Question

Two bodies of masses $1kg$ and $2kg$ are attached to the ends of a $2m$ long weightless rod. This system is rotating about an axis passing through the middle point of rod and perpendicular to length. Calculate the M.I. of system.

Answer 1

Calculate the moment of inertia of the system by assuming that the system is a combination of two particles and a rod attaching them.
We can use the definition of the moment of inertia of a system which states that the moment of inertia of a system is equal to the sum of the product of mass with the square of corresponding distance from the line of action.

Complete step by step answer:
According to the question, we have two bodies of masses $1kg$ and $2kg$. These masses are attached to the ends of a weightless rod of length $2m$. This system is rotating about an axis as shown in the figure.

Now, we know that the moment of inertia of this type of system having $n$ particles in the system will be given as-
$M.I. = \sum\limits_n {{m_n}r_n^2}$
Or we can write above equation as-
$M.I. = {m_1}r_1^2 + {m_2}r_2^2 + ....... + {m_n}r_n^2$
Now, in this system, we have two particles in the system. So, the above equation can be deduced in the following equation for a two particle system-
$M.I. = {m_1}r_1^2 + {m_2}r_2^2$ (i)
As the system (rod with two particles) is rotating about an axis which is passing through the midpoint of the rod. So,
${r_1} = {r_2} = 1m$
And from the question, we have ${m_1} = 1kg,{m_2} = 2kg$
So, putting the values of ${m_1},{m_2},{r_1}$ and ${r_2}$ in equation (i), we get-
$M.I. = 1 \times {(1)^2} + 2 \times {(1)^2} \\\ \Rightarrow M.I. = 1 + 2 \\\ \Rightarrow M.I. = 3kg{m^2} \\\$

Hence, the moment of inertia of this system (rod with two particles) is $3kg{m^2}$.

Note: Remember that the system is rotating about an axis which passes through the centre of the rod. So, the corresponding distance of the particles from a ${r_1} = {r_2} = 1m$. We have to remember the formula of moment of inertia $M.I. = \sum\limits_n {{m_n}r_n^2}$ to solve this question.