Question

From a uniform circular disc of radius $2cm$ (its centre of mass is at '$O$') a circular portion of radius $1cm$ 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 $=2cm$
The radius of removed portion, $r=1cm$
Area of whole disc, $A_1=\pi R^2=4\pi c{m}^2$
Area of removed portion, $A_2=\pi r^2=\pi c{m}^2$
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,
$x_{CM}=\frac{m_{1}x+m_{2}r}{m_1+m_2}0=\frac{\frac{3}{4}mx+\frac{1}{4}mr}{\frac{3}{4}m+\frac{1}{4}m}$
$\Rightarrow \frac{3}{4}mx=-\frac{1}{4}mr$
or $x=-\frac{1}{3}r=-\frac{1}{3}cm$
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

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