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Chapter 10: Problem 26

The shaft connecting a power plant's turbine and electric generator is a solid cylinder of mass \(6.8 \mathrm{Mg}\) and diameter \(85 \mathrm{cm} .\) Find its rotational inertia.

Short Answer

Expert verified
The rotational inertia of the shaft is approximately \(612.025 \times 10^3 \, \mathrm{kg} \cdot \mathrm{m}^2\)

Step by step solution

01

Convert Measurements to Standard Units (if necessary)

The given measurements are the mass \(m = 6.8 \, \mathrm{Mg}\) which is equal to \(6.8 \times 10^6 \, \mathrm{kg}\), and diameter \(d = 85 \, \mathrm{cm}\) which is equivalent to \(0.85 \, \mathrm{m}\). It's important to use standard units in these calculations.
02

Find the Radius

The radius 'r' of the cylinder is half its diameter. So, \(r = d/2 = 0.85 \, \mathrm{m}/2 = 0.425 \, \mathrm{m}\).
03

Calculate the Rotational Inertia

Use the formula for the moment of inertia of a solid cylinder, \(I = 0.5 * m * r^2\). Substituting the given values, we find \(I = 0.5 * 6.8 \times 10^6 \, \mathrm{kg} * (0.425 \, \mathrm{m})^2 = 612.025 \times 10^3 \, \mathrm{kg} \cdot \mathrm{m}^2\)

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