Chapter 17.1, Problem 1P

The rod's density ρ end cross-sectional area. A are constant. Express the result in terms of the rod’s total mass m.

To determine

The moment of inertia $I_y$ for the slender rod in terms of the rod’s total mass $m$.

Answer to Problem 1P

The moment of inertia $I_y$ for the slender rod in terms of the rod’s total mass $m$ is $\frac{m{l}^2}{3}$.

Explanation of Solution

Given:

The density of the rod is $\rho$.

The cross-sectional area of the rod is $A$.

Show the intersection of slender rod at the arbitrary point $\left(x,y,z\right)$ as in Figure (1).

Conclusion:

From the Figure 1,

Calculate the differential mass of the slender rod.

$dm=\rho dV$ (I)

Here, density of material is $\rho$, and the volume of the slender rod is $V$.

Substitute $Adx$ for $dV$ in Equation (I).

$dm=\rho Adx$

Calculate the entire mass of the slender rod.

$m={\displaystyle \underset{M}{\int }dm}$ (II)

Substitute $\rho Adx$ for $dm$ in Equation (II).

$\begin{array}{c}m={\displaystyle \underset{M}{\int }\rho Adx}\\ ={\displaystyle {\int }_0^l\rho Adx}\\ =\rho A{\displaystyle {\int }_0^{l}dx}\end{array}$

$\begin{array}{c}=\rho A{\left(x\right)|}_0^l\\ =\rho Al\end{array}$

Express the moment of inertia of the slender rod about the $y$ axis.

$I_y={\displaystyle {\int }_M{x}^{2}dm}$ (III)

Substitute $\rho Adx$ for $dm$ in Equation (III).

$\begin{array}{c}{I}_y={\displaystyle {\int }_M{x}^2\left(\rho Adx\right)}\\ ={\displaystyle {\int }_0^l{x}^2\left(\rho Adx\right)}\\ =\rho A{\displaystyle {\int }_0^l{x}^{2}dx}\\ =\rho A{\left[\frac{x^3}{3}\right]|}_0^l\end{array}$

$\begin{array}{c}=\rho A\frac{l^3}{3}\\ =\frac{1}{3}\rho A{l}^3\\ =\frac{1}{3}{l}^2\left(\rho Al\right)\end{array}$ (IV)

Substitute $m$ for $\rho Al$ in Equation (IV).

$\begin{array}{c}{I}_y=\frac{1}{3}{l}^2\left(m\right)\\ =\frac{m{l}^2}{3}\end{array}$

Hence, the moment of inertia $I_y$ for the slender rod in terms of the rod’s total mass $m$ is $\frac{m{l}^2}{3}$.