# Moment of Inertia of Semicircle     ## Moment of Inertia of Semicircle-Concept of Area Density and Derivation of Formula is available at Vedantu

The syllabus of the IIT JEE for the subject of Physics covers all the chapters of two previous classes, that is to say, class 11 and class 12. And hence, it is very important for the students to study all those units. And one of those units is the System of Particles and Rotational motion, which you may have studied in class 11th Physics. In which you have learned the motion of the extended body, that is to say, the system of particles. And one of the important parts of the system of Particles is that of Moment of Inertia in Semicircle.

If you have forgotten the part of the Moment of Inertia in Semicircle, then you do not have to worry at all. Vedantu provides the complete explanation of the Moment of Inertia in Semicircle for the students of IIT JEE, and that too is free of cost.

### Moment of Inertia of Semicircle

The moment of inertia plays the same role in rotational motion as the mass does in the translational motion. In other words, the moment of inertia is the measurement of resistance of the body to a change in its rotational motion. For example, if the body is at rest the larger the moment of inertia of the body the more difficult it is to put the body into the rotational motion. Similarly, the larger the moment of inertia of the body, the more difficult it is to stop its rotational motion.

The moment of inertia of the semicircle is generally expressed as I = πr4 / 4

The role that the mass plays in the translation motion, is played by the moment of inertia in the rotational motion. As mentioned above the equation for the moment of inertia is $I = \sum m_{i} r_{i}^{2}$ but for the moment of inertia in semicircle I = πr4 / 4

For finding the moment of inertia in a semicircle, it is necessary to find the moment of inertia in a full circle first. Because a semicircle is nothing but the half of the full circle, and hence for finding the moment of inertia in the semicircle, all you have to do is to divide the moment of inertia in a full circle in half, that is to say, by two.

Also, the sum of both the axis, that is to say, the x-axis and the Y-axis are to be taken for determining the moment of inertia in a semi-circle.

The law of the moment of inertia and the moment of inertia in a semicircle is very much important in our practical life. Many machines produce rotational motion such as the Steam engine, automobile engine etc. These machines produce rotational motion because they have discs called Flywheel, which have a large amount of moment of inertia. Therefore, the moment of inertia helps in calculating the angular motion of these discs.

### Finding Moment of Inertia of Semicircle

The moment of inertia of the semicircle is generally expressed as I = πr4 / 4. Here in order to find the value of the moment of inertia of a semicircle, we have to first derive the results of the moment of inertia full circle and basically divide it by two to get the required result of that moment of inertia for a semicircle. Further to determine the moment of inertia of the semi-circle, we will take the sum of both the x and y-axis.

We know that for a  full circle because of complete symmetry and uniform area distribution, the moment of inertia relative to the x-axis is equal to that of the y-axis.

Thus, Ix = Iy = ¼ πr4

M.O.I relative to the origin, Jo = Ix + Iy = ¼ πr4 + ¼ πr4 = ½ πr4

Now we need to pull out the area of a circle which gives us:

Jo = ½ (πr2) R2

Thus similarly for the semicircle, the moment of inertia of the x-axis is equal to that of the y-axis. Here, the semi-circle rotating about an axis is symmetric, and therefore, we consider these values equal. Thus, M.O.I will be half the moment of inertia of that of a full circle. Now this gives us:

Ix = Iy = ⅛ πr4 = ⅛ (Ao) R2 = ⅛ πr2) R2

Now to find the moment of inertia of the semicircle we will take the sum of both the x and y-axis.

M.O.I relative to the origin, Jo = Ix + Iy = ⅛ πr4 + ⅛ πr4 = ¼ πr4

### Concept of Area Density(σ)

Selecting and defining a tiny strip of mass with differential width. Now writing an expression for the area density for the whole semicircle and then the tiny strips of differential widths. Add all of the individual strips using integral Calculus.

Use the concept for area density, which is mass divided by area. Area density (σ) is an intensive property, meaning that it does not depend on the amount of the material, and also as long as the mass is uniform, its area density is the same whether you have chosen the entire semicircle or a small strip of differential width.

