A rigid triangular body, PQR, with sides of equal length of 1 unit moves on a flat plane. At the instant shown, edge QR is parallel to the x-axis, and the body moves such that velocities of points P and R are V_P and V_R, in the x and y directions, respectively. The magnitude of the angular velocity of the body is

The correct option is 2V_R.

1. Kinematics of a Rigid Body in Plane Motion:
For any two points $P$ and $R$ on a rigid body, their velocities are related by the relative velocity relation:

$${\mathbf{V}}_P={\mathbf{V}}_R+\mathit{\omega }\times {\mathbf{r}}_{P/R}$$

where:
• ${\mathbf{V}}_P$ is the velocity vector of point $P$,
• ${\mathbf{V}}_R$ is the velocity vector of point $R$,
• $\mathit{\omega }=\omega \mathbf{k}$ is the angular velocity vector of the body (directed perpendicular to the plane of motion along the z-axis),
• ${\mathbf{r}}_{P/R}$ is the position vector of point $P$ relative to point $R$.

2. Establishing the Coordinate System:
Let us set up a Cartesian coordinate system with point $R$ as the origin $(0,0)$.
Since the edge $QR$ is parallel to the x-axis and the triangle $PQR$ is equilateral with a side length of $1$ unit:
• Point $Q$ lies at $(-1,0)$ relative to $R$.
• The apex point $P$ is situated above the midpoint of segment $QR$.
Therefore, the coordinates of $P$ relative to $R$ are:

$${\mathbf{r}}_{P/R}=-0.5\mathbf{i}+\frac{\sqrt{3}}{2}\mathbf{j}$$

3. Substituting Known Velocities:
We are given that:
• The velocity of point $P$ is along the x-direction: ${\mathbf{V}}_P=V_P\mathbf{i}$
• The velocity of point $R$ is along the y-direction: ${\mathbf{V}}_R=V_R\mathbf{j}$
Substituting these into the rigid body kinematics equation:

$$V_P\mathbf{i}=V_R\mathbf{j}+\left(\omega \mathbf{k}\right)\times (-0.5\mathbf{i}+\frac{\sqrt{3}}{2}\mathbf{j})$$

Using the cross product relations for unit vectors ($\mathbf{k}\times \mathbf{i}=\mathbf{j}$ and $\mathbf{k}\times \mathbf{j}=-\mathbf{i}$), we compute the cross product term:

$$\left(\omega \mathbf{k}\right)\times (-0.5\mathbf{i}+\frac{\sqrt{3}}{2}\mathbf{j})=-0.5\omega \mathbf{j}-\frac{\sqrt{3}}{2}\omega \mathbf{i}$$

Now, substitute this back into the velocity equation:

$$V_P\mathbf{i}=(-\frac{\sqrt{3}}{2}\omega )\mathbf{i}+(V_R-0.5\omega )\mathbf{j}$$

4. Solving for Angular Velocity:
By equating the components on both sides of the equation:
For the $\mathbf{j}$ component:

$$0=V_R-0.5\omega$$

Rearranging the equation to find the magnitude of the angular velocity $\omega$:

$$0.5\omega =V_R\Rightarrow \omega =2V_R$$

Thus, the magnitude of the angular velocity of the rigid body is $2V_R$.