Question

the horsepower (hp) that a shaft can safely transmit varies jointly with its speed (in revolutions per minute, rpm) and the cube of its diameter. if a shaft of a certain material 2 inches in diameter can transmit 36 hp at 75 rpm, what diameter must the shaft have in order to transmit 9 hp at 55 rpm?

Explanation:

Step1: Write the joint - variation formula

Let $hp$ be the horsepower, $s$ be the speed in rpm, and $d$ be the diameter. The joint - variation formula is $hp = ksd^{3}$, where $k$ is the constant of variation.
We know that when $d = 2$ inches, $hp=36$ and $s = 75$ rpm. Substitute these values into the formula:
$36=k\times75\times2^{3}$
$36 = k\times75\times8$
$36=600k$
$k=\frac{36}{600}=\frac{3}{50}$

Step2: Find the diameter for the new situation

We want to find $d$ when $hp = 9$ and $s = 55$ rpm. Substitute $k=\frac{3}{50}$, $hp = 9$, and $s = 55$ into the formula $hp = ksd^{3}$:
$9=\frac{3}{50}\times55\times d^{3}$
First, simplify the right - hand side: $\frac{3}{50}\times55\times d^{3}=\frac{3\times55}{50}d^{3}=\frac{33}{10}d^{3}$
So, the equation becomes $9=\frac{33}{10}d^{3}$
$d^{3}=\frac{9\times10}{33}=\frac{30}{11}$
$d=\sqrt[3]{\frac{30}{11}}\approx1.4$ inches

Answer:

The diameter of the shaft must be approximately $1.4$ inches.