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.

To find the angular velocity of the rigid triangular body, we can analyze the kinematic relationship between the velocities of two points on a rigid body. For any two points $P$ and $R$ on a rigid body undergoing planar motion, their velocities are related by the equation:
$${\overrightarrow{V}}_P={\overrightarrow{V}}_R+\overrightarrow{\omega }\times {\overrightarrow{r}}_{P/R}$$
where:
- ${\overrightarrow{V}}_P$ is the velocity of point $P$, given as $V_P\hat{i}$.
- ${\overrightarrow{V}}_R$ is the velocity of point $R$, given as $V_R\hat{j}$.
- $\overrightarrow{\omega }=\omega \hat{k}$ is the angular velocity of the body.
- ${\overrightarrow{r}}_{P/R}$ is the position vector of $P$ relative to $R$.

Let us establish a Cartesian coordinate system with the origin at point $Q$. Since the edge $QR$ is parallel to the x-axis and has a length of 1 unit, the coordinates of $Q$ and $R$ are:
$$Q=(0,0)$$
$$R=(1,0)$$

Since $PQR$ is an equilateral triangle with side lengths of 1 unit, the coordinates of point $P$ are:
$$P=(1·\cos ({60}^{\circ }),1·\sin ({60}^{\circ }\left)\right)=(0.5,\frac{\sqrt{3}}{2})$$

Now, we find the relative position vector of $P$ with respect to $R$:
$${\overrightarrow{r}}_{P/R}={\overrightarrow{r}}_P-{\overrightarrow{r}}_R=(0.5-1)\hat{i}+(\frac{\sqrt{3}}{2}-0)\hat{j}=-0.5\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$$

Substitute the velocity vectors and relative position vector into the rigid body velocity formula:
$$V_P\hat{i}=V_R\hat{j}+\left(\omega \hat{k}\right)\times (-0.5\hat{i}+\frac{\sqrt{3}}{2}\hat{j})$$

Computing the cross product terms using $\hat{k}\times \hat{i}=\hat{j}$ and $\hat{k}\times \hat{j}=-\hat{i}$ gives:
$$V_P\hat{i}=V_R\hat{j}-0.5\omega \hat{j}-\frac{\sqrt{3}}{2}\omega \hat{i}$$

To solve for $\omega$, we equate the corresponding components on both sides of the equation.

Equating the $\hat{j}$ (y-axis) components:
$$0=V_R-0.5\omega$$
Rearranging the equation to solve for the magnitude of the angular velocity $\omega$ yields:
$$0.5\omega =V_R\Rightarrow \omega =2V_R$$