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So we got this as homework today, and I just don't understand it; how to start or how to do it. Explanations and/or setups would be great :)

A wheel with radius of $5$ feet is rotating at $100$ rpm.

a) Determine the angular speed of the wheel in radians per minute.

b) Determine the linear speed of a point on the circumference of the wheel in feet per minute.

Thank you so much.

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Hint: (a) What does "rpm" stands for? How many radians are there in 1 rotation. (b) If you follow that point on circumference for 1 minute, what is the length of the curve traced out by that point. –  achille hui Feb 5 '13 at 1:52
    
As it is homework, would you mind editing in the "homewrok" tag? –  Gerry Myerson Feb 5 '13 at 1:54
    
rpm means rotations per minte. I don't know... I would, if I would know how to... –  Isabell Feb 5 '13 at 2:01
    
@GerryMyerson I just added the hw tag. –  anorton Feb 5 '13 at 2:13

2 Answers 2

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This question is probably a better fit for Physics.SE. (Don't worry about that, though--if it is a better fit there, it will be moved without you having to do anything.)

EDIT: Spoiler Alert. I kinda forgot this was homework. I walk through solving the problems, so if you don't want to see it, don't look... :)

Anyway... The first part simply asks for you to find angular speed in radians per minute. This is a somewhat simple unit conversion, if you have learned about radians. Recall from trigonometry (and I guess precalc--not sure where they're introduced) that there are $2\pi$ radians in one revolution. That is, $2\pi \text{ rad} = 1\text{ rev}$.

Converting units:

$$\begin{array}{c|c} 100\text{ rev} & 2\pi \text{ rad} \\ \hline \text{min} & \text{rev} \end{array} = 200\pi\frac{\text{rad}}{\text{min}}\approx 628\frac{\text{rad}}{\text{min}}$$

Second part: This one is a little more tricky--that is, you need to know a physics formula: $$v=r\omega$$ This is one you should memorize and hold dear--it will serve you well throughout the rest of your studies of rotary motion. Now that I've said it's important, what does it mean? Well, this equation relates linear velocity and rotational velocity. $\omega$ represents rotational velocity in $\frac{\text{rad}}{\text{some time unit}}$, $r$ is the radius of the rotating body, and $v$ is linear velocity.

Using that equation, and the previous result: $$v=r\omega$$ $$v=(5\text{ feet})\left(628\frac{\text{rad}}{\text{min}}\right)\approx3141\frac{\text{feet}}{\text{min}}$$

Addendum: Because I worked that problem, I'll create a somewhat new one for you that uses similar concepts:

a) A wheel is rotating at $10\frac{\text{rad}}{\text{sec}}$. What is the speed in rpm?

b) The same wheel (still rotating at $10\frac{\text{rad}}{\text{sec}}$) has a linear velocity of $v=5\frac{\text{feet}}{\text{sec}}$. What is the wheel's radius?

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Thank you so much!! Helped me a lot! –  Isabell Feb 5 '13 at 2:17
    
It asked for radians per minute, not seconds and feet per minute. Regards –  Amzoti Feb 5 '13 at 2:19
    
@Sabeth No problem. :) I do recommend that you try to solve the example problems I added at the end of my answer; solving those will help ensure you have a firm grasp on the concepts involved. Also, a general reminder: On StackExchange, we encourage people who ask questions to "accept" an answer. You can only accept one answer per question, and it makes that answer jump to the top of the answer set. It gives 15 points of reputation to the answerer, and ~2 points of reputation to yourself. Accepting an answer is done by clicking the green checkmark by the side of the answer. –  anorton Feb 5 '13 at 2:23
    
@Amzoti Oh wow... it's amazing how one can look at something and yet completely miss it. Thanks for catching my mistake! :) –  anorton Feb 5 '13 at 2:24
    
@anorton: no problem, I did the same thing. Also note, it has different units on both answers. Regards –  Amzoti Feb 5 '13 at 2:26

Hints:

Given a fixed speed, $v$ and radius, $r$, then:

$$v = \omega r, \omega = \frac{v}{r}, \omega = \frac{2 \pi}{time-of-1-revolution}$$

Be careful to watch for units and conversions!

Regards

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Nice hints, Amzoti! –  amWhy May 4 '13 at 0:30

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