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  1. I have a gyroscope that is spinning in the xy plane along the z axis.
  2. I then roll its spinning axis by some angle theta

Now I know the gyroscope will resist my attempting to change its axis of rotation.

The question is:

  1. When I perform work to change the axis of rotation.
  2. Where will that work be transferred ? ... will the rotation speed of the disk slow down?

To clarify: see the diagram below, the axis of rotation is around the Z axis - the disk is in the x-y plan. Now say the gyroscope is in a frame that allows it the disk's rotation axis to rotate in x-z plane -- which means the pivot is in the yaxis -- so assume there is a frame around the disk and there is a rod along the y axis on wither side of the disk attached to the frame. So, if I apply a fore in the -x direction at point 'a' the gyroscope will rotate and I will have to move the force a distance delta-x so clearly work is being done (lets assume no friction). The question is where does this work go?

                      /|\ z-axis
               | 'a'   |
               |       | 
             / | \     |______\ x-axis
             \ _ /      \     /
               |         \
               |         _\| y-axis


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There really isn't, any mechanical work involved. It is similar to what happens if you push really hard against a wall: you get tired, but without doing any actual work.

See, when you apply torque to your gyroscope, you do it applying a couple, i.e. two displaced, equal but opposite forces. You may e.g. push the top of the rotation axis to the right, and the bottom to the left.

When you do this, the gyroscope will start tilting perpendicularly to the forces applied, e.g. in the front-back plane in the example above. If you apply your couple in the right direction, you can tilt the axis of rotation almost effortlessly. But the movement is so counter-intuitive that you normally apply most of the force in the wrong direction, "against the wall," and have the feeling that what you are doing is really hard.

Think of the typical example:

There is clearly no energy going anywhere, as the center of mass remains at a constant height. That's because the pull of gravity is compensated by the tension of the rope. Just effortless changing of the axis of rotation...

EDIT: I am going to restate the same as above, for the example given by the OP, just adding an extra point 'b'

                  /|\ z-axis
           | 'a'   |
           |       | 
         / | \     |______\ x-axis
         \ _ /      \     /
           |         \
           | 'b'     _\| y-axis

If you apply a force F parallel to the x axis at a, the whole gyroscope will start accelerating in that direction, and the work you do will go into kinetic energy of the whole system. To prevent this from happening, what you do is apply a force F in the +x direction at a, and a force F in the -x direction at b, so the total force is zero, and the center of mass remains at rest. What you do have is a torque, which in this case would be pointing in the -y direction.

I see from your question that you are expecting that, when applying this torque, the gyroscope changes its axis of rotation in the x-z plane. But it doesn't. See, the angular momentum of your gyroscope is pointing in the z direction, and you apply a torque in the y direction, so angular momentum, i.e. axis of rotation, will change in the y-z plane. So all movement happens perpendicular to the direction in which you are applying your force, and thus no work is done.

If you wanted your axis of rotation to move in the x-z plane, the forces you apply at a and b should be pointing in the +/-y directions. But movement is always perpendicular to the direction the force is applied, so no actual work is done.

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Hi, I just updated the question with a diagram, hope that helps. – user1172468 Sep 28 '12 at 17:18
@Jamie, thanks for the update ... wanted to point out that I've said in my question that the gyroscope is in a frame that is hinged on the y axis (it will pivot in the x-z plane) -- the axis of rotation is the y-axis. (I see my question may not be clear on this point -- will update) – user1172468 Sep 29 '12 at 20:25

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