Question

Four particles of masses m. 2m
masses m, 2m, 3m, 4m are placed at the corners of a square of side 'a' as shown in fig.
3m
Find out the co-ordinates of centre of mass.​

Answer 1

Answer:

Coordinates of the centre of mass of this square are ( 0.7a , 0.5a ).

Explanation:

Let the square be located at a position so that its one corner is at origin.

Therefore,

Coordinates of other corners :

• ( a , 0 ) : Another point at x-axis.
• ( 0 , a ) : Another point at y-axis.
• ( a , a ) : Remaining corner.

From the properties of centre of mass :

• Co-ordinates of centre of mass : ( { m${}_1$x${}_1$ + m${}_2$ x${}_2$ + ...... m$_n$ x$_n$ } / { m$_1$ + m$_2$ + .....m$_n$ } , [ m${}_1$y${}_1$ + m${}_2$ y${}_2$ + ...... m$_n$ y$_n$ ] / { m$_1$ + m$_2$ + .....m$_n$ } )

*Where m$_n$ are representing the mass of the body , located at ( x$_n$ , y$_n$ ).

Here,

We have four masses with the coordinates ( 0 , 0 ) , ( 0 , a ) , ( a , a ) and ( a , 0 ) .

Therefore,

= > Coordinates of centre of mass of this square : ( [ { 0 x m } + { 0 x 2m } + { 3m x a } + { 4m x a } ] / [ m + 2m + 3m + 4m ] , [ { 0 x m } + { 2m x a } + { 3m x a } + { 4m x 0 } ] / [ m + 2m + 3m + 4m ] )

= > Coordinates of centre of mass of this square : ( [ 0 + 0 + 3ma + 4ma ] / [ 10 m ] , [ 2ma + 3ma + 0 ] / [ 10 m ] )

= > Coordinates of centre of mass of this square : ( 7ma / 10m , 5ma / 10m ) i.e. ( 0.7a , 0.5a )

Hence the required coordinates of the centre of mass of this square are ( 0.7a , 0.5a ).

Answer 2

Answer:

Co-ordinates of centre of mass, Rcm =( $\frac{1}{2} ai$, $\frac{7}{10}a j$)

Explanation:

[Refer to the attached image for the co-ordinate diagram]

Given;-

  Masses are, = m, 2m, 3m, 4m

Let the given masses lie at distance, r₁, r₂, r₃, and r₄ respectively.

Let us also assume that Mass m lies at the origin with the co-ordinates of (0, 0).

Now;-

  Centre of Mass, Cm = $\frac{m1 r1 + m2r2 + m3r3 + m4r4}{m1 + m2 + m3 + m4}$ [By the formula for centre of mass for discrete system of particles]

So, putting the given values in the formula, we get;-

           Cm = $\frac{m * 0 + 2m(a)i + 4m(a)j + 3m(a i + aj)}{m + 2m + 3m + 4m}$

           Cm = $\frac{m(5a i + 10a) j}{m(10)}$

           Cm = $\frac{1}{2} ai$ + $\frac{7}{10}a j$

Hence, the co-ordinates of the centre of mass is ($\frac{1}{2} ai$, $\frac{7}{10}a j$)

Hope it helps! ;-))