Solveeit Logo

Question

Physics Question on System of Particles & Rotational Motion

Four particles, each of mass 1 kg, are placed at the corners of a square of side 1m in the X- Y plane. If the point of intersection of the diagonals of the square, is taken as the origin, the co-ordinates of the centre of mass are:

A

(1, 1)

B

(-1, 1)

C

(1,-1)

D

(0, 0)

Answer

(0, 0)

Explanation

Solution

Given: m1=m2=m3=m4=1kg{{m}_{1}}={{m}_{2}}={{m}_{3}}={{m}_{4}}=1\,kg AB=BC=CD=DA=1mAB=BC=CD=DA=1m Hence, the co-ordinates of A, B, C, D are given in the figure, from the relation for xcm{{x}_{cm}} is xcm=m1x1+m2x2+m3x3+m4x4m1+m2+m3+m4{{x}_{cm}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}+{{m}_{4}}{{x}_{4}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}} =1×0.5+1×0.5+1(0.5)+1×(0.5)1+1+1+1=0=\frac{1\times 0.5+1\times 0.5+1(-0.5)+1\times (-0.5)}{1+1+1+1}=0 ycm=m1y1+m2y2+m3y3+m4y4m1+m2+m3+m4{{y}_{cm}}=\frac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}+{{m}_{3}}{{y}_{3}}+{{m}_{4}}{{y}_{4}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}} =1×0.5+1×(0.5)+1×(0.5)+1×(0.5)1+1+1+1=\frac{1\times 0.5+1\times (-0.5)+1\times (-0.5)+1\times (0.5)}{1+1+1+1} Co-ordinates of centre of mass will be (0,0).