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Angular Acceleration and a grindstone

  1. Feb 1, 2005 #1
    HELP!!! I keep getting a solution that is not part of the choices. Where am I messing up?

    A grindstone, originally rotating at 126 rad/s undergoes a constant angular acceleration so that it makes 20.0 rev in the first 8.00s. What is the angular acceleration?


    a) 0.313 rad/s2
    b) 0.625 rad/s2
    c) 2.50 rad/s2
    d) 1.97 rad/s2
    e) 13.79 rad/s2

    My Calculations:
    20rev*2PIrad/rev = 40PI rad = theta
    t = 8 sec
    wo = 126 rad/sec

    theta = wo*t+1/2*a*t^2
    40PI = 126*8+.5*a*8^2
    40PI-126*8 = 32a
    a = -27.57 rad/sec^2
     
  2. jcsd
  3. Feb 1, 2005 #2
    Let's back up:

    A grindstone, originally rotating at 126 rad/s undergoes a constant angular acceleration so that it makes 20.0 rev in the first 8.00s. What is the angular acceleration?

    We have u = 126rads^-1
    We also have s = 20rev = 40Pi Radians
    t = 8s
    Now we're left with:
    40Pi = (126)(8) + 1/2a(64)
    40Pi = 1008 + 32a
    a = -27.57300918....rad s^-2

    Same answer here...
    Perhaps mistake in the question? (Maybe..)
     
  4. Feb 1, 2005 #3

    quasar987

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    I have -27.57 too. At any rate, none of the proposed answer have a minus sign, so they're all wrong :wink:.

    But it's quite probable that there's a mistake in the answer... because 13.79 rad/s2 is exactly half of our answer. As if the dude making the question had forgotten the 1/2 in

    [tex]\theta (t) = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2[/tex]

    which is a common mistake when one plugs all the numerical values and solve for the answer...
     
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