Question

Two particles are moving in different circles in same plane with different angular velocities as shown in figure. At $t=0$, initial positions of particles $A$ and $B$ are shown by dots on the respective circles. Initial distance between particles is $1\text{m}$. Particle $A$ move anticlockwise in the first circle whereas $B$ moves clockwise in the second circle. Angle described (rotated) by $A$ and $B$ in time ${}^{\prime }{t}^{\prime }$ are ${\theta }_A=\left(\frac{\pi }{2}t\right)$ and ${\theta }_B=\left(\pi t\right)$ respectively. Here $\theta$ is in radian and $t$ is in second. Radius of each circle is shown in diagram. At time $t=2$ s, the angular velocity of the particle $A$ with respect to the particle $B$ is

• A. $\frac{5\pi }{6}$ rad/s
• B. $\frac{3\pi }{2}$ rad/s
• C. $\frac{\pi }{3}$ rad/s
• D. $\frac{\pi }{2}$rad/s

Answer 1

The correct option is A $\frac{5\pi }{6}$ rad/s

At time $t=2\text{s}$, position of $A$ and $B$ are
${\theta }_A=\left(\frac{\pi }{2}t\right)$$=\left(\frac{\pi \times 2}{2}\right)$=$\pi$ rad
${\theta }_B=\left(\pi t\right)$$=(\pi \times 2)$=$2\pi$ rad

$v_A={\omega }_1{r}_1=\frac{d{\theta }_A}{dt}{r}_1=\frac{\pi }{2}\left(1\right)=\frac{\pi }{2}\text{m/sec}$
$v_B={\omega }_2{r}_2=\frac{d{\theta }_B}{dt}{r}_2=2\pi \text{m/sec}$

Distance $AB=3\text{m}$
$\omega =\frac{v_A+v_B}{AB}=\frac{\pi /2+2\pi }{3}=\frac{5\pi }{6}\text{rad/s}$