Question

Two particles $A$ and $B$ are moving in a horizontal place anticlockwise on two different concentric circles with different constant angular velocities $2\omega$ and $\omega$ respectively. Find the relative velocity (in $ms$) of $B$ w.r.t $A$ after time $t=\pi /\omega$. They both start at the position as shown in figure (Take $\omega =3radsec,r=2m$)

Answer 1

Given,

Angular velocity $\omega =2rad/s$

Radius, $r=2m$ .

Time, $t=\frac{\pi }{\omega }$

Radius of A and B is $rand2r$

Angular velocity of A and B is $2\omega and\omega$

Tangential velocity of A and B is.

$V_A=V\sin \left(2\omega t\right)\hat{i}+V\cos \left(2\omega t\right)j$

$V_A=\left(2\omega \right)r[\sin \left(2\omega t\right)\hat{i}+\cos \left(2\omega t\right)j]$

At time,$t=\frac{\pi }{\varpi }$, $V_A=2\omega r[\sin \left(2\pi \right)\hat{i}+\cos \left(2\pi \right)\hat{j}]=2\omega r\hat{j}$

$V_B=V\sin \left(2\varpi t\right)\hat{i}+V\cos \left(2\varpi t\right)j$

$V_B=\omega \left(2r\right)[\sin \left(\varpi t\right)\hat{i}+\cos \left(\varpi t\right)j]$

AT time,$t=\frac{\pi }{\varpi }$ ,$V_B=2\omega r[\sin \left(\pi \right)\hat{i}+\cos \left(\pi \right)\hat{j}]=2\omega r(-\hat{J})$

Relative velocity $V_A-V_B=2\omega r\hat{j}-2\omega r(-\hat{j})=4\omega r\hat{j}=4\times 3\times 2=24m{s}^{-1}$

Hence, Relative velocity is $24m{s}^{-1}$