Question

From a uniform circular disc of radius $2\, cm$ (its centre of mass is at '$O$') a circular portion of radius $1\,cm$ is removed such that the shift in centre of mass is maximum. The disc is now rotated by an angle $\theta$ about an axis perpendicular to its plane and passing through '$O$'. If the magnitude of displacement of new centre of mass is $\frac{1}{\sqrt{3}} cm$, then the $\theta$ is

• A. $30^{\circ}$
• B. $45^{\circ}$
• C. $60^{\circ}$
• D. $120^{\circ}$

Answer 1

Answer: (D)

The situation given can be shown as.

Here, $R$ be the radius of circular disc $=2 \,cm$
The radius of removed portion, $r=1 \, cm$
Area of whole disc, $A_{1}=\pi R^{2}=4 \pi \, cm ^{2}$
Area of removed portion, $A_{2}=\pi r^{2}=\pi \,cm ^{2}$
As, $\frac{1}{4}$ th area of the disc is removed, so remaining
area of the disc is $\frac{3}{4}$ th initial area.
Similarly, remaining mass, $m_{1}=\frac{3}{4} \,m$
Mass of removed portion, $m_{2}=\frac{1}{4} \,m$
Let $x$ be the maximum shift in centre of mass.
So, initially centre of mass along $X$-axis of complete disc,
$x_{C M}=\frac{m_{1} x+m_{2} r}{m_{1}+m_{2}} 0=\frac{\frac{3}{4} m x+\frac{1}{4} m r}{\frac{3}{4} m+\frac{1}{4} m}$
$\Rightarrow \frac{3}{4} m x=-\frac{1}{4} m r$
or $x=-\frac{1}{3} r=-\frac{1}{3} c m$
i.e., the centre of mass of remaining portion will shift to the left of origin at $\frac{1}{3} cm$.
Now, the disc is rotated by angle $\theta$,
so the centre of mass will also shift by angle $\theta$ as shown

$\Delta O P Q$ is isoscales.
So, a perpendicular drawn from $0$ ,
on $P Q$ divide the angle $\theta$ and length $P Q$ in equal parts i.e.,
$R Q=\frac{1}{2} P Q=\frac{1}{2}\left(\frac{1}{\sqrt{3}}\right) cm $
$\left(\right.$ given, $\left.P Q=\frac{1}{\sqrt{3}}\right)$
From $\triangle O R Q$.
$\sin \frac{\theta}{2}=\frac{R Q}{O Q}=\frac{\frac{1}{2} \times \frac{1}{\sqrt{3}}}{\frac{1}{3}}$
$\Rightarrow \sin \frac{\theta}{2}=\frac{\sqrt{3}}{2}=\sin 60^{\circ}$
$\Rightarrow \theta=120^{\circ}$