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Circular motion

  1. Mar 29, 2005 #1
    An exhaust fan rotates at 9.0 x 10^2 revolutions per minute. Find the acceleration of a point on a blade at a distance of 25cm from the axis of rotation.

    I don't know where to begin. Can someone help?
     
  2. jcsd
  3. Mar 29, 2005 #2

    dextercioby

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    How about:what theory would have to use to get it solved...?Formulas and a reasoning for using them...

    Open your book.

    Daniel.
     
  4. Mar 29, 2005 #3
    Ok I think the ? has to do with uniform circular motion. The only equation I am given is a (sub) c = v^2 divided by r
    A (sub) c = centripetal acceleration
    v = velocity
    r = radius
     
  5. Mar 29, 2005 #4

    dextercioby

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    [itex] a_{cp}=\omega^{2} r [/itex] is the formula that u need.

    U have been give the frequency.Find the angular velocity "omega"...

    Daniel.
     
  6. Mar 29, 2005 #5
    isnt that the same formula i just gave you?
     
  7. Mar 29, 2005 #6

    dextercioby

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    Not really.There's a close connection between the 2,though.So do it.

    Daniel.
     
  8. Mar 29, 2005 #7
    that is the problem I don't know how
     
  9. Mar 29, 2005 #8

    dextercioby

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    [tex] a_{cp}=(2\pi\nu)^{2}r [/tex]

    U know "r" and u know "nu",now find the acceleration.Convert from revs/min to Hz...

    Daniel.
     
  10. Mar 29, 2005 #9
    the question has nothing to do with frequency
     
  11. Mar 29, 2005 #10

    dextercioby

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    If you don't see it,that's bad.

    "rotates at 9.0 x 10^2 revolutions per minute"...what does that mean?

    Daniel.
     
  12. Mar 29, 2005 #11
    the answer i am supposed to get has nothing to do with frequency. all i need to know is the acceleration. no where in our notes and in our class did we talk about frequency.
     
  13. Mar 30, 2005 #12

    dextercioby

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    Incidentally,in this problem you are given the frequency & the distance between the point & the center of the circle.

    I can't do anything more.I've already told u what to do...

    Daniel.
     
  14. Mar 30, 2005 #13
    Johnf2004,
    In the given quetion you are given the frequency,f i.e. number of revolutions per minute. You can convert this into time period, T using:
    T = 2*pi/f

    Now use [tex] v = \frac{2 \pi r}{T} [/tex]
    as for an object moving with constant speed on a circular path, v = s/t
    Here, the distance,s is the circumference(2*pi*r) traversed in time t. But then t, the time is also equal to time taken for one complete revolution, hence equal to T.

    acceleration = (v^2)/r

    What dextercioby has explained is same , he started with [tex]\omega[/tex], the angular velocity
     
    Last edited: Mar 30, 2005
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