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A disk 8.0 cm in diameter is initially at rest. A small dot is painted

  1. Nov 14, 2004 #1
    A disk 8.0 cm in diameter is initially at rest. A small dot is painted on the edge of the disk. The disk accelerates at [tex]600rad/s^2 [/tex] for .5s, then coasts at a steady angular velocity for another .5s.
    What is the speed of the dot at t = 1 s?
    Through how many revolutions has it turned?

    I approached the first question by drawing a graph of the angular acceleration Vs. time. I then obtained the angular velocity by finding the area under the graph.
    I then used the formula V=rw to get the speed.
    I obtained 24m/s as my final result.

    Is this reasonable?
     
    Last edited: Nov 14, 2004
  2. jcsd
  3. Nov 15, 2004 #2

    James R

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    For constant angular acceleration:

    [tex]\omega = \omega_0 + \alpha t[/tex]
    [tex]\theta = \theta_0 + \omega_0 t + (1/2)\alpha t^2[/tex]
     
  4. Nov 15, 2004 #3

    HallsofIvy

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    Your graph of the angular acceleration should be a horizontal straight line so the area under it is just the area of a rectangle: the length of the segment on the "t" axis is 0.5 seconds and the height is 600 rad/s2. That's exactly the same as multiplying (0.5 s)(600 rad/s2)= 300 rad/s. "Graphing" works, but seems unecessary to me.
    To get the speed of the dot, note that at 300 rad/s it completes 300/(2pi) revolutions per second and each revolution has length 2pi(8) cm: the dot completes 300(8)= 2400 cm per second, exactly what you did.

    To get the number of revolutions completed you will need the second formula James R gave for the first 0.5 second, then add 150/2pi revolutions for the second 0.5 second.
     
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