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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 larger the moment of inertia of the body more difficult is to stop its rotational motion.

The moment of inertia of the semicircle is generally expressed as I = Ï€r4 / 4

The moment of inertia of the semicircle is generally expressed as I = Ï€r4 / 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 aÂ 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.

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Thus, Ix = Iy = Â¼ Ï€r4

M.O.I relative to the origin, Jo = Ix + Iy = Â¼ Ï€r4 + Â¼ Ï€r4 = Â½ Ï€r4

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

Jo = Â½ (Ï€r2) 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 = â…› Ï€r4 = â…› (Ao) R2 = â…› Ï€r2) 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 = â…› Ï€r4 + â…› Ï€r4 = Â¼ Ï€r4

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 (Ïƒ) 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= Ïƒ

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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 Î¸

y = r cos Î¸

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 = (râ‹…d Î¸) â‹… dr = r â‹… drd Î¸

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

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

Ix = oâˆ«Ï€oâˆ«R y â‹… dA

= oâˆ«Ï€oâˆ«R r sin Î¸ â‹… rdrd Î¸

= oâˆ«Ï€ sin Î¸ ( oâˆ«R r2dr) d Î¸

= oâˆ«Ï€ R3 / 3 sin Î¸ d Î¸

= R3 / 3 [cos Î¸]oÏ€

= 2R3 / 3

Meanwhile, Ix = A = Ï€ / 2 R2 y

We get,

Ï€ / 2 R2 y = 2R3 / 3

y = 4R / 3Ï€

Now in order to find the second moment of inertia:

IXX = oâˆ«Ï€oâˆ«R y2 dA

Ixx = oâˆ«Ï€ sin2Î¸ [ oâˆ«R r3dr ] d Î¸

Ixx = oâˆ«Ï€ sin2Î¸ [ r4 / 4]or d Î¸

Ixx = r4 / 4 oâˆ«Ï€ sin2 dÎ¸

Using the trigonometric identity: sin2Î¸ = 1-cos2Î¸ / 2 we calculate the integral.

Ixx =r4 / 8{ oâˆ«Ï€ (1 â€“ cos 2 Î¸ / 2) d Î¸}

Ixx ={ r4 [ Î¸ â€“ sin 2 Î¸ / 2)]oÏ€}/4

Ixx = Ï€r4 /8

FAQ (Frequently Asked Questions)

1. What is the value of Moment of Inertia of the cube?

Answer: 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?

Answer: 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?

Answer: The formula for the moment of Inertia depends upon,Â

m =Â mass,

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

â‡’ Also, the dimensional formula of the moment of inertia can be given by, M^{1} L^{2} T^{0}.

4. What is the importance of the Moment of Inertia?

Answer: 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) and how rotational motion changes when the distribution of mass changes.

5. What shape has the lowest Moment of Inertia?

Answer: For any given shape, the moment of inertia through the center of mass of that body will be the minimum, since any moment of inertia through another axis would add mr^{2}^{ }to the result.

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

Answer: Since the mass of any rotating rigid body is considered to be distributed with respect to 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.