A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be :

Question

A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be : (A) ((2a/3) , (2b/3))  (B) ((5a/3) , (5b/3))  (C) ((3a/4) , (3b/4))  (D) ((5a/12) , (5b/12))

Tardigrade Admin · Accepted Answer

x = (M (a/2) - (M/4) ×(3a/4)/M - (M)4) = ((a/2) - (3a/16)/(3)4) = ((5a/16)/(3)4) = (5a/12) y = (M (b/2) - (M/4) ×(3b/4)/M - (M)4) = (5b/12)

Q. A uniform rectangular thin sheet $A BC D$ of mass $M$ has length $a$ and breadth $b$ , as shown in the figure. If the shaded portion $H BGO$ is cut-off, the coordinates of the centre of mass of the remaining portion will be :

$(\frac{2 a}{3}, \frac{2 b}{3})$

$(\frac{5 a}{3}, \frac{5 b}{3})$

$(\frac{3 a}{4}, \frac{3 b}{4})$

$(\frac{5 a}{12}, \frac{5 b}{12})$

$x = \frac{M \frac{a}{2} - \frac{M}{4} \times \frac{3 a}{4}}{M - \frac{M}{4}}$
$= \frac{\frac{a}{2} - \frac{3 a}{16}}{\frac{3}{4}} = \frac{\frac{5 a}{16}}{\frac{3}{4}} = \frac{5 a}{12}$
$y = \frac{M \frac{b}{2} - \frac{M}{4} \times \frac{3 b}{4}}{M - \frac{M}{4}} = \frac{5 b}{12}$

Q. A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be :

(32a​,32b​)

(35a​,35b​)

(43a​,43b​)

(125a​,125b​)