted Answer

To find the revolutions per minute (rpm) of the pulley, first convert the belt speed from feet per minute to inches per minute: $902 \mathrm { ft } / \mathrm { min } \times 12 \mathrm { in } / \mathrm { ft } = 10824 \mathrm { in } / \mathrm { min }$. The circumference of the pulley is $\pi d = \pi \times 27 \mathrm { in }$. The number of revolutions per minute is the speed of the belt divided by the circumference of the pulley: $10824 \mathrm { in } / \mathrm { min } \div ( \pi \times 27 \mathrm { in }) \approx 128 \mathrm { rpm }$.

Solve the Problem 902ft/min902 \mathrm { ft } / \mathrm { min }902ft/min

Solve the Problem $902 \mathrm { ft } / \mathrm { min }$