Question

Three particles each of mass 2.5 Kg are located at the corners of a right angled triangle whose sides are 2m and 1.5 m long as shown in the figure. The CM is located at

Answer 1

Final Answer : 1/2 m

Steps and Understanding:

1) Let A be origin and By using properties of equilateral triangle, we get other points B, C

as

A= (0,0)

B = (√3/2, 3/2)

C= (√3,0)

2) Here, all 3 masses are same, so

Centre of mass position will be same as Centrifugal of triangle ABC,

So,

D = (( 0 + √3/2 + √3)/3 , (0+3/2+0)/3 )

= (√3/2 , 1/2)

So, Initial position of Centre of Mass is

R(i) = (√3/2,1/2)

3) Now, we have to remove any mass :

To make calculations easy, let us remove

Mass at point B.

Now,

Required masses : A, C.

Since, masses are same so,

Position of Centre of Mass will be same as Mid point of A and C.

$r(f) = x \: = \frac{0 + \sqrt{3} }{2} = \frac{ \sqrt{3} }{2} \\ \\ y \: = \frac{0 + 0}{2} = 0$

So,

R(f) = (√3/2,0)

5) Now,

R(f) - R(i) = (0, -1/2)

Magnitude = 1/2 m

So, shift in Centre of Mass = 1/2m .