# instantaneous velocity center

The instant center of rotation, also called the instantaneous velocity center is the point fixed to a body undergoing planar movement that has zero velocity at a particular instant of time. For example, for a rolling wheel that would be the point where the wheel touches the floor. It is known that one can calculate velocity of any point of a rotating body as if the body rotates around an axis going through the instant center of rotation. However, if one wants to calculate acceleration of a point of a rotating body this approach does not work, for example for a wheel rolling with a uniform speed on a planar surface it would obviously produce a completely wrong answer. Why does this approach work for velocity but does not work for acceleration?

Say you have a body pivoted about a point.

The velocity vector of a point located at $$\boldsymbol{r}$$ from the pivot is

$$\boldsymbol{v} = \boldsymbol{\omega} \times \boldsymbol{r}$$

What this means is the velocity is vector field (a vector that depends on position only) and that for each location there is only one velocity vector associated with that location. As a result, you can work backwards from a known location and velocity and extract the center of rotation by

$$\boldsymbol{r} =- \frac{\boldsymbol{\omega} \times \boldsymbol{v}}{\| \boldsymbol{\omega} \|^2}$$

Now acceleration of the same point is calculated from the derivative of the velocity using the product rule

$$\boldsymbol{a} = \boldsymbol{\alpha} \times \boldsymbol{r} + \boldsymbol{\omega} \times \boldsymbol{v}$$

where $$\boldsymbol{\alpha}$$ is the time derivative of $$\boldsymbol{\omega}$$.

But this calculation has two parts, the vector field part $$\boldsymbol{\alpha} \times \boldsymbol{r}$$ (called Euler's acceleration) and the part $$\boldsymbol{\omega} \times \boldsymbol{v}$$ which points towards the center of rotation.

But since the value of acceleration depends on the position and the velocity you can only recover the center of acceleration when it is identical to the center of rotation.

There just isn't enough information there to get it from $$\boldsymbol{a}$$ alone. But if you know the velocity at this point, you can recover the center of acceleration by first subtracting the velocity related terms

$$\boldsymbol{r} =- \frac{\boldsymbol{\alpha} \times ( \boldsymbol{a} - \boldsymbol{\omega} \times \boldsymbol{v})}{\| \boldsymbol{\alpha}\|^2}$$

Because that point is the instantaneous velocity center i.e. its velocity is zero. But, it doesn't mean that its acceleration is zero too. If you want to calculate the acceleration of a point like this method, you had to find the instantaneous zero acceleration center.

Instantaneous center of zero $$X$$ is the instantaneous point of zero $$X$$, right! $$X$$ $$\varepsilon$$ {Velocity, Acceleration}

Instantaneous center of rotation(ICR) is nothing but the instantaneous center of zero velocity(only) so that it's easy for us to visualize the kinematics of the rolling body which was rather difficult when it was a mixture of translation and rotation.

You can anyways only calculate the acceleration of the ICR of a body with circular symmetry easily, by calculating it's centripetal acceleration wrt the actual center of rotation(i.e. the center of a rolling sphere) because the acceleration of the point of contact due to the acceleration of the center of mass and a backward tangential acceleration exactly cancel each other. The only thing that remains is an angular(centripetal) acceleration whose magnitude is equal to $$v^2/r$$ where $$v$$ is the velocity of the center of mass and $$r$$ is the radius of the circular symmetry.

But in order to find the acceleration of any general point the only ways you can do it is either to find the instantaneous center of zero acceleration and then find the required acceleration wrt to this or to find some sort of constrained relation(for eg. the in-extensibility of a string or immobility of a point of contact) and then framing an equation by assuming a variable for the required acceleration with those that can be found relatively easily. You may also require to use many other details given in the question and then solving the system of equations.