Find the center of mass of 3 particles at vertices of an equilateral traingle, the masses of the particles are 100g , 150g and 200g respectively . Each side of equilateral triangle is 0.5 m long.
Answers
Answered by
3
AB = BC = CA = 0.5 m
L = AB cos30°
m A = 100 g, = 0.1 kg
m B = 150 g, = 0.15 kg
m C = 200 g, = 0.20 kg
So. C.O.M. = (X cm , YCM) = (0.278 m, 0.095 m)
Case 2:- assume the third vertex to be P(x,y) while the given other two vertices are A(0,0) and B(0.5,0) and since it is an equilateral triangle:
AP = BP --------(2)
using the distance formula[
]we can get the desired result by substituting values in equation (2) and equating it.
L = AB cos30°
m A = 100 g, = 0.1 kg
m B = 150 g, = 0.15 kg
m C = 200 g, = 0.20 kg
So. C.O.M. = (X cm , YCM) = (0.278 m, 0.095 m)
Case 2:- assume the third vertex to be P(x,y) while the given other two vertices are A(0,0) and B(0.5,0) and since it is an equilateral triangle:
AP = BP --------(2)
using the distance formula[
]we can get the desired result by substituting values in equation (2) and equating it.
Similar questions