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 12 hp at 75 rpm? the diameter will be (type an integer or a decimal rounded to two decimal places.)

Explanation:

Step1: Write the joint - variation formula

Let $hp$ be the horsepower, $s$ be the speed, and $d$ be the diameter. The joint - variation formula is $hp = ksd^{3}$, where $k$ is the constant of variation.
We know that when $hp = 36$, $s = 75$, and $d = 2$. Substitute these values into the formula:
$36=k\times75\times2^{3}$
$36 = k\times75\times8$
$36=600k$
$k=\frac{36}{600}=0.06$

Step2: Use the new values and the constant to find the diameter

We want to find $d$ when $hp = 12$, $s = 75$, and $k = 0.06$. Substitute into the formula $hp = ksd^{3}$:
$12=0.06\times75\times d^{3}$
$12 = 4.5d^{3}$
$d^{3}=\frac{12}{4.5}=\frac{8}{3}$
$d=\sqrt[3]{\frac{8}{3}}\approx1.39$

Answer:

$1.39$