Question

Three particles of 2kg $\left( {\vec i,\vec j} \right)$, 3kg $\left( { - 2\vec i,\vec j} \right)$ and 1kg $\left( {2\vec i,\vec j} \right)$ lie in the XOY plane of a reference frame in which distances are measured in meters. The M.I in ${\text{kg - }}{{\text{m}}^2}$ of the system about OZ axis is:
A. 45
B. 17
C. 33
D. 24

Answer 1

Calculate the distance of each particle from origin using distance formula. The moment of inertia of the system of particles is the sum of the product of mass and square of distance from the origin for each particle. The distance from the origin should have the positive sign.

Formula used:
Moment of inertia, $M.I = \sum\limits_{i = 1}^n {{m_i}r_i^2}$
Here, m is the mass and r is the distance from the origin.

Complete step by step answer:
We have given the positions of the three particles with respect to the origin O. We have to calculate the distance of these particles from origin using distance formula.Let us determine the distance of the particle from origin of mass ${m_1} = 2\,{\text{kg}}$ using the distance formula as follows,
${r_1} = \sqrt {{1^2} + {1^2}}$
$\Rightarrow {r_1} = \sqrt 2$
Similarly, we can calculate the distance of the particles of the mass ${m_2} = 3\,{\text{kg}}$ and ${m_3} = 1\,{\text{kg}}$ as follows,
${r_2} = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 1} \right)}^2}}$
$\Rightarrow {r_2} = \sqrt 5$

And,
${r_3} = \sqrt {{{\left( 2 \right)}^2} + {{\left( 1 \right)}^2}}$
$\Rightarrow {r_3} = \sqrt 5$
We have the expression for the Moment of Inertia of the system of n number of particles,
$M.I = \sum\limits_{i = 1}^n {{m_i}r_i^2}$
Here, m is the mass and r is the distance from the origin.
For the system of three particles, we can rewrite the moment of inertia as,
$M.I = {m_1}r_1^2 + {m_2}r_2^2 + {m_2}r_2^2$
Substituting ${m_1} = 2\,{\text{kg}}$, ${m_2} = 3\,{\text{kg}}$, ${m_3} = 1\,{\text{kg}}$, ${r_1} = \sqrt 2$, ${r_2} = \sqrt 5$ and ${r_3} = \sqrt 5$ in the above equation, we get,
$M.I = \left( 2 \right){\left( {\sqrt 2 } \right)^2} + \left( 3 \right){\left( {\sqrt 5 } \right)^2} + \left( 1 \right){\left( {\sqrt 5 } \right)^2}$
$\Rightarrow M.I = 4 + 15 + 5$
$\therefore M.I = 24\,{\text{kg - }}{{\text{m}}^2}$
Therefore, the moment of inertia (M.I) of the system of given three particles is $24\,{\text{kg - }}{{\text{m}}^2}$.

So, the correct answer is option D.

Note: Even if the point lies in the second, third or fourth quadrant, the distance is scalar quantity and therefore, it bears a positive sign. The given equation for the moment of inertia is valid only when the moment of inertia is to be calculated about the vertical axis perpendicular to the plane of the particles. One should never calculate the moment of inertia by calculating the centre of mass of the particles.