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Circular motion

  1. Jan 31, 2015 #1
    1. The problem statement, all variables and given/known data
    A yoyo is spun with a radius of 0.5m. It reaches a velocity of 30m/s. What is the force if the yoyo is 0.3kg?
    ----
    A clock has a hand with 0.3m long and 0.5kg. It takes 60 seconds for 1 revolution. What is its acceleration? What is the force?

    If you have any tips or instructions, please comment,

    2. Relevant equations
    Ac= v^2/r
    2(3.14)r/t
    F=ma
    3. The attempt at a solution
    I attempted to solve the first one and I got 1800 m/s and 540 Newtons, but I feel this is wayyyyyyyyyyyyyy off. Please save my soul someone.
     
  2. jcsd
  3. Jan 31, 2015 #2

    gneill

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    The 540 N looks to be in the right ballpark, so that's not so worrying. To say more we'll have to see the details of how you arrived at that value.

    I'm not sure what the 1800 m/s could pertain to. It doesn't seem to be an answer to a question posed in your problem statement.
     
  4. Jan 31, 2015 #3
    For my first equation, I plugged in A=(30)^2/.5 and got 1800 m/s^2. Then I used F=ma... F=(.3)(1800) and got 540N. I wasn't there when this was taught so I'm not sure if I'm heading in the right direction. :/
     
  5. Jan 31, 2015 #4

    haruspex

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    This is not at all clear. A yo-yo has two radii - its outer radius and the radius of its axle.
    What force? Is this perhaps the tension in the string as the yo-yo descends under gravity, when it has reached 30m/s?

    Edit: Oh I see, it may be a yo-yo but it's being treated as just mass on the end of a string of fixed length. (Are you sure that's the right interpretation? It's not what is usually meant by a spinning yo-yo.)
    But is this in the presence of gravity? If so, for the force to be constant it will be spun in a horizontal plane, and gravity will affect the tension. So my question becomes whether the force is supposed to be just the centripetal force or the whole tension.
     
    Last edited: Jan 31, 2015
  6. Jan 31, 2015 #5

    gneill

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    Okay, that's fine and makes sense now that the details are there. Your results are fine.

    Now, what's your approach for the second question?
     
  7. Jan 31, 2015 #6

    gneill

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    I believe that the yo-yo in this case just represents a weight on the end of a string, said weight undergoing circular motion with the string as the radius.
     
  8. Jan 31, 2015 #7

    haruspex

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    Yes, I realised just after your previous post that that's how it was being interpreted. But please see my follow-up question in post #4.
     
  9. Jan 31, 2015 #8
    Well for the second equation I just used the 2πR/T, so 2(3.14)(.3)/60seconds=.0314m/s. Then I plugged that into A=v^2/r, A=.0314^2/.3 and got .00328 m/s^2. Then, I plugged that it F=ma.. So F=(.5)(.00328)= .00164 N.
     
  10. Jan 31, 2015 #9

    gneill

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    Okay, the calculations look fine. You may want to make sure that the number of significant figures in your final results correspond to the number of significant figures in the given data.
     
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