Question

Obtain expression for moment of inertial disc, about an axis of rotation passing through it's center?​

Answer 1

$\huge\fcolorbox{black}{teal}{Solution:-}$

✏️ Answer:-

★ I = ½ mR²

✏️ Explanation:-

★ When we talk about the moment of inertia of a disk we can say that it is quite similar to that for a solid cylinder with any given measure of length. However, for a disk, we have to take it as a special character. Generally, it is used as a base for building the moment of inertia expression for different other shapes, such as a cylinder or a sphere.

• Meanwhile, we can also find the moment of inertia of a circular disc with respect to different situations. They are as follows---

1. Solid disk

• Here the axis of rotation is the central axis of the disk. It is expressed as;

½ MR²

2. Axis at Rim

• In this case, the axis of rotation of a solid disc is at the rim. It is given as;

3/2 MR²

3. Disc With a Hole

• Here the axis will be at the centre. It is expressed as;

½ M (a² + b²)

★ Moment Of Inertia Of A Disk rotating on its axis derivation:-

→ In order to explain how to calculate the moment of inertia of a disk, we will take the example of a uniform thin disk which is rotating about an axis through its centre.

In the figure [Have a glimpse at the figure attached above], we can see a uniform thin disk with radius r rotating about a Z-axis passing through the centre.

As we have a thin disk, the mass is distributed all over the x and y plane. Then, we move on to establishing the relation for surface mass density (σ) where it is defined as or said to be the mass per unit surface area. Since the disk is uniform, therefore, the surface mass density will also be constant where;

σ = m / A

⇒σA = m

⇒dm= σ (dA)

Now, for the simplification of the area where it can be assumed the area to be made of a collection of rings that are mostly thin in nature. The thin rings are said to be the mass increment (dm) of radius r which are at equal distance from the axis. The small area (dA) of every ring is further expressed by the length (2πr) times the small width of the rings (dr.) It is given as---

A = πr2, dA = d(πr2) = πdr2 = 2rdr

Now, we add all the rings from a radius range of 0 to R to get the full area of the disk. The radius range that is given is the value that is used in the integration of dr. If we put all these together then we get;

I = O∫^R r²σ(πr)dr

⇒I = 2 π σO∫^R r³dr

⇒I = 2 πσ r⁴/ 4 |o^R

⇒ I = 2 πσ (r⁴/ 4 – 0)

⇒I = 2 π (m / 4 )(R⁴ / 4)

⇒I = 2 π (m / π r²)(R⁴ / 4)

⇒ I = ½ mR²

This expression clearly proves all that we have discussed so far! ^^

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I hope this helps! :)

Answer 2

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⚝ Answer:-

I = ½ mR²

⚝ Explanation:-

When we talk about the moment of inertia of a disk we can say that it is quite similar to that for a solid cylinder with any given measure of length. However, for a disk, we have to take it as a special character. Generally, it is used as a base for building the moment of inertia expression for different other shapes, such as a cylinder or a sphere.

• Meanwhile, we can also find the moment of inertia of a circular disc with respect to different situations.

They are as follows-

1. Solid disk

• Here the axis of rotation is the central axis of the disk. It is expressed as;

½ MR²

2. Axis at Rim

• In this case, the axis of rotation of a solid disc is at the rim. It is given as;

3/2 MR²

3. Disc With a Hole

• Here the axis will be at the centre. It is expressed as;

½ M (a² + b²)

★ Moment Of Inertia Of A Disk rotating on its axis derivation:-

→ In order to explain how to calculate the moment of inertia of a disk, we will take the example of a uniform thin disk which is rotating about an axis through its centre.

In the figure [Have a glimpse at the figure attached above], we can see a uniform thin disk with radius r rotating about a Z-axis passing through the centre.

As we have a thin disk, the mass is distributed all over the x and y plane. Then, we move on to establishing the relation for surface mass density (σ) where it is defined as or said to be the mass per unit surface area. Since the disk is uniform, therefore, the surface mass density will also be constant where;

σ = m / A

⇒σA = m

⇒dm= σ (dA)

Now, for the simplification of the area where it can be assumed the area to be made of a collection of rings that are mostly thin in nature. The thin rings are said to be the mass increment (dm) of radius r which are at equal distance from the axis. The small area (dA) of every ring is further expressed by the length (2πr) times the small width of the rings (dr.) It is given as-

A = πr2, dA = d(πr2) = πdr2 = 2rdr

Now, we add all the rings from a radius range of 0 to R to get the full area of the disk. The radius range that is given is the value that is used in the integration of dr. If we put all these together then we get;

I = O∫^R r²σ(πr)dr

⇒I = 2 π σO∫^R r³dr

⇒I = 2 πσ r⁴/ 4 |o^R

⇒ I = 2 πσ (r⁴/ 4 – 0)

⇒I = 2 π (m / 4 )(R⁴ / 4)

⇒I = 2 π (m / π r²)(R⁴ / 4)

⇒ I = ½ mR²

This expression clearly proves all that we have discussed so far!!

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