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- Hint: Perpendicular axis theorem for the moment of inertia states that the moment of inertia of a planer object about an axis which is perpendicular to the plane is the sum of the inertia of the two other perpendicular axes through the same point in the plane of the object.
Complete step-by-step solution -
Given, the moment of inertia about one of its diameters ${\text{ = I}}$,
This implies that moment of inertia about x-axis and y-axis are equal to each other and both are equal to the moment of inertia to the of the circular disc about its diameter i.e. ${\text{I}}$.
So, we can write this as,
${{\text{I}}_{\text{x}}}{\text{ = }}{{\text{I}}_{\text{y}}}{\text{ = I}}$-----equation (1)
The axis perpendicular to the circular surface and passing through its centre is the axis which is perpendicular to the x-axis and the y-axis, hence it is z-axis.
And from the perpendicular theorem, we know that
${{\text{I}}_{\text{z}}}{\text{ = }}{{\text{I}}_{\text{x}}}{\text{ + }}{{\text{I}}_{\text{y}}}$
Let the moment of inertia about an axis perpendicular to the circular surface and passing through its centre=${{\text{I}}_{\text{c}}}$
This ${{\text{I}}_{\text{c}}}$is same as ${{\text{I}}_{\text{z}}}$, so we can write like this,
${{\text{I}}_{\text{c}}}{\text{ = }}{{\text{I}}_{\text{x}}}{\text{ + }}{{\text{I}}_{\text{y}}}$-----equation (2)
And from equation (1) we can write the values of ${{\text{I}}_{\text{x}}}{\text{& }}{{\text{I}}_{\text{y}}}$as ${\text{I}}$. So, the equation (2) can be written as follows,
${{\text{I}}_{\text{c}}}{\text{ = I + I = 2I}}$
Hence the inertia about an axis perpendicular to the circular surface and passing through its centre=${{\text{I}}_{\text{c}}} = 2{\text{I}}$
So, option (B) is the correct answer.
Note: Moment of inertia is the tendency of a particular object or body to resist angular acceleration. Moment of inertia of any object is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation. Moment of inertia is analogous to the inertia of the object. In rectilinear motion it is inertia and in rotational motion it is a moment of inertia.
Complete step-by-step solution -
Given, the moment of inertia about one of its diameters ${\text{ = I}}$,
This implies that moment of inertia about x-axis and y-axis are equal to each other and both are equal to the moment of inertia to the of the circular disc about its diameter i.e. ${\text{I}}$.
So, we can write this as,
${{\text{I}}_{\text{x}}}{\text{ = }}{{\text{I}}_{\text{y}}}{\text{ = I}}$-----equation (1)
The axis perpendicular to the circular surface and passing through its centre is the axis which is perpendicular to the x-axis and the y-axis, hence it is z-axis.
And from the perpendicular theorem, we know that
${{\text{I}}_{\text{z}}}{\text{ = }}{{\text{I}}_{\text{x}}}{\text{ + }}{{\text{I}}_{\text{y}}}$
Let the moment of inertia about an axis perpendicular to the circular surface and passing through its centre=${{\text{I}}_{\text{c}}}$
This ${{\text{I}}_{\text{c}}}$is same as ${{\text{I}}_{\text{z}}}$, so we can write like this,
${{\text{I}}_{\text{c}}}{\text{ = }}{{\text{I}}_{\text{x}}}{\text{ + }}{{\text{I}}_{\text{y}}}$-----equation (2)
And from equation (1) we can write the values of ${{\text{I}}_{\text{x}}}{\text{& }}{{\text{I}}_{\text{y}}}$as ${\text{I}}$. So, the equation (2) can be written as follows,
${{\text{I}}_{\text{c}}}{\text{ = I + I = 2I}}$
Hence the inertia about an axis perpendicular to the circular surface and passing through its centre=${{\text{I}}_{\text{c}}} = 2{\text{I}}$
So, option (B) is the correct answer.
Note: Moment of inertia is the tendency of a particular object or body to resist angular acceleration. Moment of inertia of any object is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation. Moment of inertia is analogous to the inertia of the object. In rectilinear motion it is inertia and in rotational motion it is a moment of inertia.
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