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Rotational Kinematics of a Particle

  1. Dec 5, 2007 #1
    A rigid body, starting at rest, rotates about a fixed axis with a constant angular acceleration α. Consider a particle a distance r from the axis. Express (a) the radial acceleration and (b) the tangential acceleration of this particle in terms of α, r and time t.
    c) if the resultant acceleration of the particle at some instant makes an angle of 57.0 degrees with the tangential acceleration, through what total angle has the body rotated from t=0 to that instant.

    i got radial acceleration = rω^2 = (α^2)(t^2)r
    tangential acceleration = αr

    for c i made a triangle and got cos(57deg) = cos(.99rad) = (α^2)(t^2)r/αr = αt^2

    so i plugged it into the kinematics formula
    theta = (1/2)αt^2
    theta = (1/2)cos(.99rad)
    theta = .27 radians
    but that is incorrect
  2. jcsd
  3. Dec 5, 2007 #2

    Doc Al

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    Staff: Mentor

    Why cosine? (And why change to radians?)
  4. Dec 5, 2007 #3
    will making a triangle give me any information?
    doing the same thing with tan and sin dont give me good answers

    is the resultant acceleration = sqrt([(α^2)(t^2)r]^2 + [αr]^2)
    should I be making a triangle with that?

  5. Dec 5, 2007 #4

    Doc Al

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    Staff: Mentor

    Why not? One of them will. :wink:

    Yes, but that's the same triangle.

    Just draw a right triangle with two sides being the two acceleration components.
  6. Dec 5, 2007 #5
    tan(57deg) = (α^2)(t^2)r / αr
    tan(57deg) = αt^2

    theta = (1/2)αt^2
    theta = (1/2)tan(57deg)
    theta = .7699 rad

    .7699rad * 360/2pi = 44.1 degrees

    i dont know what I was thinking initially

    but this makes more sense than what i was doing

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