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 cm2 units will be:

[AIPMT (S) 2004]

Options:

A) (3/4)ml2

B) 2ml2

C) (5/4)ml2

D) (3/2)ml2

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Answer:

Correct Answer: C

Solution:

  • Moment of inertia of the system about AX is given by

MI=mArA2+mBrB2+mCrC2

MI=m(0)2+m(l)2+m(lsin300)2

=ml2+ml24=54ml2

Alternative: Moment of inertia of a system about a line OC perpendicular to AB, in the plane of ABC is

ICO=m×0+m×(12)2+m×(12)2

ICO=ml24+ml24=ml22

According to parallel-axis theorem

IAX=ICO+Mx2

where x= distance of AX from CO, M= total mass of system

IAX=ml22+3m×(l2)2

IAX=ml22+3ml24=54ml2



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