# Trigonometrylinear and angular velocity of 2 pulleys and a belt.

#### karush

##### Well-known member
A belt contects two pulleys with radii $\displaystyle5\text { in}$ and $3\text { in}$

the $\displaystyle5\text { in}$ pulley is rotating at $\displaystyle\frac{1000\text{ rev}}{\text{min}}$

What is the linear $\displaystyle\text{v}$ in $\displaystyle\frac{\text{ft}}{\text{sec}}$ of the belt?

$\displaystyle \text{v}= \frac{1000\text{rev}}{\text{min}} \cdot\frac{\text{min}}{60\text{ sec}} \cdot\frac{10\pi \text{ in}}{\text{rev}} \cdot\frac{\text{ ft}}{12 \text{in}} =\frac{125\pi\text{ ft}}{9\text{sec}} =43.63\frac{\text{ft}}{\text{sec}}$

How many revolutions per min is the $\text{3 in}$ pulley making?

so
$\displaystyle \omega_{3in} =\frac{5}{3} \cdot\frac{1000\text{rev}}{\text{min}} \approx 1667\frac{\text{rev}}{\text{min}}$

no ans in bk on this so hope ans here is perhaps it.

#### MarkFL

Staff member
To find the linear speed of the belt, we may state (using the information about the larger pulley):

$$\displaystyle v=r\omega=\left(5\text{ in}\frac{1\text{ ft}}{12\text{ in}} \right)\left(1000\frac{\text{rev}}{\text{min}} \frac{2\pi\text{ rad}}{1\text{ rev}} \frac{1\text{ min}}{60\text{ s}} \right)=\frac{125}{9}\pi\frac{\text{ ft}}{\text{s}}$$

This agrees with your result, although I think the way I have written the intermediary steps makes it a bit clearer what is going on.

Now, to find the revolutions per minute of the smaller pulley (the second pulley), we may write (as we did in your previous topic):

$$\displaystyle \omega_2=\frac{r_1}{r_2}\omega_1$$

You have done this correctly as well.