# Accelerationg of rotating mass *along* the axis of rotation

## Main Question or Discussion Point

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.

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Simon Bridge
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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. 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

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 $\vec{r}\cdot\vec{F} = m\vec{a}$