1.9: Hemispheres
- Page ID
- 8348
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Uniform solid hemisphere
Figure I.4 will serve. The argument is exactly the same as for the cone. The volume of the elemental slice is \( \pi y^{2} \delta x = \pi (a^{2} - x^{2} ) \delta x \) and the volume of the hemisphere is \( \frac{2 \pi a^{3}}{3} \) , so the mass of the slice is
\(M \times \pi (a^{2}-x^{2}) \delta x \div (2 \pi a / 3) = \frac{3M(a^{2}-x^{2}) \delta x }{2a^{3}} \)
where \( M \) is the mass of the hemisphere. The first moment of mass of the elemental slice is \( x \) times this, so the position of the centre of mass is
\( \overline{x} = \frac{3}{2a^3} \int_0^a x(a^{2}-x^{2})dx = \frac{3a}{8} \)
Hollow hemispherical shell.
We may note to begin with that we would expect the centre of mass to be further from the base than for a uniform solid hemisphere.
Again, Figure I.4 will serve. The area of the elemental annulus is \( 2 \pi a \delta x\) (NOT \( 2 \pi y \delta x \)!) and the area of the hemisphere is \( 2 \pi a^{2} \) . Therefore the mass of the elemental annulus is
\(M \times 2 \pi a \delta x \div (2 \pi a^{2}) = M \delta x / a\)
The first moment of mass of the annulus is x times this, so the position of the centre of mass is
\( \overline{x} = \int_0^a \frac{xdx}{a} = \frac{a}{2} \)