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Determining revolutions of a wheel

  1. Mar 1, 2016 #1
    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. jcsd
  3. Mar 1, 2016 #2
    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).
     
  4. Mar 1, 2016 #3
    Ok. So I use ω=v/r, which gives me 20. But why is π in the denominator?
     
  5. Mar 1, 2016 #4
    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
     
  6. Mar 1, 2016 #5

    PeroK

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    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.
     
  7. Mar 1, 2016 #6

    PeroK

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    No. ##\omega \ne v/r##

    Not if you're working in revolutions per second.
     
    Last edited: Mar 1, 2016
  8. Mar 1, 2016 #7
    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
     
  9. Mar 2, 2016 #8

    PeroK

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    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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