find the centre of mass of three particles of masses 100g, 150g and 200 placed at the vertices of an equilateral triangle. Each side of the triangle is 0.5m long
Answers
Answer:
Explanation:
iven that,
m
1
=100g
m
2
=150g
m
3
=200g
a = 0.5 m = 50 cm
Now, according to diagram
Center of mass coordinates
X
cm
=
450
0×100+150×50+200×25
X
cm
=
450
7500+5000
X
cm
=27.8cm
X
cm
=0.28m
Now,
Y
cm
=
450
0×100+0×150+
3
×25×200
Y
cm
=
450
8660.3
Y
cm
=19.2cm
Y
cm
=0.2m
Hence, the center of mass of three particles at the vertices of an equilateral triangle is X
cm
=0.28cm and Y
cm
=0.2cm respectively
solutioniven that,
m
1
=100g
m
2
=150g
m
3
=200g
a = 0.5 m = 50 cm
Now, according to diagram
Center of mass coordinates
X
cm
=
450
0×100+150×50+200×25
X
cm
=
450
7500+5000
X
cm
=27.8cm
X
cm
=0.28m
Now,
Y
cm
=
450
0×100+0×150+
3
×25×200
Y
cm
=
450
8660.3
Y
cm
=19.2cm
Y
cm
=0.2m
Hence, the center of mass of three particles at the vertices of an equilateral triangle is X
cm
=0.28cm and Y
cm
=0.2cm respectively
solution