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Poking cubes in space

  1. Oct 28, 2013 #1
    Imagine you had a metal cube suspended in a zero gravity vacuum. If you were to poke the cube precisely in the middle of one of its faces exactly towards its center of mass, the cube would accelerate forward directly away from you with no angular acceleration.

    If you were to move your hand towards the edge of the face and poke in the SAME forward direction, now the cube would gain some amount of spin and would therefore not move away from you as fast.

    My question is: given a force applied to an object suspended in zero gravity, what portion of that force will be applied to the object's angular acceleration and what proportion to its lateral acceleration?

    I'm most interested in two-dimensional examples, so you can imagine the same experiment performed on a square with evenly distributed mass on a 2D plane, where all the force vectors being applied to the square are in that plane. With that, any force vector can be divided into two component vectors: the vector pointing directly towards (or away from) the object's center of mass from the point of contact and the vector perpendicular to it. Obviously the component vector pointing directly towards (or away from) the object's center of mass won't ever contribute to angular acceleration, so my question then becomes what percentage of a vector perpendicular to the vector between the point of contact and the object's center of mass will be applied to its angular acceleration?

    I expect the answer is as simple as "100% will be applied to angular acceleration" but I haven't convinced myself of that yet. And if it is that simple, I imagine adding friction makes things more interesting, so given coefficients of kinetic friction for angular and lateral movement, is it possible to calculate the portion of force perpendicular to the center of mass of the object that gets applied to angular acceleration versus lateral acceleration?

    Oh and a WONDERFUL THANKS to anyone able to help me with this. I've been mulling this over for a while and just need good convincin' more than anything else!
     
  2. jcsd
  3. Oct 28, 2013 #2

    AlephZero

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    Splitting the force into components like that was a good idea, but it doesn't answer your question easily. For example, you could apply a force that has no component towards the center of mass, so "splitting the force" wouldn't lead anywhere.

    The way to solve this is replace the applied force by the equivalent system of a force (in the same direction) at the center of mass, plus a moment about the center of mass. In 2D, the moment = (magnitude of the force) times (perpendicular distance to the center of mass). In 3D, the idea is the same but it's probably easier to use vectors - see https://engineering.purdue.edu/~aprakas/CE297/CE297-Ch3.pdf for example

    The force at the CM then produces linear acceleration, and the moment at the CM produces angular acceleration.
     
  4. Oct 28, 2013 #3

    A.T.

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    No. If you apply the same force over the same time, the linear veloctiy will be the same.
     
  5. Oct 28, 2013 #4

    AlephZero

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    That is wrong (as the previous replies have said) but it may seem paradoxical that it's wrong, because somehow "the same force" is providing "the same linear acceleration" plus some angular acceleration as well. The translational kinetic energy is always the same, but there is also some rotational kinetic energy, depending where the force is applied.

    The explanation is: if the force is applied off-center, the point where it is applied moves further in a given amount of time (because the object rotates as well as translates), so the work done by the force (= force times distance) is greater. That is where the "extra" rotational kinetic energy comes from.

    If you push by moving your hand at the same speed but at different positions, you will apply less force if you push "off-center", and so the translational velocity will be less.
     
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