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Circular motion of a ferris wheel

  1. Sep 26, 2005 #1
    circular motion ferris wheel

    The Ferris wheel in the figure , which rotates counterclockwise, is just starting up. At a given instant, a passenger on the rim of the wheel and passing through the lowest point of his circular motion is moving at 3.00 m/s and is gaining speed at a rate of 0.500 m/s^2. The radius of the wheel is 14.0 m. link for pic

    How do I Find the magnitude of the passenger's acceleration at this instant.

    The formula arad=v^2/r does not work. How do I solve it?

    How do I find the direction of the passenger's acceleration at this instant in degrees to the right of vertical?
     
    Last edited: Sep 26, 2005
  2. jcsd
  3. Sep 26, 2005 #2
    anyone? Help please!
     
  4. Sep 26, 2005 #3

    Päällikkö

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    [itex]\mathbf{a}_{rad}[/itex] and [itex]\mathbf{a}_{tan}[/itex] are perpendicular and components of the same acceleration [itex]\mathbf{a}[/itex]. That is [itex]\mathbf{a}_{rad} + \mathbf{a}_{tan} = \mathbf{a}[/itex].

    What sort of acceleration increases the magnitude of velocity?
     
    Last edited: Sep 26, 2005
  5. Sep 26, 2005 #4
    still completely confused
     
  6. Sep 26, 2005 #5

    Päällikkö

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    [tex]|\mathbf{a}_c| = \frac{v^2}{r}[/tex]
    [tex]|\mathbf{a}_t| = \mbox{0,500 }m/s^2[/tex]

    [tex]\mathbf{a}_c + \mathbf{a}_t = \mathbf{a}[/tex]
    [itex]\mathbf{a}_c[/itex] and [itex]\mathbf{a}_t[/itex] are perpendicular.
    What is [itex]|\mathbf{a}|[/itex]?
     
  7. Sep 26, 2005 #6
    what? what is at, and how is that 0,500 m/s^2
     
  8. Sep 26, 2005 #7

    Päällikkö

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    Sorry, I changed the indexes :)
    [itex]|\mathbf{a}_t|[/itex] is the magnitude of the tangential acceleration. That is the acceleration that changes the magnitude of velocity (you've probably noticed that in uniform circular motion the magnitude of velocity remains unchanged, even though the particle's accelerated by a centripetal force. This force only changes the direction of the particle).

    Now there's tangential acceleration involved, which is tangent to the path and thus accelerates the particle along the path. This means that the magnitude of velocity changes too.
    [tex]|\mathbf{a}_t| = \frac{d|\mathbf{v}|}{dt}[/tex]

    EDIT: Actually the latter equation's a bit dodgy as [itex]|\mathbf{a}_t|[/itex] must be equal to or greater than 0, and [tex]\frac{d|\mathbf{v}|}{dt}[/tex] can be anything.
    So, I suppose, a better way to say that would be something like
    [tex]a_t = \frac{d|\mathbf{v}|}{dt}[/tex]
    where [tex]\mathbf{a}_t = a_t \hat{\mathbf{t}}[/tex]
     
    Last edited: Sep 26, 2005
  9. Sep 26, 2005 #8

    Päällikkö

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    I drew an image for you.
    I forgot there was a problem to solve :), so this one's a more general one.
    http://img54.imageshack.us/my.php?image=img0042uf.jpg

    EDIT: [itex]\mathbf{a}[/itex] should be a bit more to the left (in the picture), as [itex]\mathbf{a} = \mathbf{a}_c + \mathbf{a}_t[/itex].
     
  10. Sep 26, 2005 #9
    Sorry, for my elementary understanding, but what is d in the formula for at?
    I know that ac=.643 m/s^2, but I can't figure out how to solve for at.
     
  11. Sep 27, 2005 #10

    Päällikkö

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    Ah, d's for derivatives:
    [tex]a = \lim \limits_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}[/tex]
    If you're not familiar with that, just ignore it.

    I'll try a different approach:
    Tangential acceleration is parallel to (tangential) velocity, which means that that is acceleration changing the magnitude of velocity. Centripetal acceleration does indeed affect the velocity, but only by turning its direction. It does not change the particle's speed (that is, the magnitude of velocity).
    Thus [itex]a_t =\mbox{0,500 }m/s^2 [/itex]

    Add the vectors and calculate the magnitude of the result-vector:
    [tex]a = \sqrt{a_c^2 + a_t^2}[/tex]
     
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