MHB Angular speed of 2 pulleys on a belt

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The larger pulley with a radius of 15 cm rotates 24 times in 36 seconds, resulting in an angular velocity of 2/3 rotations per second. The smaller pulley, with a radius of 8 cm, has an angular velocity calculated to be (8/5)π radians per second. The linear velocity of the belt connecting the two pulleys is determined to be 20π cm per second. This relationship between the pulleys is governed by the equation rω1 = Rω2, ensuring consistent motion across the belt. The calculations confirm the dynamics of the system involving both pulleys and the belt.
RobertoPink
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The two pulleys connected by a belt have a radii of 15 cm and 8 cm. The larger pulley rotates 24 times in 36 seconds.

a. Find the angular velocity of the small pulley in radians per second.

b. Find the linear velocity of a point on the belt that connects the two pulleys in centimeters per second.
 
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RobertoPink said:
The two pulleys connected by a belt have a radii of 15 cm and 8 cm. The larger pulley rotates 24 times in 36 seconds.

a. Find the angular velocity of the small pulley in radians per second.

b. Find the linear velocity of a point on the belt that connects the two pulleys in centimeters per second.

(a) $r \omega_1 = R\omega_2$

$r = 8 \, cm$, $R = 15 \, cm$, $\omega_2 = \dfrac{2}{3} \, \text{rev per sec}$

you'll need to convert $\omega_2$ to radians per second

(b) note $v = r \omega_1 = R\omega_2$
 
The larger pulley has radius 15 cm so circumference $2\pi(15)= 30\pi$ cm. The larger pulley rotates 24 times in 36 seconds so at a rate of 24/36= 2/3 rotations per second. In one second, since the larger pulley has rotated 2/3 of a rotation, the belt has moved a distance or $(2/3)(30\pi)= 20\pi$ cm.

The smaller pulley has radius 8 cm so circumference $2\pi(8)= 16\pi$. When the belt moves $20\pi$ cm, it will have moved $\frac{16\pi}{20\pi}= \frac{4}{5}$ of one rotation of the smaller pulley. The smaller pulley is turning at 4/5 rotation per second. There are $2\pi$ radians in one rotation so that is a rate of $(4/5)(2\pi)= (8/5)\pi$ radians per second.

The linear velocity of the belt is the $20\pi$ cm/sec we got earlier.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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