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Chapter 17: Problem 6

The sphere is formed by revolving the shaded area around the \(x\) axis. Determine the moment of inertia \(I_{x}\) and express the result in terms of the total mass \(m\) of the sphere. The material has a constant density \(\rho\).

Short Answer

Expert verified
The moment of inertia \(I_{x}\) for a sphere rotating around its center with mass \(m\) and density \(\rho\) is therefore given by \(I_{x} = \frac{2}{5}m r^{2}\), where \(r = \left(\frac{3m}{4\pi \rho}\right)^{1/3}\).

Step by step solution

01

Moment of Inertia Formula

Write down the moment of inertia formula for a sphere rotating about an axis through its center and along a diameter. The formula is \(I_{x} = \frac{2}{5}mr^{2}\)
02

Express mass in terms of density and radius

As the material has constant density \(\rho\), express mass in terms of density and radius using the formula \(m = \rho \frac{4}{3} \pi r^{3}\)
03

Substitute mass into Moment of Inertia formula

Replace the mass \(m\) in moment of inertia formula of Step 1 with the formula from Step 2, yielding \(I_{x} = \frac{2}{5}(\rho \frac{4}{3}\pi r^{3})r^{2} = \frac{2}{5}\rho \pi r^{5}\)
04

Simplify the equation

Now we simplify the formula, \(I_{x} = \frac{8}{5}\rho \pi r^{5}\), where \(I_{x}\) is the moment of inertia, \(\rho\) is the density, \(r\) is the radius and \(\pi\) is a constant.
05

Calculate moment of Inertia

m is equal to the product of the volume of the sphere and the density, which is: \( m = \frac{4}{3}\pi r^{3} \rho\). Solving this equation for \(r^{3}\) gives \(r^{3} = \frac{3m}{4\pi \rho}\). Substituting this into the equation of Step 4, we obtain: \(I_{x} = \frac{8}{5}\rho \pi \left( \frac{3m}{4\pi \rho}\right)^{5/3}\). This simplifies to \(I_{x} = \frac{2}{5}m r^{2}\), where \(r = \left(\frac{3m}{4\pi \rho}\right)^{1/3}\).

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