How to prove geometrically bottommost point of rolling wheel is in fact its instantaneous centre of rotation? How can we prove geometrically that net resultant velocity of any point is perpendicular to line joining bottom of wheel to the point (and wheel is Rolling without slipping on a level surface)? Hence proving velocity vectors of all the points on wheel are perpendicular to respective lines hence making that point in contact with surface as instant centre of rotation ie let point O is bottommost point of wheel and is in constact with ground now let particle P is on the wheel and is rotating with velocity v which is resultant of linear velocity of rotation about centre of mass and velocity of centre of mass how can we prove OP is perpendicular to V vector for any arbitrary P?
(Please use geometry or basic approach not calculus or matrices etc.
)
 A: First convince yourself that the movement of any rigid body can be expressed as a composition of rotation and translation (Euler's theorem + dynamics of a point paricle could do the trick).
Now focus on the bottom-most point of the wheel. It has velocity zero: 


*

*The vertical component must be zero  since the vertical projection of the movement of a point in the wheel is an harmonic movement and the bottom-most point lies in an extrem of the trajectory.

*The horizontal component must be zero since the wheel does not slide.


Since the wheel is a rigid object, its movement is decomposed in a pure translation and a pure rotation. Take for instance the movement described from the center of the wheel: all the wheel is rotating + the center of the wheel is moving forward. If instead of that, we try to describe the movement focusing in the contact point, which is stationary, the result is that the whole wheel is instantaneously rotating around the bottom-most point $O$.
The rest of it is simple algebra: Take any point on the wheel $P$, which, by the previous argument, is rotating around $O$. The segment $OP = \vec{r}$ is a radius of the circular trajectory, so $\vec{r}^2$ is constant. Then $\frac{d}{dt}{\vec{r}^2}=0\Rightarrow \vec{r}\cdot\dot{\vec{r}}=0$. The linear velocity of any point $P$ it is perpendicular to the line $OP$.
A: You combine two conditions to fix the center of rotation to the bottom contact.


*

*There is no interpenetration on the contact.
As a result, the center of rotation can only exist somewhere along the vertical line through the contact. If it where elsewhere there would be a component of velocity going through the floor, or lifting up from the contact.
From the diagram below, this constraint is that $v_y =0$, which yields $x=0$.

*There is not slip on the contact.
As a result the center of rotation must line somewhere along the ground ($y=0$) to ensure $v_x = 0$ in the diagram below. 
Combined the two conditions give use $x=0$ and $y=0$ for the contact point. For a contact where slip is allowed then only $x=0$ is imposed.


Velocity components of the rolling object, when the center of rotation is at $(x,y)$ from the contact point, and rotation is $\omega$.
$$\begin{aligned}
 v_x & = \omega\,y \\
 v_y & = \omega\,x 
\end{aligned} $$

A: This is actually a simple geometry question. You can refer to the diagram below. Note that point $P$ is a point on the wheel, $A$ is the contacting point of the wheel with the ground and all the velocity vectors labelled $v$ are of the same magnitude.

Assume that the wheel is moving at speed $v$ on the ground. That is to say, any point on the wheel has a translational velocity of $v$. The velocity of $P$ can be obtained by adding the translational velocity and the rotational velocity relative to the centre of the wheel. Note that $PM$ is tangent to circle and $OA$ is perpendicular to the ground and the velocity at $O$ is parallel to the ground. Hence, angle $MPO$ is $90$ degrees. Since $PQ$ is the angle bisector of angle $MPN$, you can easily obtain the result that $PQ$ is perpendicular to $PA$ by doing some angle chasing.
