Question

Three particles of mass $1{\rm{ kg}}$ , ${\rm{2 kg}}$ and ${\rm{3 kg}}$ are placed at $1{\rm{ cm}}$ , ${\rm{2 cm}}$ and ${\rm{3 cm}}$ from the axis of rotation respectively then moment of inertia of the system and radius of gyration of the system respectively are ______ and________ ${\rm{cm}}$.

Answer 1

Hint:In the above question, we have to find the moment of inertia of the system first and then the radius of gyration can be calculated. For calculating the radius of gyration first, we need to calculate the sum of masses of particles. Moment of inertia is resistance offered by a body to a change in its rotational motion. Moment of inertia always resists the motion. Moment of Inertia is constant for a particular rigid frame and a specific axis of rotation.

Complete step by step answer:
If a body is in rest or in the motion, it wants to be in that position. The moment of inertia of a system of particles is the sum of product of mass of particles and square of the distance from the axis of rotation; is given by;
\[I = \sum {{m_i} \cdot r_i^2} \]
Where, ${r_i}$ is the distance of particles from the axis of rotation and ${m_i}$ is the mass of the particles
In above question there are three particles and the mass of first particle is $1\,{\rm{kg}}$, mass of second particle is ${\rm{2 kg}}$ and mass of third particle is ${\rm{3 kg}}$, And the distance from the axis of rotation are $1{\rm{ cm}}$ , ${\rm{2 cm}}$ and ${\rm{3 cm}}$ respectively
So, Moment of inertia is,
$
I = \left( {1 \times {1^2}} \right) + \left( {2 \times {2^2}} \right) + \left( {3 \times {3^2}} \right)\\
= 36\,{\rm{gm}} \cdot {\rm{c}}{{\rm{m}}^2}
$…… (i)
And radius of gyration is the square root of ratio of moment of inertia to the sum of masses of particles, is given by;
$K = \sqrt {\dfrac{I}{m}} $ …… (ii)
Where I moment of inertia and m is the sum of masses of the particles.
Moment of inertia is $ = 36\,{\rm{gm}} \cdot {\rm{c}}{{\rm{m}}^2}$
And, sum of masses of particles is
$
m = {m_1} + {m_2} + {m_3}\\
= 1 + 2 + 3\\
= 6\,{\rm{gm}}
$
Substituting these values in equation (ii)
$
K = \sqrt {\dfrac{{36}}{6}} \\
= \sqrt 6 \\
= 2.449{\rm{ cm}}
$

Note: Moment of inertia can be calculated by integrating over every piece of mass, for integral form MOI is, \[I = \int\limits_{i = 1}^{i = n} {r_i^2} dm\] , where \[{r_i}\] is the distance of mass from axis of rotation.
Moment of inertia for a compound object is simply the sum of the moment of inertia for each individual object that makes up a compound object.