Question

A light rod one meter long has two point masses, 0.1 kg each, fixed at its ends. The moment of inertia of the system about a transverse axis through its center of mass is
1) $5kg{m^3}$
2) $0.25kg{m^3}$
3) $0.05kg{m^3}$
4) $0.025kg{m^3}$

Answer 1

Hint:We have all heard of inertia. Inertia is the resistance to any type of change in linear motion. Similarly, moment of inertia is the resistance to any change in angular motion. Here the moment of inertia resists angular acceleration which is defined as the summation of product of two or more masses of a particle or an object with its square of the distance from rotational axis.

Complete step by step answer:
Moment of inertia for a light rod about the transverse axis:
Here there are two masses involved so the moment of inertia is skewed.
$I = \dfrac{{{m_1}{m_2}{L^2}}}{{{m_1} + {m_2}}};$

Where:
${m_1}$= Mass of object 1;
${m_2}$= Mass of object 2;
L = Length;

Put the given values in the equation for moment of inertia $I = \dfrac{{{m_1}{m_2}{L^2}}}{{{m_1} + {m_2}}};$and solve:
$I = \dfrac{{0.1 \times 0.1 \times {1^2}}}{{0.2}};$

Calculate the above equation by multiplying values in the numerator at RHS:
$I = \dfrac{{0.01}}{{0.2}};$
The calculated value of I which is the moment of inertia is:
$I = 0.05kg{m^3};$

Final Answer:Option “3” is correct. The moment of inertia of the system about a transverse axis through its center of mass is $0.05kg{m^3}$.

Note:Moment of Inertia is different for different cases; it is generally specified w.r.t a specific rotational axis and also is dependent upon the mass distribution around the rotational axis. It can be varied depending upon the choice of axis.