i) Find the position of the centre of mass of a uniform hemispherical bowl of thickness 1 cm and inside radius 9 cm.
ii) A light handle AB of length 18 cm is attached to a point A on the rim of the bowl in such a way that the line AB passes through the centre of the hemisphere. What angle does AB make with the vertical when the bowl is hung from the end B?

Question

Accepted Answer

i) Think of the bowl as a solid hemisphere of radius 10 cm from which another solid hemisphere of radius 9 cm has been removed. Then the original 10 cm hemisphere can be treated as a composite body, consisting of the bowl and the 9 cm hemisphere.
You can put all the information in a table.

	10 cm hemisphere	= bowl	+ 9 cm hemisphere

Mass	[latex]\frac{2}{3}πρ × 10^{3}[/latex]	[latex]\frac{2}{3}πρ(10^{3} - 9^{3})[/latex]	[latex]\frac{2}{3}πρ × 9^{3}[/latex]
Distance from O to c.o.m.	[latex]\frac{3}{8} × 10[/latex]	[latex]\bar{x}[/latex]	[latex]\frac{3}{8} × 9[/latex]

Taking moments gives

[latex]\left(\frac{2}{3}πρ10^{3}\right) × \left(\frac{3}{8} × 10\right) = \left(\frac{2}{3}πρ(10^{3} - 9^{3}) × \bar{x}\right) + \left(\frac{2}{3}πρ9^{3}\right) × \left(\frac{3}{8} × 9\right).[/latex]

This can be simplified by dividing through by [latex]\frac{2}{3}ρπ[/latex] to give

[latex]\frac{3}{8} × 10^{4} = (10^{3} - 9^{3})\bar{x} + \frac{3}{8} × 9^{4}[/latex]

[latex]\bar{x} = \frac{3(10^{4} - 9^{4})}{8(10^{3} - 9^{3})}[/latex]

= 4.76 (to s.f.).

ii) The handle is light so the position of the centre of mass, G, is unchanged when it is attached. For equilibrium G must be vertically below B and the angle required is the angle ABG = α.

From the diagram [latex]tanα = \frac{OG}{OB} = \frac{\bar{x}}{28}[/latex]

⇒ α = 9.65°.

The angle between AB and the vertical is about 10°.
Note

An alternative, but equivalent, method for finding [latex]\bar{x}[/latex] is to regard the bowl as the sum of a solid hemisphere of radius 10 cm and a solid hemisphere of radius 9 cm which has negative mass.
In this case the moments equation would be written

[latex]\frac{2}{3}πρ(10^{3} - 9^{3})\bar{x} = \left(\frac{2}{3}πρ10^{3} \right) × \left(\frac{3}{8} × 10 \right) + \left(- \frac{2}{3}πρ9^{3} \right) × \left(\frac{3}{8} × 9\right). [/latex]

Question 8.13: i) Find the position of the centre of mass of a uniform hemi...

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