Wall thickness

The hollow metal ball has an outside diameter of 40 cm. Determine the wall thickness if the weight is 25 kg and the metal density is 8.45 g/cm3.

Result

h =  0.607 cm

Solution:

D=40 cm R=D/2=40/2=20 cm  m=25 kg=25⋅ 1000 g=25000 g ρ=8.45 g/cm3  m=ρV V=m/ρ=25000/8.45≐2958.5799 cm3  V=V1−V2 V=43⋅ π⋅ R3−43⋅ π⋅ (R−h)3 3⋅ V4π−R3=−(R−h)3  R−h=R3−3⋅ V4π3 h=R−R3−3⋅ V4π3=20−203−3⋅ 2958.57994⋅ 3.14163≐0.6068≐0.607 cmD=40 \ \text{cm} \ \\ R=D/2=40/2=20 \ \text{cm} \ \\ \ \\ m=25 \ kg=25 \cdot \ 1000 \ g=25000 \ g \ \\ ρ=8.45 \ \text{g/cm}^3 \ \\ \ \\ m=ρ V \ \\ V=m/ρ=25000/8.45 \doteq 2958.5799 \ \text{cm}^3 \ \\ \ \\ V=V_{1}-V_{2} \ \\ V=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ R^3-\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ (R-h)^3 \ \\ \dfrac{ 3 \cdot \ V }{ 4 \pi } - R^3=-(R-h)^3 \ \\ \ \\ R-h=\sqrt[3]{ R^3 - \dfrac{ 3 \cdot \ V }{ 4 \pi } } \ \\ h=R -\sqrt[3]{ R^3 - \dfrac{ 3 \cdot \ V }{ 4 \pi } }=20 -\sqrt[3]{ 20^3 - \dfrac{ 3 \cdot \ 2958.5799 }{ 4 \cdot \ 3.1416 } } \doteq 0.6068 \doteq 0.607 \ \text{cm}

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