MHB Angular speed of 2 pulleys on a belt

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The discussion focuses on calculating the angular speed of two pulleys connected by a belt, with radii of 15 cm and 8 cm. The larger pulley rotates 25 times in 36 seconds, resulting in an angular speed of approximately 4.36 radians per second. Since the pulleys are linked by a belt, their linear velocities are equal, allowing the use of the formula v = rω to find the angular speed of the smaller pulley. By applying the relationship between the radii and angular speeds, the angular speed of the 8 cm pulley is calculated to be approximately 8.24 radians per second. The calculations demonstrate the principles of rotational motion and the relationship between radius and angular speed.
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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 $$ 8cm $$ pulley
 
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Re: angular speed of 2 pulleys on a belt

This is how I would work the first part:

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

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:

$$v_2=v_1$$

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

$$r_2\omega_2=r_1\omega_1$$

Solve for $$\omega_2$$:

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

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

What do you find?
 
Re: angular speed of 2 pulleys on a belt

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