Question 13: Centre of Mass and Rotational Motion

Question

From a circular disc of radius 2 R , a smaller circular disc is cut with radius of the larger disc as its diameter. The centre of the hole is at a distance of $R$ from the centre of the original disc. The distance of the centre of mass of the remaining portion from the centre is:

Accepted Answer

Let the original (large) disc of radius $2R$ have mass proportional to its area
$M_1 = \pi\,(2R)^2 = 4\pi R^2,$
and the removed (small) disc of radius $R$ have
$M_2 = \pi\,R^2.$
Its center is at distance $d=R$ from the origin.  By the Ã¢ÂÂmissingÃ¢ÂÂmassÃ¢ÂÂ method, the remaining mass is
$M_{\rm rem}=M_1 - M_2 = 3\pi R^2,$
and the shift of the center of mass (away from the hole) has magnitude
$     x = \frac{M_2}{M_{\rm rem}}\;d       = \frac{\pi R^2}{3\pi R^2}\;R       = \frac{R}{3}.   $
Answer: $\displaystyle \frac{R}{3}$  (Option A).

Answer

$\frac{R}{3}$

Answer

$\frac{R}{4}$

Answer

$\frac{R}{6}$

Answer

$\frac{R}{2}$

Centre of Mass and Rotational Motion

From a circular disc of radius 2 R , a smaller circular disc is cut with radius of the larger disc as its diameter. The centre of the hole is at a distance of R from the centre of the original disc. The distance of the centre of mass of the remaining portion from the centre is:

Choose the correct answer: