Rotational Motion Question 215
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 :
Options:
A) $\frac{2}{3}Ml^{2}$
B) $\frac{13}{3}Ml^{2}$
C) $\frac{1}{3}Ml^{2}$
D) $\frac{4}{3}Ml^{2}$
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Moment of inertia of a thin rod of length l about an axis passing through centre and perpendicular to the rod $=\frac{1}{12}Ml^{2}$
Thus moment of inertia of the frame.
$\frac{ml^{2}}{12}+\frac{ml^{2}}{4}=\frac{4ml^{2}}{12}=\frac{ml^{2}}{3}$
Total M.I. $=4\times \frac{ml^{2}}{3}$
