Physics, asked by anshsrivastava72004, 4 months ago

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

Answered by dakshgamer35
0

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

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