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Calculating Torque around moving Centre of mass

  1. Apr 5, 2015 #1
    1. The problem statement, all variables and given/known data
    A man of mass M stands on a railroad car which is rounding an unbanked turn of radius R at speed v. His center of mass is height L above the car, and his feet are distance d apart. The man is facing the direction of motion . How much weight is on each of his feet?

    2. Relevant equations
    Sum of forces in the radial direction: -f1-f2 = m v^2 / R
    Sum of forces n the vertical direction: -W+ N1 + N2 = 0
    Sum of torques = 0

    3. The attempt at a solution

    f1 and f2 are the reaction forces on each foot of the man in the radial direction, N1 and N2 are the reaction forces on each of the man's feet in the vertical (up) direction and are balanced by the weight of the man.
    I need to get an extra equation from the sum of torques around some axis of rotation, but I don't understand which axis it is. The man is not spinning in any direction, but I0m trying to get an equation for the torques caused by N1 and N2 at some distance from a point through the axis of (no-)rotation. Can I calculate the torques around the tangent to the trajectory of motion that goes trough the center of mass, even though the center of mass is being accelerated and therefore isn't an inertial frame of reference? Should I go an try to sum the torques using the center of the curve as the origin of the coordinates, since that is stationary?
  2. jcsd
  3. Apr 6, 2015 #2


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    You can pick almost any axis, for example you could pick his centre of mass or one foot. You will get a different equation but after solving all the simultaneous equations the result should be the same.

    If you picked his centre of mass you would get torques due to f1, f2, N1, N2.
  4. Apr 6, 2015 #3


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    PS: It's sometimes helpful to pick an axis that has an unknown force acting through it. That way the unknown force won't feature in the equation. For example the equation that sums the torques about the centre of mass won't contain terms for "w" or "mv^2/r" because they act through the centre of mass. In short careful choice of the axis saves you a few steps when solving the simultaneous equations.
  5. Apr 6, 2015 #4
    I assume it is not necessary to consider forces f1 and f2. This would be redundant as N1 and N2 are needed. I would choose an axis which would go through the centre point between man's feet perpendicular to the radius and the direction of gravity force(so that torques of N1 N2 and centrifugal forces are non-zero). As there is no rotation about this axis, the total torque must be zero. This equation accompanied by the second law will give you the solution.
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