2004 AIPMT System of Particles and Rotational MotionMoment of Inertia of a System of Particles Medium Question 88 of 100 in System of Particles and Rotational Motion

Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side l…

Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side $l$ cm (as shown in the figure). 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 :-
A $2ml^2$
B $\dfrac{5}{4}ml^2$
C $\dfrac{3}{2}ml^2$
D $\dfrac{3}{4}ml^2$

The answer is B.

Solution

Moment of inertia of point masses: $I = \sum m r^2$, where r is the perpendicular distance of each mass from the axis AX.
Mass at A lies on the axis, so its distance $= 0$.
Mass at B: AB is perpendicular to AX, so its distance from AX $= l$.
Mass at C: its distance from AX is the horizontal distance of C from A $= l\cos 60^\circ = l\sin 30^\circ = \dfrac{l}{2}$.
$I_{AX} = m(0)^2 + ml^2 + m\left(\dfrac{l}{2}\right)^2$
$= ml^2 + \dfrac{ml^2}{4} = \dfrac{5}{4}ml^2$

Concept MI of point masses about an axis

System of Particles and Rotational Motion in past papers

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