# Accelerarion and velocity of a point in accelerated circle

I want to ask about angular motion. Suppose a circle of radius $$r$$ rotating with angular acceleration $$\alpha$$. I know that the polar coordinates of a point in the perimeter of the circle is $$(r,\theta)$$. Transforming this to cartesian, it becomes $$(r\cos\theta, r\sin \theta)$$.

I know that the angular position can be calculated using below formula:

$$\theta = 0.5\,\alpha t^2$$

I can convert it using polar to cartesian formula to get $$(x,y)$$ position of the point.

How do I calculate the x-axis and y-axis acceleration and velocity of that point every time point? I assume this is related to tangential acceleration and velocity but I am not sure how to calculate this.

If $$\omega$$ be the angular velocity, then we can also write (besides the equation you have written for $$\theta$$), $$\omega=\omega_0+\alpha t\tag{1}$$ $$\omega_0$$ is the initial angular velocity of the circle at time $$t=0$$ . The angular acceleration is different from tangential or radial acceleration, i.e., linear acceleration $$a$$, in the same lines as angular velocity is different from the linear velocity $$v$$ : $$\boldsymbol v=\boldsymbol {\omega} \times \boldsymbol r\tag{2}$$ $$\boldsymbol a=\boldsymbol {\alpha} \times \boldsymbol r\tag{3}$$ What you want to know are the $$x$$ and $$y$$ components of the linear acceleration and velocity at a point at any time $$t$$ . As you know the point, you already know $$\boldsymbol r$$ .Also, $${\alpha}$$ is given from which you can calculate $$\omega$$ from $$(1)$$ provided you know $$\omega_0$$. Finally, getting these values and given the direction of $$\boldsymbol {\alpha}$$ which is same as that of $$\boldsymbol {\omega}$$, you can calculate the desired components of $$\boldsymbol v$$ and $$\boldsymbol a$$ from $$(2)$$ and $$(3)$$.
You have the $$x$$ and $$y$$ coordinates as a function of time: $$x = r \cos(0.5 \alpha t^2)$$. Find the first and second derivative for each.