Are angular acceleration and velocity frame dependent? 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.
 A: 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!
A: First of all, how could you be sure that the same force $F$ will produce  rotation about different axes simultaneously?
When you apply some force on the body, calculating from your body fixed frame, it will have certain torque about the origin. If you change the system, then you will get different values of the torque, but here torque has all the 3 components, so you cannot apply $\tau=I\alpha$ formulae.
What you can do is to apply Euler's equation of dynamics and find $\omega_x,\,\omega_y,\,\omega_z$, at some instant $t = t_1$ this values will give you rotation about some instantaneous axes 
Now changing the coordinate system will give you different values of the $\omega$ vector's components. But to your surprise, at that very instant $t = t_1$, $\omega_x',\,\omega_y',\, \omega_z'$ will have such values that will indicate the same rotation axis (instantaneous) as we got before ,
For your convenience, you can go through Euler's Equation of dynamics, David Morin's Classical Mechanics textbook.
A: If the body is a point mass, and even if frames are stationary, still the angular velocity and accelerations may or may not come out to be the same. The question is a little too generic. 
Consider for example, the case of a uniform circular motion, if the angular velocity is measured about an axis which does NOT pass through the centre, it will be a function of time, this is because the torque of the centripetal force will NOT be zero about such an axis.
On the other hand, consider a point mass moving in a straight line. The angular velocity about axes (passing through all points on a parallel line and perpendicular to the plane of motion) will be the same. 
It mainly has to do with the expression for torque which comes from what Force the point mass is experiencing. If the torque depends on which axis is chosen, then $\omega$ and $\alpha$ will change.
If it is not a point mass, then firstly you can not measure $\omega$ and $\alpha$ about points which are not on the object/rigid body (unless its acting as the Instantaneous Centre). But, about all points on the rigid body, $\omega$ and $\alpha$ will be the same.
