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Sagot :
Answer:
1.86 cm
Explanation:
The center of gravity of the combined object x = m₁x₁ + m₂x₂ + m₃x₃/m₁ + m₂ + m₃ where m₁ = mass of sphere on left end of rod = 0.700 kg, x₁ = distance of center of mass of sphere from left end of combined object = radius of sphere = 8.00 cm, m₂ = mass of rod = 0.300 kg, x₂ = distance of center of mass of rod from left end of combined object = diameter of sphere on left end + length of rod/2 = 16.00 cm + 40 cm/2 = 16.00 cm + 20 cm = 36 cm, m₃ = mass of sphere on right end of rod = 0.580 kg, x₃ = distance of center of mass of sphere from left end of combined object = length of rod + diameter of first sphere + radius of sphere = 40 cm + 16 cm + 6.00 cm = 62 cm
x = (m₁x₁ + m₂x₂ + m₃x₃)/m₁ + m₂ + m₃
Substituting the values of the variables into the equation, we have
x = 0.700 kg × 8.00 cm + 0.300 kg × 36 cm + 0.580 kg × 62 cm/(0.700 kg + 0.300 kg + 0.580 kg)
x = 5.6 kg cm + 10.8 kg cm + 35.96 kg cm/1.580 kg
x = 52.36 kgcm/1.580 kg
x = 33.14 cm
Since the center of the rod is at x' = (40 cm + 16.00 cm + 12.00 cm)/2 = 70.00 cm/2 = 35 cm
The distance between the center of the rod and the center of gravity is x' - x = 35 cm - 33.14 cm = 1.86 cm
So, the center of gravity is 1.86 cm away from the center of the rod and closer to the 0.700 kg sphere.
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