Angular and Linear Speed (Algebra II)

Help

New member
Joined
Feb 18, 2012
Messages
1
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
 
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
....
 
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: .18.33/ ⁣/ft1///sec×12in1/ ⁣/ft×3600///sec1hr  =  791,856 in/hr\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 21π\displaystyle 21\pi inches.
. . Hence: .1 rev=21π in.\displaystyle \text{1 rev} \:=\:21\pi\text{ in.}

Therefore: .791,856/ ⁣/in1hr×1 rev21π/ ⁣/in  =  791, ⁣856 rev21π hr  =  12, ⁣002.6473 rev/hr\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 12,003 rev/hr.\displaystyle 12,003\text{ rev/hr.}
 
Top