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Acceleration of clay on a potter's wheel

  1. Oct 11, 2009 #1
    What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 63 revolutions per minute if the wheel's diameter is 34 cm? Just give 2 sig figs.


    I understand that to find VELOCITY ofo the speck of clay, I'd need to find the circumference of the circle, multiply that by 63 and divide it by time, but even then there's no time given.

    But I'm asked to find acceleration...and I'm give no value for time at all, so I don't know how to go about it?
     
    Last edited: Oct 11, 2009
  2. jcsd
  3. Oct 11, 2009 #2

    lewando

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    "per minute" is a useful time reference.
     
  4. Oct 11, 2009 #3
    Yes but it just says it's turning at that speed per minute, it doesn't say it speeds up to that from rest or give any other reference speed. How can you find acceleration if you're just told "It's going this speed"?
     
  5. Oct 11, 2009 #4
    I just don't understand how you can possibly find acceleration, when you're only told the speed something is going. That's like saying "A car is going 60 miles per hour, what is it's acceleration?" Unless it's a trick question and the answer is 0 because it's speed is constant, but that's just silly?
     
  6. Oct 11, 2009 #5
    63 rpm suggests that the magnitude of velocity is constant. Therefore to find the magnitude of acceleration you merely need to find the derivative of velocity. So what is the derivative of a constant?
     
  7. Oct 11, 2009 #6

    lewando

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    Assume steady-state rotation. The speed of the particle is constant but the velocity is not.
     
  8. Oct 11, 2009 #7
    The derivative of a constant is zero...just as I said.
     
  9. Oct 11, 2009 #8
    It's velocity isn't constant because it's changing direction, but it always stops and starts in the same place, so I still believe it would just be zero right?
     
  10. Oct 11, 2009 #9
    Yes and yes. If this problem were not referring to magnitude, the equation would be a = (v^2)/r
     
  11. Oct 11, 2009 #10

    lewando

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    The magnitude of the acceleration is not zero. If it were, the velocity would not change.
     
  12. Oct 11, 2009 #11
    No, if the magnitude of acceleration is zero, then the magnitude of velocity will not change.

    The vector acceleration does not equal zero, and therefore the vector velocity will not equal zero either.
     
  13. Oct 11, 2009 #12
    That's a bit misleading. The important thing to remember here is that velocity is a vector, and though its magnitude may remain the same, it still takes a force (An acceleration) to make an object make a turn. Otherwise it would keep going in a straight line (Newton's First Law)

    [tex]\vec a \equiv \frac{d\vec v}{dt}[/tex]

    What kind of trajectory does the speck of dirt trace out as it rotates? What do you know about the acceleration required for such a trajectory?
     
  14. Oct 11, 2009 #13

    lewando

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    If the magnitude of acceleration is zero, then the magnitude of velocity will not change AND the direction of velocity will not change.
     
  15. Oct 11, 2009 #14
    Magnitude is a scalar, NOT a vector.
     
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