Accelerarion and velocity of a point in accelerated circle

Question

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.

Answer 1

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 $(3)$.

Answer 2

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.