Centre of mass of hemisphere. Consider a hemisphere of radius R.
(a) Assuming a solid hemisphere with a mass/volume of p, find the centre of mass.
(b) Repeat, assuming the hemisphere is a thin shell with a mass/area of s.
(c) Compare the two values of zcM. Do they make sense?
The Correct Answer and Explanation is :
Let’s tackle the problem step-by-step:
(a) Centre of Mass of a Solid Hemisphere
The centre of mass (( z_{cM} )) of a solid hemisphere can be found using symmetry and calculus. For a solid hemisphere, the density (( \rho )) is uniform. The hemisphere is defined by the volume below the surface ( z = \sqrt{R^2 – x^2 – y^2} ) in the first octant. The calculation is most straightforward in spherical coordinates where ( x = r \sin \phi \cos \theta ), ( y = r \sin \phi \sin \theta ), and ( z = r \cos \phi ).
The centre of mass ( z_{cM} ) along the z-axis can be calculated by:
[ z_{cM} = \frac{1}{V} \int_V z \, dV ]
where ( V ) is the volume of the hemisphere.
Converting to spherical coordinates, and considering the hemisphere from ( \phi = 0 ) to ( \phi = \frac{\pi}{2} ) and ( \theta = 0 ) to ( 2\pi ):
[ dV = r^2 \sin \phi \, dr \, d\phi \, d\theta ]
The integral becomes:
[ \int_0^R \int_0^{\pi/2} \int_0^{2\pi} r \cos \phi \cdot r^2 \sin \phi \, d\theta \, d\phi \, dr ]
Evaluating this, we use the symmetry in ( \theta ), which integrates out to ( 2\pi ), and the relevant integrals for ( r ) and ( \phi ):
[ z_{cM} = \frac{1}{V} \cdot 2\pi \int_0^R \int_0^{\pi/2} r^3 \cos \phi \sin \phi \, d\phi \, dr ]
The result for a solid hemisphere (where ( V = \frac{2}{3}\pi R^3 )) is:
[ z_{cM} = \frac{3R}{8} ]
(b) Centre of Mass of a Thin Hemispherical Shell
For a thin shell, the mass is distributed only on the surface, so the density (( \sigma )) is uniform across the surface. The approach is similar but uses surface area instead of volume:
[ z_{cM} = \frac{1}{A} \int_A z \, dA ]
where ( A ) is the surface area of the hemisphere.
In spherical coordinates, the area element ( dA ) is:
[ dA = R^2 \sin \phi \, d\phi \, d\theta ]
The integral becomes:
[ \int_0^{\pi/2} \int_0^{2\pi} R \cos \phi \cdot R^2 \sin \phi \, d\theta \, d\phi ]
The result, with the surface area ( A = 2\pi R^2 ), is:
[ z_{cM} = \frac{R}{2} ]
(c) Comparison and Explanation
The centre of mass for a solid hemisphere (( \frac{3R}{8} )) is lower than for a thin hemispherical shell (( \frac{R}{2} )). This makes sense because the mass of the solid hemisphere is distributed throughout the volume, with more mass closer to the base compared to a thin shell where the mass is only on the surface, farther from the base.
Thus, the solid hemisphere has a centre of mass closer to its base, indicating a lower ( z_{cM} ) due to more mass being distributed near the flat face of the hemisphere. The results are intuitive, as the distribution of mass directly influences the centroid, pulling it toward the denser region, which in the case of the solid hemisphere is more spread throughout its volume.
Let me generate an image that visually represents these calculations for clarity.
The image above illustrates the centers of mass for two types of hemispheres: a solid hemisphere and a thin hemispherical shell. Each hemisphere is depicted with its center of mass clearly marked—( \frac{3R}{8} ) for the solid hemisphere and ( \frac{R}{2} ) for the thin shell. This visual representation should help clarify how the distribution of mass affects the center of mass in different geometric bodies.