Question

Two particles A and B are moving as shown in the figure. At this moment of time the angular speed of A with respect to B is

• A. $\frac{1}{r(v_a+v_b)}$
• B. $\frac{v_{a}sin{\theta }_a+v_{b}sin{\theta }_b}{r}$ in the anticlockwise direction
• C. $\frac{v_{a}sin{\theta }_a-v_{b}sin{\theta }_b}{r}$ in the clockwise direction
• D. $\frac{1}{r(v_a-v_b)}$

Answer 1

The correct option is C $\frac{v_{a}sin{\theta }_a-v_{b}sin{\theta }_b}{r}$ in the clockwise direction

Given: two particles A and B are moving as shown in the figure.

To find: the angular speed of A with respect to B at this moment

Solution: The angular speed on one particle with respect to other is given by the relative motion of first with respect to the second in direction perpendicular to the line joining them, divided by the distance between them.

Resolving Va and Vb in to horizontal and vertical components, we get $V_{ax}=V_{a}cos{\theta }_a;V_{ay}=V_{a}sin{\theta }_a;V_{bx}=V_{b}cos{\theta }_{b}and{V}_{by}=V_{b}sin{\theta }_b$

Of these components, the y components of velocity can be considered, since x components are along the same straight line and can only produce a translation.

The relative velocity of B w.r.t A will be $(v_{bY}-v_{aY})=(V_{b}sin{\theta }_b-V_{a}sin{\theta }_a)$

Hence

$\omega =\frac{v_{a_y}-v_{b_y}}{r}⟹\omega =\frac{v_a\sin \theta -v_b\sin \theta }{r}$

in anticlockwise direction.