GATE Mechanical Engineering (ME) 2019 Shift 2 Solved Paper

The given body is a rigid body; Hence the given body will not have any deformation, and the possible motion is translation and angular motion. we know that the angular motion of the object is always defined as the motion about the point. Hence, first we have to find a point parallel to both velocity ${\mathit{V}}_{\mathit{P}}$ and ${\mathit{V}}_{\mathit{R}}$.$\therefore$ Angular velocity,
$\mathit{\omega }=\frac{{\mathit{V}}_{\mathit{p}}}{\mathit{P}\mathit{S}}=\frac{{\mathit{V}}_{\mathit{R}}}{\mathit{S}\mathit{R}}$
$\mathit{P}\mathit{S}=\frac{\sqrt{3}}{2}\times$ Side of triangle $=\frac{\sqrt{3}}{2}\times 1=\frac{\sqrt{3}}{2}$
$\mathit{S}\mathit{R}=\frac{1}{2}\times$ Side of triangle $=\frac{1}{2}\times 1=\frac{1}{2}$
$\mathit{\omega }=\frac{2{\mathit{V}}_{\mathit{p}}}{\sqrt{3}}=\frac{2{\mathit{V}}_{\mathit{R}}}{1}$