Question 27: Centre of Mass and Rotational Motion

Question

From a circular disc of radius R, a square is cut out with a radius as its diagonal. The centre of mass of remaining portion is at a distance (from the centre)

Accepted Answer

Radius of the circular disc $$=R$$
Suppose, centre of mass of remaining portion is at a distance $$C_2$$ from the centre $$C$$.
Now, $$A_1\left(C C_1ight)=A_2\left(C C_2ight)$$
$$	herefore CC_2=\frac{A_1}{A_2}(CC_2)$$
For square, $$(	ext { side })^2+(	ext { side })^2=(	ext { radius })^2$$
$$\begin{array}{r}
2(	ext { side })^2=R^2 \
(	ext { side })^2=\frac{R^2}{2} \
	ext { side }=\frac{R}{\sqrt{2}}
\end{array}$$
Now, $$A_1\left(C C_1ight)=A_2\left(C C_2ight)$$
$$A_1=$$ area of the square $$=\left(\frac{R}{\sqrt{2}}ight)^2$$
$$A_2=$$ area of the remaining part of the circle
$$=\pi R^2-\left(\frac{R}{\sqrt{2}}ight)^2$$
$$\begin{aligned}
C C_2 & =\frac{\left(\frac{R}{\sqrt{2}}ight)^2}{\pi R^2-\left(\frac{R}{\sqrt{2}}ight)^2}\left(\frac{R}{2}ight) \
& =\frac{R^2}{2\left(\pi R^2-\frac{R^2}{2}ight)} \cdot \frac{R}{2} \
& =\frac{R^2}{2 \pi R^2-R^2} \cdot \frac{R}{2} \
& =\frac{R \cdot R \cdot R}{R^2(2 \pi-1) 2}=\frac{R}{4 \pi-2}\end{aligned}$$

Answer

$${R \over {(4\pi  - 2)}}$$

Answer

$${R \over {2\pi }}$$

Answer

$${R \over {\pi  - 2}}$$

Answer

$${R \over {2\pi  - 2}}$$

Centre of Mass and Rotational Motion

From a circular disc of radius R, a square is cut out with a radius as its diagonal. The centre of mass of remaining portion is at a distance (from the centre)

Choose the correct answer: