Question

A thin uniform rod of mass $10\text{kg}$ is rotating about an axis passing through the end point of the rod of length $6\text{m}$ as shown in figure. Find the moment of inertia of the rod about this axis.

• A. $24{\text{kg-m}}^2$
• B. $43.2{\text{kg-m}}^2$
• C. $12{\text{kg-m}}^2$
• D. $38{\text{kg-m}}^2$

Answer 1

The correct option is B $43.2{\text{kg-m}}^2$


Given, mass of the rod $\left(m\right)=10\text{kg}$
Length of the rod $\left(l\right)=6\text{m}$
Let us consider a mass element ${}^{\prime }d{m}^{\prime }$ of length ${}^{\prime }d{s}^{\prime }$. Let $s$ be the distance of ${}^{\prime }d{m}^{\prime }$ from ${}^{\prime }{\text{O}}^{\prime }$.
Moment of inertia about $Z-$ axis
$I_z=\int r^{2}dm$
$={\int }_0^6(s\sin {37}^{\circ }{)}^2\lambda ds$ (where $\lambda$ is linear mass density of the rod)

$\lambda =\frac{m}{l}$. So, we get
$I_z=\frac{m}{l}{\sin }^2{37}^{\circ }{\int }_0^6{s}^{2}ds$
$=\frac{10}{6}\times {\left(\frac{3}{5}\right)}^2\times {\left(\frac{s^3}{3}\right)}_0^6$
$\Rightarrow I_z=43.2{\text{kg-m}}^2$