91Ó°ÊÓ

The length of the rod =L
Mass of the rod = M
Linear mass density λ= cx²,

We have mass density λ= cx²,

Where c is constant and x is measured in meters.

so x=m

As λ is mass per unit length

kg/m=[c] m²

thus

[c]=kg/m³

The length of the rod =L
Mass of the rod = M
Linear mass density λ= cx²,

First lets draw the diagram as below

From the diagram we can get mass as

$M={∫}_{-L/2}^{L/2}\mathrm{λ}\left(x\right)dx$

Substitute the value of λ

$$\begin{gathered} M={∫}_{-\frac{L}{2}}^{L/2}c{x}^{2}dx \\ M=c{∫}_{-\frac{L}{2}}^{L/2}{x}^{2}dx \\ ⇒M={\left.c\frac{x^3}{3}\right|}_{-L/2}^{L/2} \end{gathered}$$

Solve the integration with given limits

$$\begin{gathered} ⇒M=c\left(\frac{(L/2{)}^3-(-L/2{)}^3}{3}\right) \\ ⇒M=\frac{c}{3}\left[\frac{L^3}{8}+\frac{L^3}{8}\right] \\ ⇒M=\frac{c}{3}\left[\frac{2L^3}{8}\right] \\ ⇒c=\frac{12M}{L^3} \end{gathered}$$

The length of the rod =L
Mass of the rod = M
Linear mass density λ= cx²

Draw the figure as below to understand and solve the problem

Lets consider dx a small particle with mass dm
$dm=\mathrm{λ}\left(x\right)dx$

And the moment of inertia for dx is

$dI=x^{2}dm=x^2\mathrm{λ}\left(x\right)dx\left(IX\right)$

We can get inertia upon integration

$$\begin{gathered} I_{cm}={∫}_{-\frac{L}{2}}^{L/2}(dI{)}^2 \\ ={∫}_{-\frac{L}{2}}^{L/2}{x}^2·\mathrm{λ}(x)dxâˆ\pounds \\ ⇒I_{cm}={∫}_{-\frac{L}{2}}^{L/2}{x}^{2}c(x{)}^{2}dx \\ ⇒I_{cm}=c{∫}_{-\frac{L}{2}}^{L/2}{x}^{4}dx \\ ⇒I_{cm}={\left.c\frac{x^5}{5}\right|}_{-L/2}^{L/2} \end{gathered}$$

Now substitute the values

$$\begin{gathered} ⇒I_{cm}=\frac{12M}{L^3}\left(\frac{(L/2{)}^5-(-L/2{)}^5}{5}\right) \\ ⇒I_{cm}=\frac{3}{20}M{L}^2 \end{gathered}$$