First slide

Moment of intertia

Question

Four identical thin rods each of mass M and length l form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is

Moderate

A

$\frac{4}{3} {Ml}^{2}$
B

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

$\frac{13}{3} {Ml}^{2}$
D

$\frac{1}{3} {Ml}^{2}$

Solution

The moment of inertia of each rod about an axis through the centre of frame but perpenducular to frame =
[its moment of inertia through its centre of mass and perpendicular to rod + (mass of rod) $\times$ (perpendicular distance between two axes)]
= $\frac{{Ml}^{2}}{12} + M {(\frac{l}{2})}^{2} = \frac{{Ml}^{2}}{3}$
Moment of inertia of the system = $\frac{{Ml}^{2}}{3} \times 4$
= $\frac{4}{3} {Ml}^{2}$

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