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Calculating the rotational component of a force on a Newtonian body?

  1. Jun 23, 2009 #1
    I am trying to simulate a square object affected by a force in two dimensions. I can break up the force vector into a vector whose extension intersects the center of gravity for translation and a vector perpendicular to that for rotation. My question is: how does this rotational vector relate to the actual rotation of the object?

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    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jun 24, 2009 #2

    Andy Resnick

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    The relationship between tangential force F_t (which is perpendicular to the surface) and torque T is T = r*F_t, where 'r' is the distance from the center of rotation to the applied force. The torque and angular acceleration are related by T = I*a, where I is the moment of inertia. Note that as the square turnes, if the force remains constant, the tangential component changes as well, because the angle the force makes with the surface varies.
  4. Jun 25, 2009 #3
    Thank you so much :)!
  5. Jun 25, 2009 #4

    Doc Al

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    The translational acceleration is governed by the total force, not just the component that intersects the center of mass.
  6. Jun 25, 2009 #5
    To elaborate on what Doc Al has said, the translational motion will be driven by both of your force components (the total force). The rotational motion will only be driven by the tangential component.
  7. Jun 26, 2009 #6

    D H

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    Careful! While that is correct for this, and for other simple freshman-level problems, it is not true in general. The inertia tensor for a rigid body is always constant in body-fixed coordinates for a rigid body (it's a rigid body, after all). It is not in general constant from the perspective of an inertia frame. Freshman-level problems are always constructed to ensure that the inertial frame inertia tensor is constant, in which case T = I*a is valid.
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