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I Question about torque and center of mass

  1. Jun 30, 2017 #1
    Supposing that we have a hollow sphere of mass m and radius R. Furthermore, suppose that we attach the sphere to a rod of length L and mass M.

    Now, suppose we rest the rod-sphere contraption on a pyramid a distance (X+R) from the center of mass of the sphere and a distance [(L/2)-X)] from the center of mass of the rod.

    m < M (total mass of the hollow sphere is less than the total mass of the rod)

    X, R, L, M, and, m are known

    This contraception is such that we can solve for T (torque) by finding the individual torque of each object (T_rod and T_sphere) and adding those torques together.


    What justification do we have for separating the system, finding the individual T of each object, and summing each T to find the total torque on the system?

    Please be very detailed.

    Second: How do we know what direction the disc and rod will travel (counter clockwise or clockwise)..... the disc will simply fall downward when a torque is applied directly to its center of mass (the contributing force of torque is the force of gravity)...??




    Third:

    Supposing we wanted to find the center of mass of the system to solve for the total torque on the system, how would this be done?

    Thank you to those esteemed individuals who would take the time to help a new such as myself
     
  2. jcsd
  3. Jul 1, 2017 #2
    I think a diagram of this contraption (not contraception) might help people help you. I can only speak for myself, but I found it unclear.
     
  4. Jul 1, 2017 #3

    Nidum

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    This problem has too much irrelevant information in it .

    All you actually have is a beam pivoted at an intermediate point and with an arrangement of weights acting at different distances from the pivot point .

    The beam and weights may or may not be in balance depending on the specific arrangement of weights and distances .

    If the beam and weights are initially in balance and then an additional weight is added to one side then that side will drop .
     
    Last edited: Jul 1, 2017
  5. Jul 1, 2017 #4

    Nidum

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    This is the basis of the balance calculation :

    The product of force times distance is called a torque or a moment . Let's say torque for this discussion .

    Torques have magnitude and direction . Direction can be clockwise or anticlockwise . So you can have positive and negative torques .

    For a pivoted beam and weights to be in balance the sum of all torques acting on the beam must be zero .

    For this simple problem you can say that forces acting are just the weights .
     
  6. Jul 1, 2017 #5

    I had to look up the definition of "spurious." The question and constraints do not imply any falsehood.

    Let us examine the situation.


    I apologize, but auto-correct made a fool of me. In any case, I am constructing a diagram based on the information given and in the steps that said information was given... will post soon
     
  7. Jul 1, 2017 #6
    If you split off part of the rod, then the part not connected to the pivot will just fall down as well. They are connected, however, and so you consider it part of a system. I'm not sure I entirely understand your question, but this is the best way I can answer it.

    If it's well balanced, it would be very difficult to predict which way the disc and rod will fall just as it would be difficult to predict which way a well balanced pencil will fall.

    The system won't be perfectly symmetrical. If the center of mass is very slightly off to one side, it may begin to fall in that direction. That would cause it to be more imbalanced, thus creating a positive feedback loop.

    I hope that answers your questions, but I do find them somewhat confusing.
     
  8. Jul 1, 2017 #7

    Doc Al

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    Whether the ball is attached or not, the torque on the ball exerted about the chosen axis would be the same. (The resulting motion will of course be different if the ball were detached.)
     
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