Basically what I am trying to ask is if a body has an angular velocity $\omega$ or angular acceleration $\alpha$ about an axis then will it have the same angular velocity and acceleration along any other axis?
I am really confused .
As we know torque $\tau$ of all the forces acting on the body about the earlier axis is
$$\tau=I\alpha$$
Where I is the moment of inertia of the body about that axis. ( $\alpha$ depends on $\tau$ and $I$ .
If we change the axis then the radius vector of each particle from the force will change about the new axis and hence torque will change (because $\tau=r\times F$) .
Also the moment of inertia of the body about that axis will change.
Hence $\alpha$ may change or may not ( according to me) because $\tau$ and $I$ may change.
[Physics] Are angular acceleration and velocity frame dependent
rotational-dynamics
Best Answer
This issue is a bit confusing because there are two types of angular momentum. There's spin, where a rigid body rotates about an axis through its center of mass, and there's orbital, where the center of mass of a rigid body rotates about an axis. For example, the Earth spins about its axis and rotates around the Sun. The total angular momentum can always be decomposed into a sum of these two terms.
You probably have never heard of this, because the definition of the torque is just $\mathbf{r} \times \mathbf{F}$, which doesn't have any reference to "spin or orbital". But when you write the equation $\tau = I \alpha$, you're implicitly choosing to talk about one or the other, or else $\tau$, $I$, and $\alpha$ are meaningless. (For example, the Earth takes 1 day to spin but 1 year to orbit. So you can't just say "the" angular velocity of the Earth.)
If you're talking about spin, $I$ means the moment of inertia about the center of mass, $\alpha$ means the angular acceleration about an axis through the center of mass, and $\tau$ means the torque about the center of mass. There's no meaningful way to change axes or origin because it's always the center of mass.
Now let's talk about the orbital part. The instantaneous angular velocity can always be defined, even if the object isn't moving in a circle about some point, using the equation $$\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$$ However, you can see this quantity is frame dependent: if we just move the origin, $\mathbf{r}$ will change but $\mathbf{v}$ won't, so $\boldsymbol{\omega}$ will change. As an example, if you see an airplane from the ground, its angular velocity appears very low, but if you're hovering next to it, it zips around really fast. So in this case, everything does change.
Okay, maybe neither of these examples were really what you wanted: the first was trivial, and the second didn't say much. We can get more insight by not doing the spin-orbital decomposition at all, which requires tossing out $\boldsymbol{\omega}$ and $\alpha$. The only rotational equation we have left is $$\tau = \frac{dL}{dt}$$ which expanded out is $$ \sum \mathbf{r} \times \mathbf{F} = \frac{d}{dt}\left( \sum \mathbf{r} \times \mathbf{p}\right)$$ Let's transform to another frame. To make it easy, let's just shift the origin by $\mathbf{r}_0$. Now if physics works, the resulting equation should be equivalent.
Let's confirm it. In the other frame, we have $$ \sum (\mathbf{r} + \mathbf{r}_0) \times \mathbf{F} = \frac{d}{dt}\left(\sum (\mathbf{r} + \mathbf{r}_0) \times \mathbf{p}\right)$$ If we subtract out our previous equation, we're left with $$ \sum \mathbf{r}_0 \times \mathbf{F} = \frac{d}{dt} \left( \sum \mathbf{r}_0 \times \mathbf{p} \right)$$ Since $\mathbf{r}_0$ is constant, the derivative only acts on $\mathbf{p}$, giving $\sum \mathbf{r}_0 \times \mathbf{F}$ on the right hand side. So the equation is true. Physics works!