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Work in a Rotational Motion

  1. Oct 16, 2006 #1
    A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m is rotating at 280 rev/min. It must be brought to a stop in 15.0s (a) How much work must be done to stop it?
    t=15.0s
    m=32.0kg
    r=1.20m
    ω= 29.3 rad/s after conversion
    I need to find W

    I think I need to find torque in order to find work. Would torque τ=Fr
    F=ma so F=32.0kg(9.8m/s^2)= 313.6 N τ=313.6(1.20m)=376.3
    Is this the right approach, if it is what equation should I use to find W. The equation in my book is an integral does that sound right? Thanks
     
  2. jcsd
  3. Oct 16, 2006 #2
    I'm not sure on this one, but I know that [tex]KE = \frac {1} {2} I\omega^2[/tex] where I is the moment of inertia (in the case of a thin hoop it is mr^2).

    I also imagine that Wnet = change in KE still applies in this situation, but I'm not sure.
     
    Last edited: Oct 16, 2006
  4. Oct 17, 2006 #3
    Thanks for writing that out I was able to figure it out using what you wrote.
     
  5. Oct 17, 2006 #4
    What is the required average power to do this? The answer to the previous question is 1.98e4 J.
    Ρ= τω for power
    so I have to find torque which is τ= Fd
    F=ma so τ= mad
    so Ρ= (mad)(29.3rad/s)
    First of all is this right? What I'm getting stuck at is that I don't think a=g (9.8) in this situation so I have to calculate a using ωf = ωi + αt
    0= 29.3 + α(15s)
    α= -1.95rad/s^2
    and plug that in for acceleration with distance= 1.2m

    Can anyone let me know if there's something I'm missing.
     
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