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Chapter 17: Problem 9

Determine the moment of inertia of the homogeneous triangular prism with respect to the \(y\) axis. Express the result in terms of the mass \(m\) of the prism. Hint: For integration, use thin plate elements parallel to the \(x-y\) plane and having a thickness \(d z\).

Short Answer

Expert verified
The moment of inertia of the homogeneous triangular prism with respect to the y-axis is \(\frac{m*h^2}{6}\), where \(m\) is the mass of the prism and \(h\) is its height.

Step by step solution

01

Visualize the problem

First, visualize that the triangular prism is composed of infinitesimally thin plate elements parallel to the x-y plane and having thickness \(d z\). These are at a distance \(z\) from the \(y\)-axis.
02

Calculate the moment of inertia of an infinitesimal element

Each of these elements will have a volume \(d V = S d z\), where \(S\) is the cross-sectional area of the prism at position \(z\). The mass of the element is \(d m = \rho d V\), where \(\rho\) is the mass density of the prism. Its moment of inertia with respect to the y-axis is \(d I_y = d m * z^2\). Substituting for \(d m\) gives \(d I_y = \rho * S * d z * z^2\).
03

Determine the cross-sectional area

Since the prism is triangular, the cross-sectional area decreases linearly from the base towards the apex. Thus, \(S = S_0 * (1 - z/h)\) where \(S_0\) is the base area and \(h\) is the height of the prism.
04

Integrate over the whole prism

To determine the moment of inertia of the whole prism, integrate over the entire height: \(I_y = \int_0^h \rho * S * z^2 * d z\). Substitute for S and simplify the integral.
05

Express the result in terms of the mass

Finally, express the result in terms of the mass. Note that the mass density \(\rho = m/V\), where \(V = S_0 * h / 2\) is the volume of the prism. Simplify the final formula.

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