Five masses are placed in a plane as shown in figure. The coordinates of the centre of mass are nearest to

Question

Five masses are placed in a plane as shown in figure. The coordinates of the centre of mass are nearest to <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/87fd4397a5d39aa22cda70d22d4203cd-.png" /> (A) (1.2,1.4) (B) (1.3,1.1) (C) (1.1,1.3) (D) (1.0,1.0)

Tardigrade Admin · Accepted Answer

x cm =(1 kg (0)+2 kg (2)+3 kg (0)+4 kg (2)+5 kg (1)/(1+2+3+4+5) kg ) =(17/15)=1.1 y cm =(1 kg (0)+2 kg (0)+3 kg (2)+4 kg (2)+5 kg (1)/(1+2+3+4+5) kg ) =(19/15)=1.3

Q. Five masses are placed in a plane as shown in figure. The coordinates of the centre of mass are nearest to

$(1.2, 1.4)$

$(1.3, 1.1)$

$(1.1, 1.3)$

$(1.0, 1.0)$

$x_{c m} = \frac{1 k g ( 0 ) + 2 k g ( 2 ) + 3 k g ( 0 ) + 4 k g ( 2 ) + 5 k g ( 1 )}{( 1 + 2 + 3 + 4 + 5 ) k g}$
$= \frac{17}{15} = 1.1$
$y_{c m} = \frac{1 k g ( 0 ) + 2 k g ( 0 ) + 3 k g ( 2 ) + 4 k g ( 2 ) + 5 k g ( 1 )}{( 1 + 2 + 3 + 4 + 5 ) k g}$
$= \frac{19}{15} = 1.3$

Q. Five masses are placed in a plane as shown in figure. The coordinates of the centre of mass are nearest to

(1.2,1.4)

(1.3,1.1)

(1.1,1.3)

(1.0,1.0)

xcm​=(1+2+3+4+5)kg1kg(0)+2kg(2)+3kg(0)+4kg(2)+5kg(1)​ =1517​=1.1 ycm​=(1+2+3+4+5)kg1kg(0)+2kg(0)+3kg(2)+4kg(2)+5kg(1)​ =1519​=1.3

$(1.2, 1.4)$

$(1.3, 1.1)$

$(1.1, 1.3)$

$(1.0, 1.0)$

$x_{c m} = \frac{1 k g ( 0 ) + 2 k g ( 2 ) + 3 k g ( 0 ) + 4 k g ( 2 ) + 5 k g ( 1 )}{( 1 + 2 + 3 + 4 + 5 ) k g}$
$= \frac{17}{15} = 1.1$
$y_{c m} = \frac{1 k g ( 0 ) + 2 k g ( 0 ) + 3 k g ( 2 ) + 4 k g ( 2 ) + 5 k g ( 1 )}{( 1 + 2 + 3 + 4 + 5 ) k g}$
$= \frac{19}{15} = 1.3$