Question

Two sphere each of mass $1kg$ are attached to the two ends of a rod of mass $1kg$ and length $1m$. The moment of inertia of the system about an axis passing through its centre of gravity and perpendicular to the rod will be

• A. $1kg{m}^2$
• B. $\frac{1}{2}kg{m}^2$
• C. $\frac{7}{12}kg{m}^2$
• D. $\frac{12}{7}kg{m}^2$

Answer 1

The correct option is C $\frac{7}{12}kg{m}^2$

MI of each sphere about CG( center of gravity) is $m(\frac{l}{2}{)}^2=1\times (\frac{1}{2}{)}^2=\frac{1}{4}$

MI of two spheres is $2\times \frac{1}{4}=\frac{1}{2}$

MI of the rod Is: $m\frac{l^2}{12}=1\times \frac{1^2}{12}=\frac{1}{12}$

Total MI is: $\frac{1}{2}+\frac{1}{12}=\frac{7}{12}Kg{m}^2$

Alternatively,
The given system of two spheres forms a dipole and thus the reduced mass
of the dipole is given as $\mu =\frac{m_1{m}_2}{m_1+m_2}$
Thus its moment of inertia about the rotational axis is given as $\mu r^2=\frac{1}{2}$ and the moment of inertia of the rod is
given as $\frac{1}{12}m{l}^2=\frac{1}{12}$
Thus we get the total moment of inertia as
$\frac{1}{2}+\frac{1}{12}=\frac{7}{12}Kg{m}^2$