Partiċelli tal-mument tal-inerzja u korpi riġidi – problemi u soluzzjonijiet

The moment of inertia of the particle

1. A 100-gram ball connected to one end of a cord with a length of 30 cm. What is the moment of inertia of ball about the axis of rotation AB? Ignore cord’s mass.

Magħruf:

The axis of rotation at AB

Massa ball (m) = 100 gram = 100/1000 = 0.1 kg

The distance between ball and the axis rotation (r) = 30 cm = 0.3 m

Mfittxija: Moment of inertia of ball (I)

Soluzzjoni:

Jiena = is-Sur² = (0.1 kg)(0.3 metru kwadru)²

I = (0.1 kg)(0.09 m²)

I = 0.009 kg m²

2. A 100-gram ball, m₁, and a 200-gram ball, m₂, connected by a rod with a length of 60 cm. The mass of the rod is ignored. The axis of rotation located at the center of the rod. What is the moment of inertia of the balls about the axis of rotation?

Magħruf:

Massa tal-ballun 1 (m₁) = 100 gram = 100/1000 = 0.1 kg

The distance of ball 1 and the axis of rotation (r₁) = 30 cm = 30/100 = 0.3 m

Mass of ball (m₂) = 200 gram = 200/1000 = 0.2 kg

il distanza of ball 2 and the axis of rotation (r₂) = 30 cm = 30/100 = 0.3 m

Meħtieġ: moment of inertia of the balls

Tweġiba:

Jiena = m₁ r₁² +m₂ r₂²

I = (0.1 kg)(0.3 m)² + (0.2 kg)(0.3 m)²

I = (0.1 kg)(0.09 m²) + (0.2 kg)(0.09 m²)

I= 0.009 kg m² + 0.018 kg m²

I= 0.027 kg m²

3. A 200-gram ball, m₁ and a 100-gram ball, m_2, connected by a rod with length of 60 cm. Ignore rod’s mass. The axis of rotation located at ball m₂. What is the moment of inertia of the balls. Ignore rod’s mass.

Magħruf:

Massa tal-ballun 1 (m₁) = 200 gram = 200/1000 = 0.2 kg

The distance between ball 1 and the axis of rotation (r₁) = 60 cm = 60/100 = 0.6 m

Massa tal-ballun 2 (m₂) = 100 gram = 100/1000 = 0.1 kg

The distance between ball 2 and the axis of rotation (r₂) = 0 metri

Meħtieġ: Moment of inertia of the balls

Soluzzjoni:

Jiena = m₁ r₁² +m₂ r₂²

I = (0.2 kg)(0,6 m)² + (0.2 kg)(0)²

I = (0.2 kg)(0.36 m²) + 0

I = 0.072 kg m²

4. The mass of each ball is 100 gram, connected by cord. The length of cord is 60 cm and the width of cord is 30 cm. What is the moment of inertia of the balls about the axis of rotation. Ignore cord’s mass.

Magħruf:

Mass of ball = m₁ = m₂ = m₃ = m₄ = 100 gram = 100/1000 = 0.1 kg

The distance between ball and the axis of rotation (r₁) = 30 cm = 30/100 = 0.3 m

The distance between ball 2 and the axis of rotation (r₂) = 30 cm = 30/100 = 0.3 m

The distance between ball 3 and the axis of rotation (r₃) = 30 cm = 30/100 = 0.3 m

The distance between ball 4 and the axis of rotation (r₄) = 30 cm = 30/100 = 0.3 m

Magħruf: Mument ta 'inerzja

Soluzzjoni:

Jiena = m₁ r₁² +m₂ r₂² +m₃ r₃² +m₄ r₄²

I = (0.1 kg)(0.3 m)² + (0.1 kg)(0.3 m)² + (0.1 kg)(0.3 m)² + (0.1 kg)(0.3 m)²

I = (0.1 kg)(0.09 m²) + (0.1 kg)(0.09 m²) + (0.1 kg)(0.09 m²) + (0.1 kg)(0.09 m²)

I = 0.036 kg m²

The moment of inertia of rigid object

5. What is the moment of inertia of a 2-kg long uniform rod with length of 2 m. The axis of rotation located at the center of the rod.

Magħruf:

Mass of rod (M) = 2 kg

The length of rod (L) = 2 m

Mfittxija: Mument ta 'inerzja

Soluzzjoni:

The formula of the moment of inertia when the axis of rotation located at the center of long uniform rod :

I = (1/12) M L²

I = (1/12) (2 kg)(2 m)²

I = (1/12) (2 kg)(4 m²)

I = (1/12)(8 kg m²)

I = 8/12 kg m²

I = 2/3 kg m²

6. What is the moment of inertia of a 2-kg long uniform rod with a length of 2 m? The axis of rotation located at one end of the rod.

Magħruf:

Mass of rod (M) = 2 kg

The length of rigid rod (L) = 2 m

Mfittxija: Mument ta 'inerzja

Soluzzjoni:

The formula of the moment of inertia when the axis of rotation located at one end of the rod :

I = (1/3) M L²

I = (1/3) (2 kg)(2 m)²

I = (1/3) (2 kg)(4 m²)

I = (1/3)(8 kg m²)

I = 8/3 kg m²

7. A 10-kg solid cylinder with a radius of 0.1 m. The axis of rotational located at the center of the solid cylinder, shown in the figure below. What is the moment of inertia of the cylinder?

Magħruf:

Mass of solid cylinder (M) = 10 kg

Radius of cylinder (L) = 0.1 m

Mfittxija: Il-mument ta 'inerzja

Mfittxija: Il-mument ta 'inerzja

Soluzzjoni:

The formula of moment inertia when the axis of rotation located at the center of cylinder :

I = (1/2) M R²

I = (1/2) (10 kg)(0.1 m)²

I = (1/2) (10 kg)(0.01 m²)

I = (1/2)(0.1 kg m²)

I = 0.05 kg m²

8. A 20-kg uniform sphere with the length of 0.1 m. The axis of rotation located at the center of the sphere shown in the figure below.

Magħruf:

Mass of sphere (M) = 20 kg

The radius of sphere (L) = 0.1 m

Mfittxija: a moment of inertia

Soluzzjoni:

The formula of the moment of inertia when the axis of rotation located at the center of the sphere :

I = (2/5) M R²

I = (2/5)(20 kg)(0.1 m)²

I = (2/5)(20 kg)(0.01 m²)

I = (2/5)(0.2 kg m²)

I = 0.4/5 kg m²

I = 0.08 kg m²

9. A 2-kg rectangular thin plate with a length of 0.5 m and width of 0.2 m. The axis of rotation located at the center of the rectangular plat shown in the figure below. What is the moment of inertia of the rectangular?

Magħruf:

Mass of rectangular plat (M) = 2 kg

The length of plat (a) = 0.5 m

The width of plat (b) = 0.2 m

Meħtieġ: Mument ta 'inerzja

Soluzzjoni:

Formula of moment of inertia when the axis of rotation located at the center of plat :

I = (1/12) M (a² + b²)

I = (1/12)(2)(0.5² + 0.2²)

I = (2/12)(0.25 + 0.04)

I = (1/6)(0.29)

I = 0.29/6 kg m²