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Another Circular motion question

  1. Apr 6, 2005 #1
    thanks so much for any help..work and formula used would be appreciated...

    A record of diameter 30cm roates on a turntable at 33.3r/min.

    a.) How fast is the outside edge of the record moving?

    b.) how many times as fast would it move if the frequency were raised to 78 r/min.?
     
  2. jcsd
  3. Apr 6, 2005 #2

    a) if the whole record moves at 33.3 revs per min, then a point on the outside edge does as well, which means that point has to go around the circumfrence (C) of the record within the minute, your speed= C/min.

    b) same idea as part a, change the frequecny, and then compare.
     
  4. Apr 6, 2005 #3
    so i did this out..for A.) i got 3138m/s
    for B.) i got 7351m/s, therefore it's 2.34 times as fast...is that correct?
     
  5. Apr 7, 2005 #4
    ok ya, i mean it had to go around the circumfrence 33.3 times per minute, but thats what you did, so good. Those are the numbers i got, except your units are wrong, its cm/min not m/s.
     
  6. Apr 7, 2005 #5

    dextercioby

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    Gale,i get double for the first number...

    [tex]v=\omega R \ [m \ s^{-1}] [/tex]

    R=0.3m

    [tex] \omega=2\pi \nu=\left(2\pi \ \mbox{rad}\right) \left(\frac{33.3}{60} \ Hz \right) \simeq \frac{200\pi}{180} \mbox{rad} \ s^{-1} [/tex]

    Ergo

    [tex] v\simeq \frac{\pi}{3} m \ s^{-1} = \frac{6000\pi}{3} cm \ (min)^{-1}[/tex]

    which is double that the # you referred to in post #4.

    Daniel.


    EDIT:As Gale pointed out,the radius is only 0.15m.So that explains the incorrect result i got.
     
    Last edited: Apr 7, 2005
  7. Apr 7, 2005 #6
    you got double because you let R=.3 whereas .3 is the diameter.
     
  8. Apr 7, 2005 #7

    dextercioby

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    ****.:redface: :rofl: :tongue2:

    Daniel.
     
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