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Physics rotational motion

  1. Apr 13, 2016 #1
    1. The problem statement, all variables and given/known data
    A solid sphere, starting from rest, rotates with constant angular acceleration α about a fixed axis passing through its center. Consider a particle within this sphere that lies at a perpendicular distance r from this axis. If, at some instant, the linear acceleration of this particle makes an angle of 50° with respect to the direction of its tangential acceleration, what is the total angle that the sphere has turned through up until that instant?

    2. Relevant equations
    tangential acceleration = αr
    centripetal acceleration = ω^2r

    3. The attempt at a solution
    Since the angular acceleration is constant, the tangential acceleration would also remain constant and I don't see how the angle could be changing.
     
  2. jcsd
  3. Apr 13, 2016 #2

    Simon Bridge

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    Definition of acceleration.
    Note: angular acceleration is constant but not zero... who says the centripetal acceleration is constant?
    You are given the final angle... to make sense of the problem statement you need to make a decision about the initial angle and maybe other things too. You'll need to look through your notes for the right context.
     
    Last edited: Apr 13, 2016
  4. Apr 13, 2016 #3

    haruspex

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    It would be well to consider that as a vector equation. In that view, what varies?
     
  5. Apr 14, 2016 #4
    The angular velocity would be changing. But considering those as vectors, the direction would be constantly changing.
     
  6. Apr 14, 2016 #5

    haruspex

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    Exactly so. Therefore the tangential acceleration is changing (in direction).
    Starting from rest at t=0, what is its tangential acceleration at time t, and what is its radial acceleration at that time?
     
  7. Apr 14, 2016 #6
    Since its starting from rest, it would be zero at t=0. Then after that, the tangential acceleration is given by rα or (Δω)/(Δt) and the centripetal acceleration is given by (ω^2)r
     
  8. Apr 14, 2016 #7

    haruspex

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    Yes, but express ω in terms of t.
     
  9. Apr 14, 2016 #8
    ω in terms of t, so (Δθ)/(Δt)
     
  10. Apr 14, 2016 #9

    haruspex

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    Use the fact that α is constant.
     
  11. Apr 14, 2016 #10
    right, so that allows us to use the equation ω=ω(initial) + αt, and since its staring from rest, the ω(initial is zero).
     
  12. Apr 14, 2016 #11

    haruspex

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    Right. So you know the tangential acceleration at time t and the radial acceleration at time t. So in what direction is the overall linear acceleration at time t?
     
  13. Apr 14, 2016 #12
    In the direction of the rotation, between the tangential and centripetal acceleration vectors
     
  14. Apr 14, 2016 #13

    haruspex

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    You can write it as an explicit formula, in terms of r, t and alpha.
    It might help if I ask you this first: if there is a vertically upward force U acting on a body and a force R acting horizontally to the right, at what angle to the horizontal is the net force?
     
  15. Apr 14, 2016 #14
    A 45 degree angle?
     
  16. Apr 14, 2016 #15
    i have the same problem and have no clue how to start
     
  17. Apr 14, 2016 #16

    haruspex

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    No, it depends on R and U. Do you know how to draw a force diagram, with two forces at right angles, the lengths of the lines representing the magnitudes of the forces? What line represents the resultant? What angle does it make to the other lines?
     
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