A circular hole of radius ((a/2)) is cut out of a circular disc of radius 'a' as shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be :

Question

A circular hole of radius ((a/2)) is cut out of a circular disc of radius 'a' as shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be : (A) (1/6)a (B) (10/11)a (C) (5/6) a (D) (2/3) a

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9a4f9148502a2e2c0937af48ab7b9306-.png" /> Let σ be the uniform mass density of disc then x COM =((σ π a 2) a -σ π(( a 2/4)) × (3 a /2)/σ π a 2-(σ π a 2)4) =(a-(3 a/8)/1-(1)4)=(5 a/6)

Q. A circular hole of radius $(\frac{a}{2})$ is cut out of a circular disc of radius 'a' as shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be :

$\frac{1}{6} a$

$\frac{10}{11} a$

$\frac{5}{6} a$

$\frac{2}{3} a$

Let $σ$ be the uniform mass density of disc then
$x_{COM} = \frac{( σπ a ^{2} ) a - σπ ( \frac{a ^{2}}{4} ) \times \frac{3 a}{2}}{σπ a ^{2} - \frac{σπ a ^{2}}{4}}$
$= \frac{a - \frac{3 a}{8}}{1 - \frac{1}{4}} = \frac{5 a}{6}$

Q. A circular hole of radius (2a​) is cut out of a circular disc of radius 'a' as shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be :

61​a

1110​a

65​a

32​a