The macro-scale area density is given in this equation: dm/da= σ

### Derivation of Formula for Moment of Inertia of Semicircle

1. Here in order to derive the moment of inertia of a semicircle we define the coordinates using the polar system. We get:

z = r sin θ

y = r cos θ

2. Now we have to determine the differential area by finding the area of the element. It is given as:

ABCD is a sector with area = (r⋅d θ) ⋅ dr = r ⋅ drd θ

The centroid of this elemental area from x-axis = y sin θ

3. Using this we can find the first moment of inertia about the x-axis. We get:

$I_{x} = \int_{0}^{\pi} \int_{0}^{R} y . dA$

= $\int_{0}^{\pi} \int_{0}^{R} r sinθ . rdrdθ$

= $\int_{0}^{\pi} sin θ (\int_{0}^{R} r^{2}dr ) dθ$

=$\int_{0}^{\pi} \frac{R^{3}}{3} sin θ d θ$

= $\frac {R^{3}} {3} {cos θ _{0}^{\pi}}$

= $\frac{2R^{3}}{3}$

Meanwhile, Ix = A = π / 2 R2 y

We get,

π / 2 R2 y = 2R3 / 3

y = 4R / 3π

Now in order to find the second moment of inertia:

$I_{xx} = \int_{0}^{\pi}\int_{0}^{R} y^{2} dA$

$I_{xx} = \int_{o}^{R} sin^{2} θ ( \int_{0}^{R} r^{3} dr ) dθ$

$I_{xx} = \int_{}^{} sin^{2} θ (\frac{r^{4}}{4} )_{0}^{r} d θ$

$I_{xx} = \frac{r^4} {4}\int_{0}^{\pi} sin^2 dθ$

Using the trigonometric identity: $sin^2θ = \frac{1-cos^2θ } {2}$ we calculate the integral.

$I_{xx} = \frac{r^4}{8} {\int_{0}^{\pi}\frac{(1 – cos 2 θ ) }{2} d θ}$

$I_{xx} =\frac{( r^{4} θ– \frac{sin2θ}{2})_{0}^{\pi})}{4}$

$I_{xx} = \frac{πr^{4}}{8}$

### Benefits of having the Explanation of the Moment of Inertia in a Semicircle from the Vedantu

The subject of science requires the students a great understanding of the topic. But since science is a practical subject, its theory sometimes becomes rather difficult to understand for the students of the IIT JEE. And also the science subject has a different set of vocabulary as well, and the same goes for the IIT JEE topic moment of inertia in a semi-circle.

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### A Brief Overview of the Moment of Inertia

The product of the mass of the object and its Velocity is the Vector quantity, known as Linear Momentum. Now, the rotational counterpart of this Linear Momentum is called “Angular Momentum”. And the Moment of Inertia is the quantity, to find out the torque, that is to say, the moment of force needed for the desired angular momentum. In the calculation of the angular Momentum, the Moment of Inertia is applied. The moment of inertia is also sometimes termed as rotational inertia or the angular mass, as well.

For the rigid body, the moment of inertia depends on quite many factors, such as the mass of the rigid body, the shape of a rigid body, distribution of the mass is both the axis of the rotation and the position and rotation of the axis of rotation.

$I = \sum m_{i} r_{i}^{2}$ is the equation for the moment of inertia.

## FAQs on Moment of Inertia of Semicircle

1. What is the value of moment of inertia of the cube?

Moment of inertia is the property of the mass of the rigid body that defines the total net torque needed for a desired or required angular acceleration about an axis of rotation.

2. How can we increase the moment of inertia of the body?

The moment of inertia is mainly the calculation of the required force to rotate an object. This value can be increased or decreased by a corresponding increase in the radius from the axis of rotation, the moment of inertia increases thus decreasing the speed of rotation.

3. On what factors the moment of inertia of any given body depends?

The formula for the moment of Inertia depends upon,

m =  mass,

r = distance of the mass of which mi is to find out about the axis of the rotation.

⇒ Also, the dimensional formula of the moment of inertia can be given by, M1 L2 T0.

4. What is the importance of the moment of inertia?

Rotational inertia is important in physics that involves the mass in rotational motion. It is used to calculate the angular momentum which also allows us to explain (via conservation of angular momentum) how rotational motion changes when the distribution of mass changes.

5. What shape has the lowest moment of inertia?

For any given shape, the moment of inertia through the centre of mass of that body will be the minimum, since any moment of inertia through another axis would add mr2 to the result.

6. What is the significance or importance of the radius of gyration?

Since the mass of any rotating rigid body is considered to be distributed concerning the axis of rotation, we have defined a new parameter known as the radius of gyration. It is that distance whose square when multiplied with the mass of that body gives us the moment of inertia of the body about that given axis.

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