# Two identical discs of same radius R are rotating about their axes

Problem: Two identical discs of same radius R are rotating about their axes in opposite directions with the constant angular speed $\omega$. The discs are in the same horizontal plane. At time t=0, the point P and Q are facing each other as shown in the figure. The relative speed between the two points P and Q is $v_r$. In one time period (T) of rotation of the discs, $v_r$ as a function of time is best represented by  (IIT JEE 2012) 1. 2. 3. 4. Solution: The relative velocity of the point P w.r.t. the point Q is given by \begin{align} \label{pca:eqn:1} \vec{v}_r=\vec{v}_P-\vec{v}_Q. \end{align} It is easy to see that $|\vec{v}_P|=|\vec{v}_Q|=\omega R$ and angle traversed in time $t$ is $\omega t$. Thus, velocities of P and Q are \begin{align} &\vec{v}_P=\omega R(-\sin\omega t\,\hat{\imath}-\cos\omega t\,\hat{\jmath}), \nonumber\\ &\vec{v}_Q=\omega R(\sin\omega t\,\hat\imath-\cos\omega t\,\hat\jmath). \nonumber \end{align} Substitute $\vec{v}_P$ and $\vec{v}_Q$ in above equation to get $\vec{v}_r=-2\omega R\sin\omega t\,\hat\imath$ and thus $v_r=2\omega R|\sin\omega t|$. We encourage you to draw $\vec{v}_{P}$ and $\vec{v}_{Q}$ at time $t=0$, $T/4$, $T/2$, and $3T/4$ and see the values of $v_r$ at these instants.