Question

A uniform rod of mass $2\text{kg}$ and length
$3\text{m}$ is bent in the form of an equilateral triangle. The moment of inertia of the triangle about a vertical axis perpendicular to the plane of the triangle and passing through the center is

• A. $0.3{\text{kg m}}^2$
• B. $0.1{\text{kg m}}^2$
• C. $0.5{\text{kg m}}^2$
• D. $1.0{\text{kg m}}^2$

Answer 1

The correct option is A $0.3{\text{kg m}}^2$

Given:
$m=2\text{kg};L=3\text{m}$

Now, the rod is bent in the form of an equilateral triangle. So, mass and length of each side becomes,

$M_0=\frac{M}{3}=\frac{2}{3}\text{kg}l=\frac{L}{3}=1\text{m}$


From $△EOC$

$\tan {30}^{\circ }=\frac{d}{\left(\frac{l}{2}\right)}$

$\Rightarrow d=\left(\frac{l}{2}\right)\tan {30}^{\circ }=\frac{1}{2\sqrt{3}}$

Now, moment of inertia of the rod $\text{EG}$ about an axis passing through $\text{O}$ and perpendicular to the plane is,

$(I_O{)}_{EG}=(I_O{)}_{com}+M_0{d}^2\left(\text{from parallel axis theorem}\right)$

$(I_O{)}_{EG}=\frac{1}{12}{M}_0{l}^2+M_0{d}^2$

$(I_O{)}_{EG}=M_0(\frac{1}{12}+\frac{1}{12})=\frac{M_0}{6}$

Due to symmetry, other two sides will also have the same moment of inertia about $\text{O}.$

Now, $(I_O{)}_{EG}=(I_O{)}_{EF}=(I_O{)}_{FG}=I^{{}^{\prime }}$

Now, moment of inertia of the system about $\text{O}$ is,

$I=(I_O{)}_{EG}+(I_O{)}_{EF}+(I_O{)}_{FG}=3I^{{}^{\prime }}$

$\Rightarrow I=3I^{{}^{\prime }}=3\times \frac{M_0}{6}=\frac{1}{3}[\because M_0=\frac{2}{3}]$

$\therefore I=0.33\approx 0.3{\text{kg m}}^2$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{A}\right)$ is the correct answer.