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Swing Arm Velocity on Moving body

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data

    What is the Average Velocity of the outer end of a 0.5m swing arm weighing 2.75 lbs (1.247 kg). with the weight evenly distributed the length of the arm. The arm extends horizontally at right angles to the moving body it is attached to, pivoting on a bearing at the inner end, with a small roller ball bearing holding up the outer end above a smooth flat surface. The inner end of the arm pivots on the edge of a 9 pound (4.082 kg) body, on linear bearings, that is accelerating at a rate of 533.33 ft/s (162.56 m/s) in a straight line for a total length of 1/2" (0.0127 m). When the body reaches the end of it's travel, and is brought to an abrupt stop, it is moving at a velocity of 2.02 m/s. The arm is held at an angle of 115.0 degrees (2.007 Rad) to the center line of the direction of travel of the body by a flange to prevent it from moving any further backwards. When the body stops, the arm is free to move
    forward for a total of 50 degrees (0.873 Rad) towards the front of the body before it reaches it's own bump stop. There is a total external counter force to the forward motion of the arm of 51.80 Newtons.

    What is the final equivalent linear velocity of the outer end of the arm and what is the average equivalent linear velocity?

    2. Relevant equations

    Mass in slugs: M=W/32
    Newtons Second Law: F=M*A
    Newtons Second Law of Rotation: Torque=Iα
    Moment of Inertia: I=1/3ML2
    Torque: τ=rFSin∅
    Angular Velocity: ω2=ω02+2α∅

    3. The attempt at a solution

    Total body force F = 4.082 kg * 162.56 m/s = 667.2 Newtons

    Total force on arm F = 1.254 kg * 162.56 m/s = 203.867 Newtons

    Moment of Inertia of Arm around the end of the arm I = 1/3 * 1.254 kg * 0.5 m = 0.1045 kg m2

    Torque of arm t = 0.5 m * (203.867 N - 51.800 N) * Sin 0.873 Rad = 68.909 N m

    Torque = Iα SO
    α = torque / I = 68.909 N m / 0.1045 kg m2 = 659.369 rad/s2

    Max. Angular Velocity at end of swing = ω2
    = (2.02 m/s / 0.5 m)2 + (2 * 659.369 rad/s2 * 0.873 rad) = 34.1634 rad/s

    Equivalent Max. Linear Velocity = 34.1634 rad/s * 0.5 m = 17.0817 m/s

    Ave. Angular Velocity = ((2.02 m/s / 0.5 m) + 34.1634 rad/s) / 2 = 19.1017 rad/s

    Equivalent Ave. Linear Velocity = 19.1017 rad/s * 0.5m = 9.55 m/s
     
  2. jcsd
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