Accelerationg of rotating mass *along* the axis of rotation

[birulami]

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

Thanks,
Harald.

 

[Simon Bridge]

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 m that I would like to accelerate along its axis of rotation by a. Does this count as a "change in its state of motion"?

yes. Newton's laws.

Will it resist the acceleration more that just F=m\times a.

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. a_z=F_z/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 F_r=ma and another perpendicular to that giving rise to an angular acceleration by rF_t=Iα

 

[birulami]

Yes, that was what I was after. The m\times a should have been m\cdot a. And yes, the force should point at the center of mass as to not tilt the axis of rotation.

Thanks,
Harald.

 

[Simon Bridge]

Since m is a scalar, and a is a vector, it should be just m\vec{a} ... 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 \vec{r}\wedge\vec{F}=I\vec{\alpha} and \vec{r}\cdot\vec{F} = m\vec{a}

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.