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( P.19) Tangential Acceleration

  1. Jun 8, 2014 #1
    A circular disk of radius 0.010 m rotates with a constant angular speed of 5.0 rev/s. What is the acceleration of a point on the edge of the disk?
    answer is 9.9 m/s ^2

    My attempt:
    The problem is asking me for the tangential acceleration of the disk.
    At = rα

    so to solve the above equation i need the value of α

    the problem just gave me the velocity , so can i use that velocity in lie of acceleration??? why not!?
     
  2. jcsd
  3. Jun 8, 2014 #2

    Nathanael

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    If the angular speed is constant, then α (angular acceleration) is zero, right?

    (The derivative of a constant is zero.)
    (In other words, angular speed is not changing.)

    Therefore the tangential acceleration is zero.

    The question is not asking for the tangential acceleration, it just asked for the acceleration.


    What other acceleration could it be talking about?


    EDIT:
    It could be asking for tangential acceleration, but that's not what it's specifically asking for, it's just asking for whatever acceleration there is (whether it's in the tangent direction or some other direction)
     
    Last edited: Jun 8, 2014
  4. Jun 8, 2014 #3
    it is asking for LINEAR acceleration right?
    or CENTRIPETAL acceleration
    I would totally say that because the disk is in constant speed and the acceleration of the whole disk is 0 m/s^2 but the answer is 9.9 m/s^2
     
    Last edited: Jun 8, 2014
  5. Jun 8, 2014 #4

    Nathanael

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    It's asking for centripetal acceleration :)
     
  6. Jun 8, 2014 #5
    aha,OK let me see now...
     
  7. Jun 9, 2014 #6
    so I got for centripetal acceleration = rw^2
    I converted 5.0 rev to rads so I got 31.414=5 rad/s

    then I plug in

    = (0.010 m) (31.415 /s)^2
    = 9.9 m/s^2

    Question
    When I do the convertion from 5.0 to radians. Do I need to convert that 5.0 rev to radians??

    for example i had
    = (5.0 rev/s) (2 pi radians/1 rev)
    = (5.0 rev/s) [(2) (3.1415 rad/1 rev)]
    = (5.0 rev/s) (6.283 rad/1 rev)
    = 31.415 rad/s (final answer) (do I really need to put say 1 rev = 2 pi rad?
     
  8. Jun 9, 2014 #7

    Nathanael

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    Yes. The centripetal acceleration formula is:

    [itex]a_c=\frac{v^2}{r}=\frac{(ωr)^2}{r}=rω^2[/itex]

    If ω was measured in anything other than radians (like revolutions) then [itex]ωr=v[/itex] would not be true.
    (and so the above formula would not simplify to [itex]a_c=rω^2[/itex])
     
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