Basic Physics

Menu

Moment of inertia particles and rigid body – problems and solutions

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.

Moment of inertia particles and rigid body – problems and solutions 1Known :

The axis of rotation at AB

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

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

Wanted: Moment of inertia of ball (I)

Solution :

I = m r2 = (0.1 kg)(0.3 m)2

I = (0.1 kg)(0.09 m2)

I = 0.009 kg m2

Read :  Area expansion – problems and solutions

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?

Moment of inertia particles and rigid body – problems and solutions 2Known :

Mass of ball 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

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

Wanted : moment of inertia of the balls

Jawab :

I = 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

Read :  Rotation of rigid bodies – problems and solutions

3. A 200-gram ball, m1 and a 100-gram ball, m2, 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.

Moment of inertia particles and rigid body – problems and solutions 3Known :

Mass of ball 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

Mass of ball 2 (m2) = 100 gram = 100/1000 = 0.1 kg

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

Wanted : Moment of inertia of the balls

Solution :

I = 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

Read :  Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions

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.

Moment of inertia particles and rigid body – problems and solutions 4Known :

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

Known : Moment of inertia

Solution :

I = 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

Read :  Motion on the horizontal surface without the friction force – application of Newton's law of motion problems and solutions

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.

Moment of inertia particles and rigid body – problems and solutions 5Known :

Mass of rod (M) = 2 kg

The length of rod (L) = 2 m

Wanted: Moment of inertia

Solution :

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

Read :  The magnitude of net torque – problems and solutions

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.

Moment of inertia particles and rigid body – problems and solutions 6Known :

Mass of rod (M) = 2 kg

The length of rigid rod (L) = 2 m

Wanted: Moment of inertia

Solution :

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

Read :  Fluid statics – problems and solutions

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?

Moment of inertia particles and rigid body – problems and solutions 7Known :

Mass of solid cylinder (M) = 10 kg

Radius of cylinder (L) = 0.1 m

Wanted: The moment of inertia

Wanted: The moment of inertia

Solution :

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.

Moment of inertia particles and rigid body – problems and solutions 8Known :

Mass of sphere (M) = 20 kg

The radius of sphere (L) = 0.1 m

Wanted: a moment of inertia

Solution :

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

Read :  Perfectly elastic collisions in one dimension – problems and solutions

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?

Known :

Moment of inertia particles and rigid body – problems and solutions 9Mass of rectangular plat (M) = 2 kg

The length of plat (a) = 0.5 m

The width of plat (b) = 0.2 m

Wanted : Moment of inertia

Solution :

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

Leave a Reply

Your email address will not be published.

This site uses Akismet to reduce spam. Learn how your comment data is processed.