A rigid circular disc of radius r (in m) is rolling without slipping on a flat surface as shown in the figure below. The angular velocity of the disc is ω (in rad/s-1). The velocities (in m/s-1) at points 0 and A, respectively, are:

The correct answer is:

$r\omega \hat{i}\text{and}0\hat{i}$

Step-by-Step Explanation:

1. Analyzing the System from the Image:
Based on the provided illustration, we observe the following details:
- A rigid circular disc of radius $r$ is rolling on a flat horizontal surface.
- The center of the disc is denoted by point $O$, and the point of contact with the horizontal flat ground is denoted by point $A$.
- A Cartesian coordinate system is shown on the left, where the horizontal unit vector is $\hat{i}$ (directed to the right) and the vertical unit vector is $\hat{j}$ (directed upwards).
- The circular arrow labeled with angular velocity $\omega$ indicates that the disc is rotating in the clockwise direction.

2. Condition for Pure Rolling (Rolling Without Slipping):
For a body rolling without slipping on a stationary flat surface, the point of contact with the ground must be instantaneously at rest relative to the ground.
Therefore, the velocity of the point of contact $A$ is zero:
${\overrightarrow{v}}_A=0\hat{i}$

3. Determining the Velocity at the Center (Point O):
The velocity of any point on a rolling body can be related to the velocity of its center of mass $O$ using the relation:
${\overrightarrow{v}}_A={\overrightarrow{v}}_O+\overrightarrow{\omega }\times {\overrightarrow{r}}_{A/O}$
Let us write each term in vector form:
- Let the translation velocity of the center $O$ be ${\overrightarrow{v}}_O=v_O\hat{i}$.
- Since the disc rotates clockwise in the 2D plane, its angular velocity vector points into the page:
$\overrightarrow{\omega }=-\omega \hat{k}$
- The position vector of the contact point $A$ relative to the center $O$ is downwards:
${\overrightarrow{r}}_{A/O}=-r\hat{j}$

Now, substitute these vector terms into the velocity equation:
${\overrightarrow{v}}_A=v_O+(-\omega \hat{k})\times (-r\hat{j})$
Using the cross product relation $\hat{k}\times \hat{j}=-\hat{i}$, we simplify:
${\overrightarrow{v}}_A=v_O\hat{i}+r\omega (\hat{k}\times \hat{j})$
${\overrightarrow{v}}_A=(v_O$

Since the disc is rolling without slipping:
${\overrightarrow{v}}_A=0\Rightarrow v_O-r\omega =0\Rightarrow v_O=r\omega$
Thus, the velocity of the center $O$ is:
${\overrightarrow{v}}_O=r\omega \hat{i}$

Therefore, the velocities at points $O$ and $A$ are $r\omega \hat{i}$ and $0\hat{i}$ respectively.