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

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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)

Two identical discs of same

  1. Two identical discs of same
  2. Two identical discs of same
  3. Two identical discs of same
  4. Two identical discs of same

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}

Two identical discs of same
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.

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