Step 1

We are given the expression for the inclination of triangular prism is $z = \frac{{ - h}}{a}\left( {x - a} \right)$.

We are asked to determine the moment of inertia of the homogenous triangular prism with respect to the y-axis.

 
Step 2

Consider a small elemental segment of thickness $dz$ and width $x$. It is shown as:

We have the equation for the inclination of triangular prism is $z = \frac{{ - h}}{a}\left( {x - a} \right)$ or $ - \frac{{za}}{h} + a = x$.

 
Step 3

The expression to calculate the mass of the elemental plate segment is,

\[\begin{array}{c} dm = \rho dv\\ dm = \rho \left( {xb} \right)dz\\ dm = \rho b\left( { - \frac{{za}}{h} + a} \right)dz\\ dm = \frac{{ - \rho ab}}{h}\left( {z - h} \right)dz \end{array}\] ... (1)
 
Step 4

The expression to calculate the mass moment of inertia for the elemental plate segment about its centroid is,

\[d{I_C} = \frac{{\left( {dm} \right){x^2}}}{{12}}\] ... (2)
 
Step 5

The expression to calculate the mass moment of inertia for the elemental plate segment about the y-axis is,

\[d{I_y} = d{I_C} + \left( {dm} \right){d^2}\] ... (3)

Here, d is the distance of centroid of elemental plate from y-axis.

 
Step 6

The expression to calculate the distance of centroid of elemental plate from y-axis by triangular relation is,

\[\begin{array}{c} {d^2} = {z^2} + {\left( {\frac{x}{2}} \right)^2}\\ {d^2} = {z^2} + {\left[ {\frac{{\left( {\frac{{ - az}}{h} + a} \right)}}{2}} \right]^2}\\ {d^2} = {z^2} + {\left[ {\frac{{\left( { - az + ah} \right)}}{{2h}}} \right]^2}\\ {d^2} = {z^2} + \frac{{{a^2}}}{{4{h^2}}}{\left( {z - h} \right)^2} \end{array}\] ... (4)
 
Step 7

On combining equations (1), (2), (3), and (4) we get:

\[\begin{array}{l} d{I_y} = \left[ {\frac{{\left( {dm} \right){x^2}}}{{12}}} \right] + \left[ {{z^2} + \frac{{{a^2}}}{{4{h^2}}}{{\left( {z - h} \right)}^2}} \right]dm\\ d{I_y} = \left[ {\frac{1}{{12}}{{\left( {\frac{{ - za}}{h} + a} \right)}^2}\left( {dm} \right)} \right] + \left[ {{z^2} + \frac{{{a^2}}}{{4{h^2}}}{{\left( {z - h} \right)}^2}} \right]dm\\ d{I_y} = \left[ {{z^2} + \frac{{{a^2}}}{{3{h^2}}}{{\left( {z - h} \right)}^2}} \right]\left[ {\frac{{ - \rho ab}}{h}\left( {z - h} \right)} \right]dz\\ d{I_y} = \left[ {\frac{{ - \rho {a^3}b}}{{3{h^3}}}{{\left( {z - h} \right)}^3} + \frac{{\rho ab}}{h}{z^2}\left( {z - h} \right)} \right]dz \end{array}\]
 
Step 8

On integrating the above expression, we get:

\[\begin{array}{c} \int {d{I_y}} = \int\limits_0^h {\left[ {\frac{{ - \rho {a^3}b}}{{3{h^3}}}{{\left( {z - h} \right)}^3} + \frac{{\rho ab}}{h}{z^2}\left( {z - h} \right)} \right]} dz\\ {I_y} = ab\rho \int\limits_0^h {\left[ {\frac{{{a^2}}}{{3{h^3}}}{{\left( {h - z} \right)}^3} + {z^2} - \frac{{{z^3}}}{h}} \right]} dz\\ {I_y} = ab\rho \left[ {\frac{{{a^2}}}{{12{h^3}}}{{\left( {h - z} \right)}^4} + \frac{{{z^3}}}{3} - \frac{{{z^4}}}{{4h}}} \right]_0^h\\ {I_y} = ab\rho \left[ {\frac{{{a^2}}}{{12{h^3}}}{{\left( {h - h} \right)}^4} + \frac{{{h^3}}}{3} - \frac{{{h^4}}}{{4h}} - \frac{{{a^2}}}{{12{h^3}}}{{\left( {h - 0} \right)}^4}} \right] \end{array}\] \[{I_y} = \frac{{ab\rho }}{h}\left( {\frac{{{a^2}{h^2}}}{{12}} + \frac{{{h^4}}}{{12}}} \right)\]... (5)
 
Step 9

The expression to calculate the density of the triangular prism is,

\[\begin{array}{l} \rho = \frac{m}{V}\\ \rho = \frac{m}{{\left( {\frac{1}{2}ahb} \right)}} \end{array}\]
 
Step 10

Substitute the value of $\rho $ in equation (5), we get:

\[\begin{array}{c} {I_y} = \frac{{ab\left[ {\frac{m}{{\left( {\frac{1}{2}ahb} \right)}}} \right]}}{h}\left( {\frac{{{a^2}{h^2}}}{{12}} + \frac{{{h^4}}}{{12}}} \right)\\ {I_y} = \frac{{2m}}{{{h^2}}}\left( {\frac{{{a^2}{h^2}}}{{12}} + \frac{{{h^4}}}{{12}}} \right)\\ {I_y} = \frac{m}{6}\left( {{a^2} + {h^2}} \right) \end{array}\]