Question

Three particles, each of mass $200{\text{ g}}$ , are kept at the corner of an equilateral triangle of side ${\text{10 cm}}$.Find the moment of inertia of the system about an axis
(a) joining two of the particles and
(b) passing through one of the particles and perpendicular to the plane of the particles

Answer 1

To solve this question we will use concepts of moment of inertia and perpendicular axis theorem. There are three particles each of mass situated at corners of the equilateral triangle of the side. For the first part of the question we take an axis passing through two particles so the moment of inertia across these particles will be zero. The resultant moment of inertia will be the moment of inertia of the remaining particle. For the second part we also use a similar method.

Complete step by step answer:
Let us consider the particles to be A , B ,and C and mass was given as mass $= 200{\text{ g}} = 0.2kg$ $(1kg = 1000g)$.
Side of the equilateral triangle was given ${\text{ = 10 cm}} = 0.1\,m$.

(a) Let us take the axis passing through B and C. Moment of inertia I is the rotational analogue of mass. In rotation of a body about a fixed axis, the moment of inertia plays a similar role as mass does in linear motion. It is a measure of inertia of the body for rotational motion (rotational inertia).

The moment of inertia of a body about a given axis of rotation is defined as the sum of the product of the masses of the particles in the body and square of the distance from the axis of rotation.
$I = M{R^2}$
The perpendicular distances of the center of masses of particles B and C are zero as the axis is passing through them itself.The perpendicular distance of particle A from axis is given by
${\text{AR}} = \dfrac{{\sqrt 3 }}{2} \times 0.1$
$\Rightarrow {\text{AR}} = 0.05\sqrt 3 \,m$
This is the perpendicular distance of the center of mass of A from the axis.Now resultant moment of inertia will be
$I = M{R^2}$
$\Rightarrow I = \left( {0.2} \right){\left( {0.05\sqrt 3 } \right)^2}$
$\therefore I = 1.5 \times {10^{ - 3}}\,kg{m^2}$

Therefore,the moment of inertia of the system about an axis joining two of the particles
is $1.5 \times {10^{ - 3}}\,kg{m^2}$.

(b) Let the axis be passing through A ,now as the perpendicular distance of center of mass from A to axis will be 0 .Therefore the resultant moment of inertia will be the sum of moment inertias of particles B and C .
$I = M{R^2} + M{R^2}$
where R is the distance of B and C from A (as the axis is passing through .
As the particles are on an equilateral triangle they are equidistant from each other.
$R = side = 0.1m$
$\Rightarrow I = 2M{R^2}$

\therefore I= 4 \times {10^{ - 3}}\,kg{m^2}$$ **Therefore,the moment of inertia of the system about an axis passing through one of the particles and perpendicular to the plane of the particles is $4 \times {10^{ - 3}}\,kg{m^2}$.** **Note:** Moment of inertia is nothing but the property by which a body opposes changes while rotating about an axis. The factors on which moment of inertia depends are the mass of the body, the shape and size and also mass distribution manner of the given body and the axis of rotation position.