Question 720: Centre of Mass and Rotational Motion

Question

Consider a disc rotating in the horizontal plane with a constant angular speed $\omega $ about its centre
O. The disc has a shaded region on one side of the diameter and an unshaded region on the
other side as shown in the figure. When the disc is in the orientation as shown, two pebbles P
and Q are simultaneously projected at an angle towards R. The velocity of projection is in the y - z
plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on
the disc before the disc has completed ${1 \over 8}$ rotation, (ii) their range is less than half the disc radius,
and (iii) $\omega $ remains constant throughout. Then
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l25415rw/2fd36b7e-6400-4e4a-a8ac-0e998f64abe2/bcbfdbc0-bf4d-11ec-b997-87cc5d16a405/file-1l25415rx.png?format=png)

Accepted Answer

$\left(\frac{1}{8}ight)^{\mathrm{th}}$ rotation of the disc with a constant $\Omega$ implies that the disc has turned through
$ \frac{1}{8} 	imes 360^{\circ}=45^{\circ} $
The orientation of the disc now is
For the particle thrown from $Q$ towards $R$, will have only the velocity given to it with respect to disc i.e., it possess a velocity only in the Y-Z plane and covers a range along OR. Hence, it can be seen to land on the disc in the unshaded region. Actually point $O$ of the disc has zero velocity and $Q$ being very close to $O$, will have only the velocity given to it w.r.t disc.
For the particle thrown from P , will have an additional velocity $=R \Omega$ along +X direction i.e., it will cover a range $=\frac{R}{2}$ along PO as well as a distance $=(R \omega) \frac{T}{8}=\frac{R}{8} 	imes 2 \pi=\frac{\pi R}{4}$ along $+X$ direction
The resultant displacement of particle P is thus obtained as
$\begin{aligned} \Rightarrow \quad d & =\frac{R}{2} \sqrt{1+\frac{\pi^2}{4}} \ \Rightarrow \quad \pi^2 & =10 \ d & =\frac{\mathrm{R}}{4} \sqrt{14} \ 	an 	heta & =\frac{\pi \mathrm{R} / 4}{\mathrm{R} / 2}=\frac{\pi}{2}\end{aligned}$
In the diagram shown,
$ \begin{aligned} P S & =R \cos 	heta \ & =R 	imes \frac{2}{\sqrt{14}} \ & =\frac{2 R}{\sqrt{14}} \end{aligned} $
Since, $d>$ PS $\quad$ PS $=\frac{R}{7} \sqrt{14}$
Hence, the particle from P also lands in the unshaded region.
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Answer

P lands in the shaded region and Q in the unshaded region.

Answer

P lands in the unshaded region and Q in the shaded region.

Answer

Both P and Q land in the unshaded region.

Answer

Both P and Q land in the shaded region.

Centre of Mass and Rotational Motion

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