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Accelerationg of rotating mass *along* the axis of rotation

  1. Sep 2, 2012 #1
    Says Wikipedia: "The moment of inertia is a measure of an object's resistance to any change in its state of rotation".

    Now consider a rotating mass [itex]m[/itex] that I would like to accelerate along its axis of rotation by [itex]a[/itex]. Does this count as a "change in its state of motion"? Will it resist the acceleration more that just [itex]F=m\times a[/itex]. And if yes, how much?

    Thanks,
    Harald.
     
  2. jcsd
  3. Sep 2, 2012 #2

    Simon Bridge

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    yes. Newton's laws.
    no. this is a linear acceleration - regular inertia is all you need.

    Lets be sure I understand you: something is freely rotating about its center of mass - the rotation takes place in the x-y plane so the angular momentum points in the +z direction ... the z axis is the axis of rotation.

    To accelerate the object in the +z direction, you apply an unbalanced force in the z direction through the center of mass. az=Fz/m is correct.

    An arbitrary force applied to a free body will have a component through the center of mass giving rise to a linear acceleration by Fr=ma and another perpendicular to that giving rise to an angular acceleration by rFt=Iα
     
  4. Sep 2, 2012 #3
    Yes, that was what I was after. The [itex]m\times a[/itex] should have been [itex]m\cdot a[/itex]. And yes, the force should point at the center of mass as to not tilt the axis of rotation.

    Thanks,
    Harald.
     
  5. Sep 2, 2012 #4

    Simon Bridge

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    Since m is a scalar, and a is a vector, it should be just [itex]m\vec{a}[/itex] ... don't worry about it ;)

    I could have said that, for an arbitrary force F at position vector r from the center of mass, then [itex]\vec{r}\wedge\vec{F}=I\vec{\alpha}[/itex] and [itex]\vec{r}\cdot\vec{F} = m\vec{a}[/itex]

    You realize that the Earth is a rotating body being accelerated by an unbalanced force acting through it's center of mass?

    Anyway, knowing how a general vector works on a rigid body should help you now.
     
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