Chapter B, Problem B.1P

A thin plate with a mass m is cut in the shape of an equilateral triangle of side a. Determine the mass moment of inertia of the plate with respect to (a) the centroidal axes AA' and BB’, (b) the centroidal axis CC' that is perpendicular to the plate.

To determine

(a)

The mass moment of inertia of the plate about $AA’\text{ and }BB\text{'}$ axis.

Answer to Problem B.1P

We got the MI as $I_{AA\text{'}}=I_{BB\text{'}}=\frac{m{a}^2}{24}$

Explanation of Solution

Given information:

Mass $m$

Sides $a$

Concept used:

Following formula is used-

1. $dm=\rho tdA$
2. MI of stripabout its base $d{I}_{AA\text{'}}=x^{2}dm$

Calculation:

Diagram,

(a)               (b)

MI is given as below:

$\begin{array}{l}d{I}_{AA\text{'}}=2x^{2}dm\\ I_{AA\text{'}}=2{\int }_0^{a/2}{x}^{2}dm=2{\int }_0^{a/2}{x}^{2}ptdA\\ I_{AA\text{'}}=2pt{\int }_0^{a/2}{x}^{2}ydx=2pt{\int }_0^{a/2}{x}^2\sqrt{3}(0.5a-x)dx\\ I_{AA\text{'}}=2\sqrt{3}pt{\int }_0^{a/2}(0.5a{x}^2-x^3)dx\\ I_{AA\text{'}}=2\sqrt{3}pt{\left[\frac{0.5a{x}^3}{3}-\frac{x^4}{4}\right]}_0^{a/2}=2\sqrt{3}pt\left[\frac{0.5a{(a/2)}^3}{3}-\frac{{(a/2)}^4}{4}\right]\\ I_{AA\text{'}}=2\sqrt{3}pt\left[\frac{a^4}{192}\right]\\ I_{AA\text{'}}=\frac{m{a}^2}{24}\text{ (}\because \text{m=ptA=pt}\frac{1}{2}a\times a\sqrt{3}/2=\text{pt}\frac{\sqrt{3}}{4}{a}^2)\end{array}$

The MI about BB’ will be same due to symmetry.

Conclusion:

We got the MI as $I_{AA\text{'}}=I_{BB\text{'}}=\frac{m{a}^2}{24}$

To determine

(b)

Find the mass moment of inertia of the plate about $CC\text{'}$ axis.

Answer to Problem B.1P

We got the MI as $I_{CC\text{'}}=\frac{m{a}^2}{12}$

Explanation of Solution

Given information:

Mass $m$

Sides $a$

Concept used:

Following formula is used-

1. Parallel axis theorem $I_{CC\text{'}}=I_{AA\text{'}}+I_{BB\text{'}}$

Calculation:

Diagram,

(a)                 (b)

MI is given as below:

$\begin{array}{l}d{I}_{AA\text{'}}=2x^{2}dm\\ I_{AA\text{'}}=2{\int }_0^{a/2}{x}^{2}dm=2{\int }_0^{a/2}{x}^{2}ptdA\\ I_{AA\text{'}}=2pt{\int }_0^{a/2}{x}^{2}ydx=2pt{\int }_0^{a/2}{x}^2\sqrt{3}(0.5a-x)dx\\ I_{AA\text{'}}=2\sqrt{3}pt{\int }_0^{a/2}(0.5a{x}^2-x^3)dx\\ I_{AA\text{'}}=2\sqrt{3}pt{\left[\frac{0.5a{x}^3}{3}-\frac{x^4}{4}\right]}_0^{a/2}=2\sqrt{3}pt\left[\frac{0.5a{(a/2)}^3}{3}-\frac{{(a/2)}^4}{4}\right]\\ I_{AA\text{'}}=2\sqrt{3}pt\left[\frac{a^4}{192}\right]\\ I_{AA\text{'}}=\frac{m{a}^2}{24}\text{ (}\because \text{m=ptA=pt}\frac{1}{2}a\times a\sqrt{3}/2=\text{pt}\frac{\sqrt{3}}{4}{a}^2)\end{array}$

The MI about BB’ will be same due to symmetry

MI about CC’

$\begin{array}{l}{I}_{CC\text{'}}=I_{AA\text{'}}+I_{BB\text{'}}\\ I_{CC\text{'}}=\frac{m{a}^2}{24}\text{+}\frac{m{a}^2}{24}\\ I_{CC\text{'}}=\frac{m{a}^2}{12}\end{array}$

Conclusion:

We got the MI as $I_{CC\text{'}}=\frac{m{a}^2}{12}$