Question

10
2
Particle X and particle Y are initially positioned at
(-a, 0) and (0, -b) respectively. At t = 0, the
particles X and Y start moving with uî and vi
respectively. Maximum angular speed of the line
joining the particles is​

Answer 1

Angular speed of the line joining the particles is $\frac{ub-va}{a^{2}+b^{2}}$​

Explanation:

In given question particle X is point A moving with velocity u, particle Y at point B moving with velocity v.

From figure

$\sin \theta = \frac{b}{\sqrt{a^{2}+b^{2}}}$

$\cos \phi = \frac{a}{\sqrt{a^{2}+b^{2}}}$

1. Component of velocity of particle X  perpendicular to line joining A and B is $u\sin \theta$.

2. Component of velocity of particle Y  perpendicular to line joining A and B is $v\sin \phi$.

3. Relative velocity of particle X with respect to Y perpendicular to line joining AB

$V_{XY}=u\sin \theta - v\sin \phi$

4. Angular speed with respect to Y  

$\omega =\frac{Tangential speed}{perpendicular distance}$

Where tangential speed is  $V_{XY}=u\sin \theta - v\sin \phi$

and Perpendicular distance XY is $\sqrt{a^{2}+b^{2}$

5. So $\omega = \frac{u\sin \theta -v\sin \phi}{\sqrt{a^{2}+b^{2}}}$

on putting value $\sin \theta$ and $\sin \phi$ in above equation

6. We get $\omega =\frac{ub-va}{a^{2}+b^{2}}$