Question

A thin uniform wire is bent to form two equal sides An and AC where AB=AC=5the third side BC =6is twice the density distance of centre of mass from A is

Answer 1

Given : AB = 5 cm; AC = 5cm ; BC = 6cm
Cartesian Coordinates:
A = (0,0)
B = (5,0)
C = (x,y) (Say)
 AC = 5
⇒ x² + y² = 25 (Distance Formula)  ------- i
Also BC = 6 cm
⇒(x-5)² + (y - 0)² = 6² ------- ii
On Solving i and ii
We Get x = 1.4 ; y = 4.8
⇒ C = (1.4,4.8)
Mid pt Formula = [(x₁ + x₂)/2 , (y₁ + y₂)/2]
So, For Rod AC  = C.O.M = Mid pt of AC (x₁,y₁) = (0.7 , 2.4)
Similarly ,For Rod AB = C.O.M (x₂,y₂)= (2.5,0)
For Rod BC C.O.M (x₃,y₃)= (3.2,2.4)
Mass of AC m₁ = m
Mass of AB m₂ = m
Mass of BC m₃ = 2m (∵Density is increased by 2 times)
Centre of Mass of 3 particle system
(Xcm,Ycm) = {[(m₁x₁ + m₂x₂ + m₃x₃)/(m₁ + m₂ + m₃)] , {[(m₁y₁ + m₂y₂ + m₃y₃)/ (m₁ + m₂ + m₃)]}
(Xcm,Ycm) ={[m(0.7) + m(2.5) + 2m(3.2)/(m + m + 2m)] , [m(2.4) + m(0) + 2m(2.4)/(m + m + 2m)]}
(Xcm , Ycm) = (2.4, 1.8)
There For Centre of Mass(C.O.M) From A = (2.4,1,4)