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Torque relative to an inertial frame

  1. Feb 24, 2010 #1
    I have a thought experiment I cannot resolve. Maybe someone smarter than I can resolve this. Suppose we have a long, thin rod, rotating in the counter-clockwise direction around a pivot axis which is at the end of the rod with an initial angular velocity equal to w. The axis is connected to a second object which we will denote as the slider. The slider can slide along a straight, linear track. We assume no friction in this experiment. Initially, we prevent the slider from moving, but at some point in time when the rotator has passed its zero degree position with respect to the x-axis of an x-y coordinate system, we release the mechanism which holds the slider stationary. Immediately, the slider begins to accelerate in the positive y-direction. This is because of the y-component of the centrifugual reactive force acting on the axis which is attached to the slider. This force is equal and opposite to the centripetal force which acts on the center of mass of the rod. With respect to an observer in the frame of the slider, he would observe the angular velocity of the rod will slow down due to the gravitational field in his frame caused by the increasing acceleration of the slider. The initial angular velocity w of the rod will therefore decrease and this decrease must also be observed by an observer in a laboratory inertial frame. What torque acting on the center of mass of the rod with respect to an inertial frame would cause the angular velocity of the rod to decrease? Obviously, there is no "gravitational field" to slow it down.
  2. jcsd
  3. Feb 25, 2010 #2
    Since there has been no response to this, I offer a possible explanation. If I am wrong, I am willing to listen and learn. I propose the angular velocity does decrease with respect to a laboratory frame, but it is not due to a torque. How could this be? We must include the concept of curvature. In plane vector geometry the curvature is defined as the magnitude of dt/ds, where t is the tangent vector and s is the displacement of an object. It is also derived as 1/r, where r is the radius of the circle of curvature. If we take the simple case of a rod rotating around a vertical axis that is attached to the earth, then the centripetal force can be expressed by:

    Fcen = vel sq/r m

    Here, 1/r can be substituted for the curvature k, or:

    Fcen = vel sq k m

    Now consider the thought experiment. When the rod rotates when the slider is prevented from moving, the rod is essentially rotating around an axis which is fixed to the mass of the earth, so the above equations apply. But when the slider is allowed to move, the tension in the rod does not remain constant, but is at a maximum when the slider is not moving, and is at a minimum when the rod has rotated 90 degrees and is only interacting with the mass of the slider. We can then conclude that k is not constant. K measures the tendency of an object to turn or "twist" as it moves. Since k is decreasing the tendency for the rod to twist decreases, and this is manifested as a decrease in the angular velocity of the rod with respect to a laboratory frame. Here is an analogous situation. Imagine a car with a mass of 100 kg with a tangential linear velocity of 10 m/s going around a circular track with a radius of 20 meters. The centripetal force would be:

    Fcen = 10 sq/20 100

    The curvature would be 1/20.

    Now imagine the same car going at the same tangential linear velocity of 10m/s but around a track with a radius of 100 meters. Its curvature would be 1/100, less than the first case. Even though the car's tangential linear velocity is the same, the angular velocity will be less. This is because of the relation:

    w = v/r or w = vk.

    Clearly, as k decreases the angular velocity of the car decreases. I propose this is what is happening to the rod to account for its decrease in angular velocity.
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