Angular and Linear Speed (Algebra II)

[Help]

A cylinder on a printing press has a 21 in. diameter. The linear speed of a point on the cylinder's suface is 18.33 ft/s. What is its angular speed in revolutions per hour?

The wheel of a bicycle has a 67 cm diameter. The linear speed during a race was 40.423 km/h. What is its angular speed in revolutions per hour?

Please help; I have tried to solve these for over 2 hours and I still can't get the right answer. Thanks so much!

These are my steps for question 1:
R=10.5in=0.875ft
V=18.33ft/s=65388ft/h
W=65388ft/h/0.875ft=75414.86 rev per hour.

However the answer is 12003 rev per hour.

Again please help!! I am desperate for a solution

 

[pappus]

A cylinder on a printing press has a 21 in. diameter. The linear speed of a point on the cylinder's suface is 18.33 ft/s. What is its angular speed in revolutions per hour?

The wheel of a bicycle has a 67 cm diameter. The linear speed during a race was 40.423 km/h. What is its angular speed in revolutions per hour?

Please help; I have tried to solve these for over 2 hours and I still can't get the right answer. Thanks so much!

These are my steps for question 1:
R=10.5in=0.875ft
V=18.33ft/s=65388ft/h
W=65388ft/h/0.875ft=75414.86 rev per hour. <--- 1 revolution corresponds to 1 circumference of the cylinder

However the answer is 12003 rev per hour.

Again please help!! I am desperate for a solution

....

 

[soroban]

Hello, Help!

Okay, here's the solution in baby-steps.
Hope you can use this for a template on future probems.

A cylinder on a printing press has a 21-inch diameter.
The linear speed of a point on the cylinder's surface is 18.33 ft/s.
What is its angular speed in revolutions per hour?

The speed of the point is: .\(\displaystyle \dfrac{18.33\,\text{ft} }{1\,\text{sec}}
\)

This equals: .\(\displaystyle \dfrac{18.33\,\rlap{/\!/}\text{ft}}{1\,\rlap{///}\text{sec}} \times \dfrac{12\,\text{in}}{1\,\rlap{/\!/}\text{ft}} \times \dfrac{3600\,\rlap{///}\text{sec}}{1\,\text{hr}} \;=\;791,856\text{ in/hr} \)

The circumference of the cylinder is \(\displaystyle 21\pi\) inches.
. . Hence: .\(\displaystyle \text{1 rev} \:=\:21\pi\text{ in.}\)

Therefore: .\(\displaystyle \dfrac{791,856\,\rlap{/\!/}\text{in}}{1\,\text{hr}} \times \dfrac{1\text{ rev}}{21\pi\,\rlap{/\!/}\text{in}} \;=\;\dfrac{791,\!856\text{ rev}}{21\pi\text{ hr}} \;=\;12,\!002.6473\text{ rev/hr}\)

The angular speed is approximately \(\displaystyle 12,003\text{ rev/hr.}\)