The MOI = I = kmr^2, where r is the radius of gyration. The pivot point marks the geometric point relative to the body where the radius of gyration is pinned. That is to say, the body of mass m rotates around that pivot point...we sometimes call the vector perpendicular to the radius and the plane of rotation the axis of rotation. And that's the axis you are asking about.
Where that axis is depends on how the body is rotating. And how the body is rotating depends on what torque got it to rotating in the first place and where the pivot point is pinned down.
For example, we can spin a rod of length L and mass m from one end or from the middle. It depends on where the pivot is pinned down and where we apply the twisting force F over r which is the radius.
If we pin down the one end and apply the force F to the other end, so that r = L, then the MOI = kmr^2 = mL^2/3 ; where k = 1/3 [See source.] But if we do the same rod by putting the pivot in the middle r = L/2, we find that moi = mL^2/12 where k = 1/3 again but kr^2 = 1/3 * L^2/4 gives the 1/12.
As you can see, for KEa = 1/2 Iw^2 for that rod, the angular kinetic energy for the rod spinning around one end would be greater than that same rod rotating around its middle, both at the same angular speed w. That results because the MOI for the end spin is greater than the moi < MOI for the spin around the middle.
Ek = 1/2*J*ω^2 is the energy stored inside a body which is rotating around its axis of rotation ...if the axis of rotation is not passing through the body's center of mass the formula doesn't change....what changes is the moment of inertia J according to Steiner theorem
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The MOI = I = kmr^2, where r is the radius of gyration. The pivot point marks the geometric point relative to the body where the radius of gyration is pinned. That is to say, the body of mass m rotates around that pivot point...we sometimes call the vector perpendicular to the radius and the plane of rotation the axis of rotation. And that's the axis you are asking about.
Where that axis is depends on how the body is rotating. And how the body is rotating depends on what torque got it to rotating in the first place and where the pivot point is pinned down.
For example, we can spin a rod of length L and mass m from one end or from the middle. It depends on where the pivot is pinned down and where we apply the twisting force F over r which is the radius.
If we pin down the one end and apply the force F to the other end, so that r = L, then the MOI = kmr^2 = mL^2/3 ; where k = 1/3 [See source.] But if we do the same rod by putting the pivot in the middle r = L/2, we find that moi = mL^2/12 where k = 1/3 again but kr^2 = 1/3 * L^2/4 gives the 1/12.
As you can see, for KEa = 1/2 Iw^2 for that rod, the angular kinetic energy for the rod spinning around one end would be greater than that same rod rotating around its middle, both at the same angular speed w. That results because the MOI for the end spin is greater than the moi < MOI for the spin around the middle.
Ek = 1/2*J*ω^2 is the energy stored inside a body which is rotating around its axis of rotation ...if the axis of rotation is not passing through the body's center of mass the formula doesn't change....what changes is the moment of inertia J according to Steiner theorem
The axis of rotation.