The moment of inertia of a thin uniform circular disc about one of its diameters is ${\text{I}}$. The moment of inertia about an axis perpendicular to the circular surface and passing through its centre:
(A). $\sqrt {\text{2}} {\text{I}}$
(B). $2{\text{I}}$
(C). $\dfrac{{\text{I}}}{2}$
(D). $\dfrac{{\text{I}}}{{\sqrt 2 }}$
Answer
Verified
488.1k+ views
- 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.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
Find the value of x if the mode of the following data class 11 maths CBSE
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
Trending doubts
10 examples of friction in our daily life
The correct order of melting point of 14th group elements class 11 chemistry CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
State and prove Bernoullis theorem class 11 physics CBSE
What organs are located on the left side of your body class 11 biology CBSE