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Torque about an accelerating axis

  1. Sep 17, 2008 #1
    How are torques computed about an axis when the axis (and the attached body) are both accelerating at the same rate? For example, if a car is in free fall with the doors open (with the rear end of the car pointing towards the ground) will the doors shut themselves due to torque? Or, is the torque zero because the axis and door(s) are moving uniformly?

    Also, in a more general sense, if several objects are in the same accelerating reference frame, can a "pseudo" form of Newton's laws (adjusted for the fictitious forces) hold for all objects in the frame?

    This isn't homework, just a curiosity.
  2. jcsd
  3. Sep 18, 2008 #2

    Shooting Star

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    The torque about any point O due the force F acting through P is defined as OPXF. We use this all the time when studying the motion of wheels rolling down inclined planes etc. if O is the door hinge and P is any point on the door, then it may seem a bit counter intuitive to notice that the vector OP remains constant, since both P and O fall at the same rate, and therefore the torque is constant. However, the angular momentum about any instantaneously stationary horizontal axis passing through O is also increasing due to the increasing speed of the point P, and there is no contradiction.

    The car door in your example will not swing shut since the whole car is in free fall, and effectively there is no gravity in the car frame.

    In mechanical problems, in general we have to make one of the following choices:

    1. Select an initial frame and consider only the “real” forces.
    2. Select a non-inertial frame and consider not only the “real” forces, but suitably defined pseudo or fictitious forces.

    In the falling car example, the addition of the force (–mg ) to a particle of mass m makes the frame compatible with Newton’s laws of motion. Another example is a rotating frame, where you have to take into account the centrifugal and the Coriolis forces, and then start to use Newton's laws again.
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