Question

three particles each of mass 2kg are at the corners of an equilateral triangle side √3cm if one particle is removed find the shift in the center of the mas

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 .

Answer 2

Answer:

hope dis helps u

Explanation:

if u r satisfied with my answer please mark it as brainliest...