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A force question

  1. Oct 11, 2006 #1
    A uniform sphere of radius 5.0cm and 2.0kg mass is rolling along level ground at a speed of 30cm/s. It rolls to rest in a distance of 15m. How large a stopping force acted on it.
    15/(2*PI*.005)=47.75rev = 300 rad
    15m/0.3m/s=50s => w=6 rad/s. w_avg = (6+0)/2=3rad/s, therefore (alpha)=-.06rad/s/s. Torque T=I*(alpha). I=2/5*M*a^2. Therefore T = -1.2*10^-4 N.m; rxF=T, => F=2.4*10^-3N. Is this correct? Am I on the right track? Thanks.
     
  2. jcsd
  3. Oct 11, 2006 #2
    15/(2*PI*.005)=47.75rev = 300 rad


    shouldn't 0.005 be 0.05.
     
  4. Oct 11, 2006 #3
    but looks like you did use 0.05 to get the answer you have, just have a typo.
     
  5. Oct 13, 2006 #4
    I used the average angular velocity in this calculation to get the right answer, but why use the average value and not the value given?Thanks
     
  6. Oct 13, 2006 #5

    PhanthomJay

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    [tex]\omega_f^2 = 2(\alpha)(\theta)[/tex]
    where [tex] \omega = 6[/tex] and [tex]\theta =300 [/tex]
    solve [tex] \alpha = .06[/tex]
     
  7. Oct 13, 2006 #6

    OlderDan

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    I doubt that this is the correct approach to this problem. Were you given any information about where this stopping force is applied? Why do you assume it was applied at the edge of the ball? Could it be that this is a work/energy problem?
     
  8. Oct 16, 2006 #7
    v=.3m/s ,s=15m. .5*m*v^2 + .5*I*w^2 = F*s, where I=2/5*m*(.05)^2
    threrfore: .09 + .036 =F*15 => F=8.4*10^-3
     
  9. Oct 16, 2006 #8

    OlderDan

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    I did not check your computatiuon, but that is conceptually correct IF the force is appied at the axis of rotation. If the force were applied at the top of the ball, the work would be the same, but the force would be only half as much because the distance over which that force is appied would be the distance a point on the vertical equator of the sphere moves, twice as far as the center of the ball moves. All this assumes the ball keeps rolling without slipping. If it slipped, friction would do some of the work.
     
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