Question

In Exercises 19 and 20, use cylindrical coordinates to verify the given formula for the moment of inertia of the solid of uniform density. Cylindrical shell: \(I_{z}=\frac{1}{2} m\left(a^{2}+b^{2}\right)\) \(0

Short answer

The use of cylindrical coordinates, bounding conditions, and evaluation of the integral confirm that the given formula for the moment of inertia is valid for a cylindrical shell of uniform density.

Step-by-step solution

Understand the problem

The problem is asking to verify the given formula for the moment of inertia of a solid of uniform density. The formula provided is \(I_{z}=\frac{1}{2} m\left(a^{2}+b^{2}\right)\), where \(m\) is the mass of the object, \(I_{z}\) is the moment of inertia around the z-axis, and \(a\) and \(b\) are the inner and outer radii of the cylindrical shell.

Apply the Bounded Cylinder Density Distribution

Assume the bounded cylinder has uniform density \(D\), then its volume \(V\) is \(\pi b^{2}h - \pi a^{2}h\) and mass \(m\) is \(D V = D(\pi b^{2}h - \pi a^{2}h)\).

Establish Two Integral Expressions

Establish two integral expressions for the moment of inertia \(I_{z}\), where \(I_{z} = \int_{a}^{b} \int_{0}^{h} \int_{0}^{2\pi} r^3 D d\theta dz dr\) and \(I_{z} = \frac{1}{2} m(a^{2}+b^{2})\).

Calculate the Integral

Evaluate the integral as follows: \(I_{z} = D \int_{a}^{b} \int_{0}^{h} \int_{0}^{2\pi} r^3 d\theta dz dr = 2\pi D \int_{a}^{b} [(\frac{1}{2} r^4)h]_{a}^{b} dr = D \pi h (\frac{1}{2} (b^4 - a^4)) = \frac{1}{2} D (\pi b^{2}h - \pi a^{2}h) (a^{2}+b^{2})\).

Compare With Given Formula

Since \(m = D V = D(\pi b^{2}h - \pi a^{2}h)\), \(I_{z} = \frac{1}{2} m(a^{2}+b^{2})\), therefore the calculation matches the given formula for the moment of inertia.