Question

In the diagram shown below all three rods are of equal length L and equal mass M. The system is rotated such that rod B is the axis. What is the moment of inertia of the system?

• A. $\frac{M{L}^2}{6}$
• B. $\frac{4}{3}M{L}^2$
• C. $\frac{M{L}^2}{3}$
• D. $\frac{2}{3}M{L}^2$

Answer 1

The correct option is A $\frac{M{L}^2}{6}$

Moment of inertia of system
$=M.IofA+M.I.ofB+M.IofC$
$M.IofA=M.T$ through centre and perpendicular to length $=\frac{1}{12}M{L}^2$
M.I. of $C=M.I.$ of $A=\frac{1}{12}M{L}^2$
$M.I$ of $B=0$
(moment of mass about an axis passing through its own position is zero)
$\therefore TotalM.I.=\frac{1}{12}M{L}^2+\frac{1}{12}M{L}^2=\frac{1}{16}M{L}^2$.