Question

Four particles of masses 1kg, 2kg, 3kg and 4kg are placed at the four vertices A, B, C and D of a square of side 1m. Find the position of centre of mass of the particles.

Answer 1

The position of centre of mass of the particles is ( 0.7, 0.5 ).

We know if (x1,y1 ) , (x2,y2) , (x3,y3) and (x4,y4) are the coordinates of the masses m1 , m2 , m3 and m4 respectively, then coordinates of centre of mass:

X = (m1x1 + m2x2 + m3x3 + m4x4 ) / ( m1 + m2 + m3 + m4 )

Y = (m1y1 + m2y2 + m3y3 + m4y4 ) / ( m1 + m2 + m3 + m4 )

Here m1 = 1Kg

m2 = 2Kg

m3 = 3Kg

m4 = 4Kg

is kept on 4 vertices of square of side 1m.

• Let vertices be ( 0 , 0 ) , ( 0 ,1 ) , ( 1 , 1 ) and ( 1 , 0 )

Then the coordinates of centre of mass of the particles will be,

• X = 1x0 + 2x0 + 3x1 + 4x1 / 1 + 2 + 3 + 4= 7/10 = 0.7

• Y= 1x0 + 2x1 + 3x1 + 4x0 / 1 +2 +3 + 4 = 5/10 = 0.5

Intuitively,  

• total mass in the lower part of the square = 5Kg

• and total mass in the upper part of the square = 5Kg.

Therefore, y coordinate of center of mass will be in the middle ie, 0.5.

Now,

• total mass on the left side of the square is 3Kg
• and total mass on the right side of the square is 7Kg.

Therefore the center of mass should be on the right side. Therefore x coordinate should be greater than 0.5., here X = 0.7

Hence verified.