Rotational Motion Question 57
Question: Three particles, each of mass m grams situated at the vertices of an equilateral triangle ABC of side 1 cm . The moment of inertia of the system about a line AX perpendicular to AB and in the plane of ABC, in gram $ -cm^{2} $ units will be:
[AIPMT (S) 2004]
Options:
A) $ (3/4)ml^{2} $
B) $ 2ml^{2} $
C) $ (5/4)ml^{2} $
D) $ (3/2)ml^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
- Moment of inertia of the system about AX is given by
$ MI=m _{A}r_A^{2}+m _{B}r_B^{2}+m _{C}r_C^{2} $
$ MI=m{{(0)}^{2}}+m{{(l)}^{2}}+m{{(l\sin 30^{0})}^{2}} $
$ =ml^{2}+\frac{ml^{2}}{4}=\frac{5}{4}ml^{2} $
Alternative: Moment of inertia of a system about a line OC perpendicular to AB, in the plane of ABC is
$ I _{CO}=m\times 0+m\times {{( \frac{1}{2} )}^{2}}+m\times {{( \frac{1}{2} )}^{2}} $
$ \therefore $ $ I _{CO}=\frac{ml^{2}}{4}+\frac{ml^{2}}{4}=\frac{ml^{2}}{2} $
According to parallel-axis theorem
$ I _{AX}=I _{CO}+Mx^{2} $
where $ x= $ distance of $ AX $ from CO, $ M= $ total mass of system
$ I _{AX}=\frac{ml^{2}}{2}+3m\times {{( \frac{l}{2} )}^{2}} $
$ I _{AX}=\frac{ml^{2}}{2}+\frac{3ml^{2}}{4}=\frac{5}{4}ml^{2} $
