Find the position of centre of mass of a uniform disc of radius R from which a hole of radius r is cut out. The centre of the hole is at a distance R / 2 from the centre of the disc.

Question

Find the position of centre of mass of a uniform disc of radius R from which a hole of radius r is cut out. The centre of the hole is at a distance R / 2 from the centre of the disc. (A) (R r2/2(R2-r2)) towards right of O (B) (R r2/2(R2-r2)) towards left of O (C) (2 R r2/(R2+r2)) towards right of O (D) (2 R r2/(R2+r2)) towards left of O

Tardigrade Admin · Accepted Answer

For a disc, mass is proportional to surface area. So Mass of cut portion: m1=k π r2, Mass of remaining portion: m2=k π(R2-r2) Apply m1 x1=m2 x2, where x1=R / 2, and solve to get x2=(R r2/2(R2-r2))

Q. Find the position of centre of mass of a uniform disc of radius $R$ from which a hole of radius $r$ is cut out. The centre of the hole is at a distance $R /2$ from the centre of the disc.

$\frac{R r ^{2}}{2 ( R ^{2} - r ^{2} )}$ towards right of $O$

$\frac{R r ^{2}}{2 ( R ^{2} - r ^{2} )}$ towards left of $O$

$\frac{2 R r ^{2}}{( R ^{2} + r ^{2} )}$ towards right of $O$

$\frac{2 R r ^{2}}{( R ^{2} + r ^{2} )}$ towards left of $O$

For a disc, mass is proportional to surface area. So
Mass of cut portion: $m_{1} = kπ r^{2}$ ,
Mass of remaining portion: $m_{2} = kπ (R^{2} - r^{2})$
Apply $m_{1} x_{1} = m_{2} x_{2}$ , where $x_{1} = R /2$ , and solve to get
$x_{2} = \frac{R r ^{2}}{2 ( R ^{2} - r ^{2} )}$

Q. Find the position of centre of mass of a uniform disc of radius R from which a hole of radius r is cut out. The centre of the hole is at a distance R/2 from the centre of the disc.

2(R2−r2)Rr2​ towards right of O

2(R2−r2)Rr2​ towards left of O

(R2+r2)2Rr2​ towards right of O

(R2+r2)2Rr2​ towards left of O