# Determining revolutions of a wheel

1. Mar 1, 2016

### brake4country

1. The problem statement, all variables and given/known data
The picture to this problem is very basic so I don't think I have to draw one here. It's a picture of a clown juggling on a tightrope. The problem is:
If the radius of the wheel on Puncho's unicycle is 0.25 m, and Puncho is riding with a velocity of 10 m/s, how many revolutions does the wheel make each second?
A: 20/π

2. Relevant equations
C=2πr; d=vt

3. The attempt at a solution
I substituted 2πr for d and got:
2πr=vt
(2)π(.25m) = 10m/s (t)
t = 1/20 s

I know the question is asking for revolutions/1 second but I don't know where to go from here after I find a value for t. Any suggestions? Thanks.

2. Mar 1, 2016

### Aniruddha@94

Why have you taken 2πr=vt ? Is v the velocity of the centre of the wheel or the velocity of a point on the circumference of the wheel(,i.e. tangential velocity)? Also, the wheel appears to move along the rope without slipping (can you use this?)
The question asks for revolutions/s . These are units of ω (angular velocity).

3. Mar 1, 2016

### brake4country

Ok. So I use ω=v/r, which gives me 20. But why is π in the denominator?

4. Mar 1, 2016

### Aniruddha@94

You got 20 in what units? Try to convert it to rev/s
Also, how did you get 20? v/r is 10/0.25= 40

5. Mar 1, 2016

### PeroK

You are nearly there. You seem to have lost $\pi$ from your equation for $t$. If one revolution takes $t$ seconds, then how many revolutions per second is that? If you don't see this immediately, try taking $t = 0.1, 0.2, 0.5, \dots$ and try to see what's happening.

6. Mar 1, 2016

### PeroK

No. $\omega \ne v/r$

Not if you're working in revolutions per second.

Last edited: Mar 1, 2016
7. Mar 1, 2016

### brake4country

I think I got it. The MCAT doesn't require this type of problem but I wanted to attempt it anyway.

So, ω=v/r and f=ω/2π. Substituting we get: v/r/2π=40/2π

Answer: 20/π

8. Mar 2, 2016

### PeroK

That's correct, but your first approach with

$f = v/d$

Was quicker. There's no need to use the intermediate radians per second.

In your first post you simply dropped the $\pi$ from your equation.

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