# TrigonometryAngular speed of 2 pulleys on a belt

#### karush

##### Well-known member
two pulleys connected by a belt have 15cm and 8cm radius

The larger pulley rotates $25$ times in $36$ sec,

Find the angular speed of each pulleey in radians per second.

the 15cm pulley has circumferce of $30\pi$ so

$\displaystyle\frac{25\text { rev}}{36 \text {sec}} \cdot\frac{30\pi\text{ cm}}{ rev} =\frac{750\text{ cm\pi}}{36\text {sec}} =\frac{65.5\text{ cm}\text{ rad}}{\text{sec}}$

not sure how to get the v of the $$\displaystyle 8cm$$ pulley

Last edited:

#### MarkFL

Staff member
Re: angular speed of 2 pulleys on a belt

This is how I would work the first part:

$$\displaystyle \frac{25\text{ rev}}{36\text{ s}}\cdot\frac{2\pi\text{ rad}}{1\text{ rev}}=\frac{25}{18}\pi\frac{\text{rad}}{\text{s}}$$

Angular speed should have units of radians/time.

Since the pulleys are connected by a belt, then the linear velocity of the outer edge of each pulley will be the same:

$$\displaystyle v_2=v_1$$

Using, $$\displaystyle v=r\omega$$, we may state:

$$\displaystyle r_2\omega_2=r_1\omega_1$$

Solve for $$\displaystyle \omega_2$$:

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

Now let $$\displaystyle r_1=15\text{ cm},\,r_2=8\text{ cm},\,\omega_1=\frac{25}{18}\pi\frac{ \text{rad}}{\text{s}}$$

What do you find?

#### karush

##### Well-known member
Re: angular speed of 2 pulleys on a belt

$\displaystyle\frac{15}{8}\cdot\frac{25}{18}\pi \text{ = } \frac{125}{48}\pi\ \frac{\text{rad}}{s}$