Question

Three odd-shaped blocks of chocolate have the following masses and center-of-mass coordinates: (1) 0.310 kg , ( 0.200 m , 0.310 m ); (2) 0.410 kg , ( 0.110 m , -0.380 m );(3) 0.210 kg , ( -0.290 m , 0.630 m ).

Part A

Find the x-coordinate of the center of mass of the system of three chocolate blocks.

Part B

Find the y-coordinate of the center of mass of the system of three chocolate blocks.

Answer 1

Given,

mass and coordinates of first mass is 0.310 kg and ( 0.200 m, 0.310m)

Let first mass be m₁ and coordinates be ( x₁,y₁)

hence

m₁ = 0.310 kg

x₁ = 0.200

y₁ = 0.310

Similarly,

mass and coordinates of second mass is 0.410 kg and ( 0.110 m, -0.380m)

Let second mass be m₂ and coordinates be ( x₂,y₂)

hence

m₂ = 0.410 kg

x₂ = 0.110

y₂ = -0.380

Similarly

mass and coordinates of third mass is 0.210 kg and ( -0.290 m, 0.630m)

Let third mass be m₃ and coordinates be ( x₃,y₃)

hence

m₃ = 0.210 kg

x₃ = -0.290

y₃ = 0.630

Now

x-cordinate of center of mass is given by

=> = ( m₁*x₁ + m₂*x₂ + m₃*x₃ ) / ( m₁ + m₂ + m₃ )

         = ( 0.310*0.200 + 0.410*0.110 + 0.210*(-0.290)) / ( 0.310 + 0.410 + 0.210 )

         = ( 0.062 + 0.0451 - 0.0609 ) / ( 0.93 ) = 0.0462 / 0.93

         = 0.0497

or   = 0.0497 m

Similarly

y-cordinate of center of mass is given by

=> = ( m₁*y₁ + m₂*y₂ + m₃*y₃ ) / ( m₁ + m₂ + m₃ )

         = ( 0.310*0.310 + 0.410*(-0.380) + 0.210*0.630) / ( 0.310 + 0.410 + 0.210 )

         = ( 0.0961 - 0.1558 + 0.1323 ) / ( 0.93 ) = 0.0726 / 0.93

         = 0.0781 m

or    = 0.0781 m