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Homework Help: Space Curvilinear Motion question. Very hard

  1. Aug 27, 2010 #1
    1. The problem statement, all variables and given/known data

    Diagram from picture attached
    R= 200 + 50sin (2pi nt), n=2
    R= 200+ 50sin(4pi*t) mm
    ThetaDot = 120 rev/minute
    = 12.566 rad/s
    Gamma(Y) = 30 degrees

    Calculate acceleration when velocity is max

    2. Relevant equations
    ar= rDotdot - r*(thetaDot)^2
    aTheta= r*thetaDotdot + 2*rDot*thetaDot
    az = zDotdot

    3. The attempt at a solution

    At max velocity, cos(4pi*t) = 1
    so t=0, 0.25, or 0.5
    so sin(4pi*t) will always = 0

    Resolve R to get the radius at max velocity
    r = Rsin(theta)
    = (0.2+ 0.05sin(4pi*t)) * sin 30 , sin(4pi*t) = 0
    =0.2 * sin 30 = 0.1 m

    Differentiate R and we will get oscillation velocity
    dR/dt= 0.2pi*cos(4pi*t)


    Resolve the velocity to get Vr
    Vr= [0.2pi*cos(4pi*t)] * sin 30
    =0.1pi*cos(4pi*t)
    = 0.1pi

    Differentiate dR/dt would get the oscillation acceleration

    d2R/dt2= -0.8*pi^2*sin(4pi*t)
    at max velocity, -0.8*pi^2*sin(4pi*t) = 0

    So all there is left is aTheta
    aTheta= r*thetaDotdot + 2*rDot*thetaDot
    = 0 + 2 * (0.1pi) * 12.566
    = 7.895

    a of ball = aTheta
    =7.895 m/s^2

    But the real answer is 17.66!! I asked my friends, my senior, and even my TUTOR cant answer the question! This is just an exercise, not an assignment.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Aug 16, 2011 #2
    You have to add in the acceleration towards the centre of the circle. That works out at [itex]\dot{\theta}[/itex]2Rwhich is 12.572x0.1= 15.8. Then take the resultant of the 2 components, [itex]\sqrt{7.8952+15.82}[/itex] = 17.66.
     
    Last edited: Aug 16, 2011
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