A Mechanics Problem Involving Suspension of a Uniform Solid Hemisphere From a Fixed Point

The following is a useful Mechanics 3 level problem involving suspension of a rigid body (a uniform solid hemisphere) from a fixed point. It uses the result that when a rigid body is suspended freely from a fixed point and rests in equilibrium, its centre of mass is vertically below the point of suspension. This problem also uses a result about the location of the centre of mass of a uniform solid hemisphere which I derive in my suggested solution below.

The Problem

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 Suggested solution

The centre of mass G of the hemisphere must be located vertically below the point of suspension A, as shown in the following diagram:

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The diagram also shows the angle \theta which we need to find and the radius r of the hemisphere. If we knew the length XG, i.e., the distance between the centre of mass of the hemisphere and its base, we could solve the problem using simple trigonometry.

To find the length XG we imagine a uniform solid hemisphere of radius R with the centre of its base at the origin:

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Since the hemisphere is uniform we can ignore density (mass per unit volume) in calculating its centre of mass, and use only volumes. We imagine dividing the hemisphere into discs of infinitesimal width dx with centre of mass at x. The volume of each disc is \pi y^2 dx. The centre of mass \bar{x} of the hemisphere is then found from the equation

\bar{x} \int_{x=0}^R \pi y^2 dx = \int_{x=0}^R x \pi y^2 dx

Using y^2 = R^2 - x^2 in this we get

\bar{x} = \frac{\int_{x=0}^Rx\pi(R^2 - x^2)dx }{\int_{x=0}^R\pi(R^2 - x^2)dx}

= \frac{\int_{x=0}^R(xR^2 - x^3)dx }{\int_{x=0}^R(R^2 - x^2)dx}

= \frac{\big[R^2\frac{x^2}{2} - \frac{1}{4}x^4\big]_0^R}{\big[R^2x - \frac{1}{3}x^3\big]_0^R}

= \frac{3}{8}R

Therefore the centre of mass of a uniform solid hemisphere lies on its central axis at a distance from its base of \frac{3}{8} of its radius. It follows that the distance XG in the suspension problem is \frac{3r}{8} and we have the situation illustrated in the following diagram:

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Then

\tan\theta = \frac{r}{\frac{3r}{8}} = \frac{8}{3}

so the required angle is

\theta = \arctan\frac{8}{3} = 69.44^{\circ}. (Answer)