Disc Partitioned - 1

When a disc is rolling on the ground then we know that its instantaneous axis of rotation is the point on disc in contact with the ground. Considering that point We can find the velocity of any point of the disc by using $v = ω \times r$ . Where $r$ is the position vector of any point of the disc with the point of contact as the origin and ω is the angular velocity vector. According to question,

$|v|<|V|$

$\Rightarrow |ω \times r| < |ω \times R|$ Because $V = ω \times R$ where R is the position vector of COM of disc. As angle between ω and r (radius vector) is always $\frac{\pi}{2}$ , we can write it as : $|ω| \times |r| < |ω| \times |R| \because sin(\frac{\pi}{2}) = 1$

$\Rightarrow |r| < |R|$

Thus we can make an arc of radius R from the point of contact and the area of the disc inside that arc can easily be calculated by splitting it into two segments of equal area.

The final answer is $\frac{2\pi}{3} - \frac{\sqrt{3}}{2}$ square meters.

If we adopt a coordinate system $Oxyz$ with $Ox$ horizontal, $Oy$ vertically upwards, and $O$ at the centre of the disc, then the angular velocity of the disc is $\mathbf{\omega} \; = \; \left( \begin{array}{c} 0 \\ 0 \\ -\frac{V}{R}\end{array}\right)$ (since the disc is rolling, the point of contact $A$ must be instantaneously at rest, and this determined the magnitude of the angular velocity). Thus the velocity of a point on the disc with coordinates $(x,y,0)$ is $\mathbf{v} \; =\; \left(\begin{array}{c} V \\ 0 \\ 0 \end{array} \right) + \mathbf{\omega} \times \left(\begin{array}{c} x \\ y \\ 0 \end{array} \right) \; = \; \left(\begin{array}{c} V + \frac{V}{R}y \\ -\frac{V}{R}x \\ 0 \end{array} \right)$ and hence $|\mathbf{v}| < V$ precisely when $x^2 + (y+R)^2 < R^2$ . Thus the required region of the disc is the intersection of the two discs $x^2 + y^2 < R^2 \qquad \qquad x^2 + (y+R)^2 < R^2$ By standard circle-thinking, this region has area $2\Big[\tfrac12 R^2 \tfrac{2\pi}{3} - \tfrac12 R^2 \sin\tfrac{2\pi}{3}\Big] \; = \; \Big[\tfrac23\pi - \tfrac12\sqrt{3}\Big]R^2$ making the desired answer $\tfrac23\pi - \tfrac12\sqrt{3} = \boxed{1.22837}$ .