If there is sphere rolling on the horizontal surface without slipping. Then we know the instantaneous axis of rotation is the point which is in contact with surface. Now my doubt is, why the angular velocity taken from centre of sphere and IAOR is same ?
Please someone help me.
Put yourself at rest with the centre of the sphere of radius $R$. The value of the velocity of the contact point of the sphere and the ground is $$v= R \omega\tag{1}$$ for some $\omega$. If next you put yourself at rest with the instantaneous contact point, the value of the velocity of the centre of the sphere must be $-v$ (since the contact point is at rest with you now). However you know that, instantaneously, the motion of the sphere is a rotation around the instantaneous contact point so that it must be $$-v = R \omega'\tag{2}$$ for some $\omega'$. Comparing (1) and (2), you see that $$\omega'= -\omega\:.$$
It is a law of rigid body motion that the angular velocity vector is shared with an entire body in motion. This is a consequence of that fact that all distances between particles in the body remain constant.
In fact, it is the definition of rotational velocity
For a rigid body in motion, at any instant, there is always a unique vector $\vec{\omega}$ such that the linear velocity vector of any point located a distance $\vec{r}$ from the axis of rotation is found by $$\vec{v} = \vec{\omega} \times \vec{r}$$
The reverse is found similarly, whereas if a point has linear velocity vector $\vec{v}$ and a body is rotating by $\vec{\omega}$ the rotation axis is located at $$ \vec{r}_{\rm IAOR} = -\vec{r}= \frac{\vec{\omega} \times \vec{v} }{ \| \vec{\omega} \|^2 } $$
You can prove the above using the vector triple product identities
$$ \require{cancel} \vec{\omega} \times \vec{v} = \vec{\omega} \times ( \vec{\omega} \times \vec{r} ) = \vec{\omega} ( \cancel{\vec{\omega} \cdot \vec{r}} ) - \vec{r} ( \vec{\omega} \cdot \vec{\omega} ) =- \vec{r} \| \vec{\omega} \|^2 $$
If you imagine two points in space rotating around each other with no other points of reference, both will calculate the same value for orbital/angular momentum from their own perspective for the other point.
I believe what you're asking is similar. The IAOR is continuously changing such that it's relation to the center of the body is analogous to the relationship a fixed point of a rigid body has with the body's center (in terms of radians). The only difference is the signs would be flipped.
It's just a matter of which point you're using as a reference, both give you the same magnitude quantity just as both points in space see the other rotating around itself at the same rate.
