the horsepower (hp) that a shaft can safely transmit varies jointly with its speed (in revolutions per…

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 2 hp at 5 rpm?\nthe diameter will be \n(type an integer or a decimal rounded to two decimal places.)

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 2 hp at 5 rpm?\nthe diameter will be \n(type an integer or a decimal rounded to two decimal places.)

Answer

Explanation:

Step1: Set up the joint - variation equation

Let $H$ be the horsepower, $S$ be the speed in rpm, and $d$ be the diameter of the shaft. The joint - variation equation is $H = kSd^{3}$, where $k$ is the constant of variation. We know that when $d = 2$ inches, $H = 36$ hp, and $S=75$ rpm. Substitute these values into the equation: $36=k\times75\times2^{3}$.

Step2: Solve for the constant $k$

First, calculate $2^{3}=8$. Then the equation becomes $36 = k\times75\times8$. $36=600k$. $k=\frac{36}{600}=0.06$.

Step3: Use the value of $k$ to find the new diameter

We want to find $d$ when $H = 2$ hp and $S = 5$ rpm. Substitute $k = 0.06$, $H = 2$, and $S = 5$ into the equation $H=kSd^{3}$. $2=0.06\times5\times d^{3}$. $2 = 0.3d^{3}$. $d^{3}=\frac{2}{0.3}=\frac{20}{3}\approx6.67$. $d=\sqrt[3]{\frac{20}{3}}\approx1.89$ inches.

Answer:

$1.89$