MHB Linear and angular velocity of 2 pulleys and a belt.

[karush]

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]

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

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

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. (Bandit)

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

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

You have done this correctly as well. (Clapping)