First slide

Moment of inertia

Question

A thin rod of length 4l and mass 4m is bent at the points as shown in figure. What is the moment of inertia of the rod about the axis passing through point O and perpendicular to the plane of the paper?

Moderate

A

$\frac{{Ml}^{2}}{3}$
B

$\frac{10 {Ml}^{2}}{3}$
C

$\frac{{Ml}^{2}}{12}$
D

$\frac{{MI}^{2}}{24}$

Solution

Total $MI = I_{1} + I_{2} + I_{3} + I_{4}$
$= 2 I_{1} + 2 I_{2}$
$= 2 (I_{1} + I_{2}) [I_{i} = I_{3}; I_{1} = I_{4}]$
Now,
$I_{2} = I_{3} = \frac{{Ml}^{2}}{3} - - - - (1)$
Using parallel axis theorem, we have
$I = I_{CM} + {Mx}^{2} and x = \sqrt{l^{2} + {(\frac{l}{2})}^{2}}$
$I_{1} = I_{4} = \frac{{Ml}^{2}}{12} + M {[\sqrt{l^{2} + {(\frac{l}{2})}^{2}}]}^{2} = \frac{4 {Ml}^{2}}{3} - - - - (2)$
Putting all values of (1) & (2) in $MI = I_{1} + I_{2} + I_{3} + I_{4}$ , we get
$I = \frac{10 {Ml}^{2}}{3}$

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