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Angular acceleration and force with constant angular velocity?

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data

    A mass M=2kg moves at a constant speed constrained by a string to move in a circle with radius .5m on a frictionless table. The tension in the string is 5N.

    What is the tangential speed?
    What is the angular velocity?
    What is the angular acceleration?
    How much work has the rope done when the mass has gone through angle [itex]\pi[/itex]?

    2. Relevant equations

    a_c = v^2/r
    [itex]\omega[/itex]=v_t/r

    3. The attempt at a solution

    I calculated that the tangential speed is equal to sqrt(1.25)m/s.
    I also calculated that the angular velocity is (.5)(sqrt(1.25) m/s counterclockwise from the positive x-axis.

    But how do I do the angular acceleration?

    Is the force equal to zero, because the force is to the center and the direction is tangential?
     
  2. jcsd
  3. Nov 3, 2011 #2

    gneill

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    Staff: Mentor

    Is the mass speeding up or slowing down, or does it have a constant speed?
     
  4. Nov 3, 2011 #3
    Constant speed
     
  5. Nov 3, 2011 #4

    gneill

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    Staff: Mentor

    So if the speed is constant, what does that say about the angular acceleration? Is the rotation rate changing?
     
  6. Nov 3, 2011 #5
    ω is constant, but it is uniformly changing direction. So there has to be an angular acceleration.

    edit --

    but since angular acceleration = tangential accel. / r -- would angular acceleration = 0?
     
  7. Nov 3, 2011 #6

    gneill

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    Staff: Mentor

    ω is not changing direction. It's a vector that's perpendicular to the plane of motion, i.e., it lies along the rotational axis. Angular acceleration, [itex]\alpha[/itex] is the rate of change of ω, and if ω is constant, [itex]\alpha[/itex] is zero.

    On the other hand, since the linear velocity vector is changing with time, there IS linear acceleration. This acceleration is in fact the centripetal acceleration (always pointing towards the center of motion).
     
  8. Nov 3, 2011 #7
    Ok thanks a lot! I really appreciate the help. I just didn't have a firm grasp on the concept.
     
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