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Circular path-) finding period

  1. Jul 22, 2009 #1
    Circular path---) finding period

    1. The problem statement, all variables and given/known data

    A 50-g weight tied to a string is twirled in uniform circular motion. If the circumference of the circular path is 207.3 cm and the weight completes 67.1 revolutions per minute, what is the period of the motion, in seconds?

    2. Relevant equations

    V= 2* pi* r / T
    V= velocity
    r= radius
    T = period

    3. The attempt at a solution

    ok so i converted the circumference to the radius first convert to meters = 2.07300 meters

    Then divide that by pi to get the diamater then by 2 to get the radius so i arrive at .330095 as the radius.

    Where i am stuck is how to convert the revolutions per minute to the velocity and im sure the weight comes into play . Any tips would be great thanks.
     
  2. jcsd
  3. Jul 22, 2009 #2

    Nabeshin

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    Re: Circular path---) finding period

    Think about the information you are given. What does the fact that the weight completes 67.1 revolutions per minute mean? What are the units on this, and what are the units on period?
     
  4. Jul 22, 2009 #3
    Re: Circular path---) finding period

    the units on period is seconds but i dont see how i can use revolutions and the weight to obtain the velocity
     
  5. Jul 22, 2009 #4

    Nabeshin

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    Re: Circular path---) finding period

    And what are the units on the 67.1 revolutions per minute? Do you know what these units correspond to? You're trying very hard to apply a formula which isn't the best formula to apply for this given problem.
     
  6. Jul 22, 2009 #5
    Re: Circular path---) finding period

    because trying to find period i would have t = 2 * pi * r / v

    so the meters from radius and velocity cancel leaving seconds.
    But i yet to see how weight comes into play
     
  7. Jul 22, 2009 #6
    Re: Circular path---) finding period

    And what is the relation between the revolutions per minute and the period?
    The magic of it all is that the weight doesn't come into play. :)
     
  8. Jul 22, 2009 #7
    Re: Circular path---) finding period

    that was the equation provided by the instructor =(
     
  9. Jul 22, 2009 #8

    Nabeshin

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    Re: Circular path---) finding period

    Ok, the units on period are seconds but it's actually seconds/revolution, right?

    Now, you have a piece of information about revolutions/minute...
     
  10. Jul 22, 2009 #9
    Re: Circular path---) finding period

    2pi radians = 1 revolution so thats 60 seconds per 2 pi radiants so thats 10.679 radiants
     
  11. Jul 22, 2009 #10

    Nabeshin

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    Re: Circular path---) finding period

    Why are you converting into radians? Okay... let's see, you have the following:
    [tex]\frac{67.1 Revolutions}{minute}[/tex]

    Your answer is a period, which is going to be of the form
    [tex]\frac{seconds}{revolution}[/tex]

    So, your mission is basically to convert the first into the form of the second...
     
  12. Jul 22, 2009 #11
    Re: Circular path---) finding period

    Be careful, you're mixing a lot of things up.

    This is what's relevant for you:

    [tex]1 \frac{revolution}{minute}= \frac{2\pi radians}{60 seconds} = \tfrac{1}{60}\frac{revolutions}{second}[/tex]

    What you calculated just has my head spinning. @@
     
  13. Jul 22, 2009 #12
    Re: Circular path---) finding period

    i got .117 for the period

    67.1 * 2pi / 60 * 1/60
     
  14. Jul 22, 2009 #13
    Re: Circular path---) finding period

    That is incorrect. Remember, the period is in [tex]\frac{revolutions}{second}[/tex]
    And NOT in [tex]\frac{radians}{second}[/tex]
    You don't need to substitute the revolutions with [tex]2\pi[/tex]

    Just for reference, I got [tex]V\approx 1.456 \tfrac{m}{s}[/tex]
    What's the textbook's answer?
     
    Last edited: Jul 22, 2009
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