Question

A particle is thrown from point B as shown in figure. At this moment of time, the angular speed of particle with respect to an axis passing through point A and perpendicular to plane of figure is

• A. $\frac{\left(v_b\sin 2{\theta }_b\right)}{r}$
• B. $\frac{(-v_b)}{r}$
• C. $\frac{\left(v_b\sin {\theta }_b\right)}{r}$
• D. $\frac{\left(v_b\right)}{r}$

Answer 1

The correct option is C $\frac{\left(v_b\sin {\theta }_b\right)}{r}$

We know that
$\omega =\frac{V_{BA\perp }}{r_{BA}}$
By resolving the velocity $\left(v_b\right)$ in horizontal and vertical components

$v_{bx}=v_b\cos {\theta }_b,v_{by}=v_b\sin {\theta }_b$

Of these components, the $y$ components of velocity can be considered, since $x$ components are along the line joining $A$ and $B$ and can only produce a translation.

Relative velocity of particle at $B$ with respect to $A$, along $y-$ axis $V_{BA\perp }=v_b\sin {\theta }_b$
Angular velocity of $B$ with respect to $A$

$\omega =\frac{V_{BA\perp }}{r_{BA}}=\frac{v_b\sin {\theta }_b}{r}$

Hence option C is the correct answer.