Question 719: Centre of Mass and Rotational Motion

Question

Two identical discs of same radius R are rotating about their axes in
opposite directions with the same constant angular speed $\omega $. The
discs are in the same horizontal plane. At time t = 0, the points 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
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l253pwrf/e0b97c1e-5121-4e85-8649-96f0433caf1e/83e2eeb0-bf4c-11ec-b997-87cc5d16a405/file-1l253pwrg.png?format=png)

Accepted Answer

The relative velocity of the point $P$ w.r.t. the point $Q$ is given by
$ \vec{v}_r=\vec{v}_P-\vec{v}_Q $ .........(1)
It is easy to see that $\left|\vec{v}_Pight|=\left|\vec{v}_Qight|=\omega R$ and angle traversed in time $t$ is $\omega t$. Thus, velocities of $P$ and $Q$ are
$ \begin{aligned} & \vec{v}_P=\omega R(-\sin \omega t \hat{\imath}-\cos \omega t \hat{\jmath}) \ & \vec{v}_Q=\omega R(\sin \omega t \hat{\imath}-\cos \omega t \hat{\jmath}) \end{aligned} $
Substitute $\vec{v}_P$ and $\vec{v}_Q$ in equation (1) to get $\vec{v}_r=$ $-2 \omega R \sin \omega t \hat{\imath}$ and thus $v_r=2 \omega R|\sin \omega t|$.
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m03gvlg1/fc7da263-3f52-4cee-bbc2-eef78298497a/c2433c10-5f85-11ef-b175-bb7f2dd2f87a/file-jaoe38c1m03gvlg2.png?format=png)

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Centre of Mass and Rotational Motion

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