Question

A uniform cube with mass 0.500 kg and volume 0.0270 m3 is sitting on the floor. A uniform sphere with radius 0.400 m and mass 0.800 kg sits on top of the cube. How far is the center of mass of the two-object system above the floor?

Answer 1

Given : A uniform cube with mass 0.500 kg and volume 0.0270 m3 is sitting on the floor.

A uniform sphere with radius 0.400 m and mass 0.800 kg sits on top of the cube.

To Find : How far is the center of mass of the two-object system above the floor?

Solution:

A uniform cube with mass 0.500 kg and volume 0.0270 m³

Hence side of cube =  ∛0.0270  = 0.3 m

center of mass of cube will be at center of cube hence at an height of 0.3/2

= 0.15 m   above ground

m₁ = 0.5 kg  x₁  = 0.15 m

A uniform sphere with radius 0.400 m and mass 0.800 kg sits on top of the cube.

center of mass of sphere will be at center of cube hence at an height of 0.3  + 0.4 = 0.7 m above ground

m₂ = 0.8 kg  x₂  = 0.7 m

Center of mass  =  ( m₁  x₁    +  m₂ x₂) / (m₁ + m₂ )

=  ( 0.5 * 0.15  + 0.8 * 0.7 ) / ( 0.5 + 0.8)

= ( 0.075 + 0.56) / (1.3)

=  0.488

center of mass of the two-object system above the floor =  0.488 m

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Answer 2

First of all, imagine the diagram in 2D (as attached) !

Now, let side of cube be d :

$\therefore \: d = \sqrt[3]{0.0270} = 0.3 \: m$

Now, the centre of mass of the cube will be located as the centre (i.e. at half the height of the cube) :

$\therefore \: y_{1} = \dfrac{d}{2} = 0.15 \: m$

Now, we also know that centre of mass of sphere is at its centre (see diagram):

$\therefore \: y_{2} = r + d$

$\implies\: y_{2} = 0.4 + 0.3$

$\implies\: y_{2} = 0.7 \: m$

Now, centre of mass of system :

$\therefore \: \bar{y} = \dfrac{ \sum(m_{i}y_{i}) }{ \sum( m_{i})}$

$\implies\: \bar{y} = \dfrac{ (0.5 \times 0.15) + (0.8 \times 0.7)}{(0.5 + 0.8)}$

$\implies\: \bar{y} = \dfrac{0.56 + 0.075}{1.3}$

$\implies\: \bar{y} = 0.488 \: m$

So, centre of mass is 0.488 metres above ground.