Question

The linear mass density of a thin rod AB of length L varies from A to B as $\lambda \left(x\right)={\lambda }_0[1+(x/L\left)\right]$, Where $x$ is the distance from $A$. If $M$ is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is:

• A. $\left(\frac{2}{5}\right){\mathrm{ML}}^2$
• B. $\left(\frac{5}{12}\right){\mathrm{ML}}^2$
• C. $\left(\frac{7}{18}\right)M{L}^2$
• D. $\left(\frac{3}{7}\right)M{L}^2$

Answer 1

The correct option is C

$\left(\frac{7}{18}\right)M{L}^2$

Step 1: Given Data:
Linear mass density $\lambda \left(x\right)={\lambda }_0[1+(x/L\left)\right]$

Where $x$ is the distance from $A$

The mass of the rod $M$

Step 2: Formula used:

Linear mass density can be given as-

$\lambda (massdensity)=\frac{mass}{length}$

Mass moment of inertia $I=M{L}^2$

Step 3: Calculation of mass and moment of inertia

Let us assume a segment of length $dx$ at a distance of $x$ from one end and its mass can be calculated as-

$\begin{array}{rcl}dm& =& {\lambda }_0\left[1+\left(\frac{x}{L}\right)\right]dx\\ {\int }_0^{M}dm& =& {\int }_0^L{\lambda }_0\left[1+\left(\frac{x}{L}\right)dx\right]\\ M& =& 3{\lambda }_0\frac{L}{2}\dots \dots \dots \dots \left(1\right)\end{array}$

Mass moment of inertia is calculate as-

$\begin{array}{rcl}dI& =& dm{x}^2\\ \int dI& =& dm{x}^2\\ I& =& {\int }_0^L{\lambda }_0\left[1+\left(\frac{x}{L}\right)\right]dx{x}^2\\ I& =& \frac{7{\lambda }_0{L}^3}{12}\end{array}$

Now from $\left(1\right)$,

$$\begin{gathered} {\lambda }_0=\frac{2M}{3L} \\ I=\frac{7}{12}\left(\frac{2M}{3L}\right){\left(L\right)}^3 \\ I=\frac{7M{L}^2}{18} \end{gathered}$$

Hence, option C is the correct answer.