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1.9: Hemispheres

Last updated

Mar 5, 2022
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- 1.8: Hollow Cone
- 1.S: Centers of Mass (Summary)

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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

$\overline{x} = \int_0^a \frac{xdx}{a} = \frac{a}{2}$

Uniform solid hemisphere

Hollow hemispherical shell.

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