# Moment of inertia

1. Moment of inertia of the particle

Review a rotating particle. The particle with mass m is given the force F so that the particle rotates about the axis O. The particle is r apart from the axis of rotation. First the particle is rest (v = 0). After moved by the force of F, the particles move with a certain speed so that the particles have tangential acceleration. The relationship between force (F), mass (m), and the tangential acceleration of particles are expressed by equation 3:

F = m a_{tan } (Equation 3)

The particles rotate so that the particles have angular acceleration. The relationship between tangential acceleration and angular acceleration is expressed by equation 4:

a_{tan} = r α (Equation 4)

Substitute a in equation 3 with a in equation 4:

F = m r α

r F = m r^{2 }α

τ = (m r^{2}) α (Equation 5)

r F is the moment of force and m r^{2} is the moment of inertia of particle. Equation 5 states the relationship between the moment of force, the moment of inertia, and the angular acceleration of rotating particles. Equation 5 is the equation of Newton’s second law for rotating particles.

The moment of inertia of a particle is the product of the mass of the particle (m) and the square of the distance between particles with the axis of rotation (r^{2}).

I = m r^{2 }(Equation 6)

I = moment of inertia of particle, m = mass of the particle, r = distance between particle with the axis of rotation. Equation 6 is used to determine the moment of inertia of a particle.

3.2 Sample problems of the moment of inertia of the particle

Sample problem 1.

A particle with a mass of 2 kg was tied to a rope 0.5 meters long and then rotated. What is the moment of inertia of particles when rotating?

Solution:

I = m r^{2}

I = (2 kg) (0.5 m)^{2}

I = 0.5 kg m^{2}

Sample problem 2.

Two particles, each having a mass of 2 kg and 4 kg, are connected by a lightweight wire, where the wire length is 2 meters. Ignore the mass of the wire. Determine the moment of inertia of the two particles, if:

a) The axis of rotation is located between the two particles

Solution:

I = m_{1 }r_{1}^{2} + m_{2} r_{2}^{2}

I = (2)(12) + (4)(12)

I = 2 + 4

I = 6 kg m^{2}

b) The axis of rotation is at a distance of 0.5 meters from the particle with a mass of 2 kg

I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2}

I = (2)(0.5^{2}) + (4)(1.5^{2})

I = (2)(0.25) + (4)(2.25)

I = 0.5 + 9

I = 9.5 kg m^{2}

c) The axis of rotation is at a distance of 0.5 meters from the particle with a mass of 4 kg

I = m_{1 }r_{1}^{2} + m_{2} r_{2}^{2}

I = (2)(1.52) + (4)(0.5^{2})

I = (2)(2.25) + (4)(0.25)

I = 4.5 + 1

I = 5.5 kg m^{2}

Based on the results of the calculations above, the moment of inertia is strongly influenced by the location of the axis of rotation. Although the shape and size are the same, because the location of the axis of rotation is different, the moment of inertia is also different.

Particles near the axis of rotation have smaller the moment of inertia, whereas particles that are far from the axis of rotation have greater the moment of inertia. If we assume that the two particles above are rigid objects, then each particle that is near the axis of rotation has a smaller moment of inertia than the moment of inertia of the particle farther away from the axis of rotation.

3.3 Moment of inertia of a homogeneous rigid body

Rigid objects are composed of many particles. The moment of inertia of a rigid body is the total number of moments of inertia of each particle that builds the object.

I = Σ m r^{2}

I = m_{1} r_{1}^{2 }+ m_{2} r_{2}^{2 }+ m_{3} r_{3}^{2 }+ ….. + m_{n} r_{n}^{2}

To determine the moment of inertia of a rigid object, we need to review the object when it rotates because the location of the axis of rotation affects the moment of inertia of the object. The moment of inertia (I) of each particle also depends on the mass (m) of the particle and the square of the distance (r^{2}) of the particle from the axis of rotation. The mass of all the particles builds the object is the mass of the object.

The problem is, the distance of each particle from the axis of rotation varies.

Review the decrease in the formula of the moment of inertia of a thin ring with fingers R and mass M. If the axis of rotation is located at the center of the ring, all particles of the thin ring are r from the axis of rotation. The moment of inertia of a thin ring equals the moments of inertia of all particles:

I = Σ m r^{2}

I = m_{1} r_{1}^{2 }+ m_{2} r_{2}^{2 }+ m_{3} r_{3}^{2 }+ ….. + m_{n} r_{n}^{2}

The mass of all particles = mass of thin ring (M)

I = M (r_{1}^{2 }+ r_{2}^{2 }+ r_{3}^{2 }+ ….. + r_{n}^{2})

Each particles of the thin ring are r from the axis of rotation so that r_{1 }= r_{2} = r_{3} = rn = R

The equation of the moment of inertia of the thin ring:

I = M R^{2}

I = moment of inertia of the thin ring, M = mass of thin ring, R = radius of the thin ring.

What if the axis of rotation is not located at the center of the ring? If the axis of rotation is not located at the center of the ring, the moment inertia formula of the thin ring cannot be derived using the above method because the distance of each particle from the axis of rotation is different. Deriving the formula of the moment of inertia for a problem like this is not discussed in this topic.

The formula of the moment of inertia for some rigid objects

3.4 Sample problem of the moment of inertia of a homogeneous rigid body

Sample problem 1:

Rod cylinder AB has a mass of 3 kg when rotated through B,

the moment of inertia is 27 kg m^{2}. What is the moment of inertia if

rotated through C?

Solution:

Solution:

The formula of the moment of inertia for the homogeneous solid rod (the axis of rotation is at the center of the rod):

I = 1/12 M L^{2}

27 = (1/12) M L^{2}

(27)(12) = M L^{2}

324 = M L^{2}

The formula of the moment of inertia of a homogeneous solid rod (the axis of rotation on the edge of the rod):

I = 1/3 M L^{2}

I = 1/3 (324)

I = 108 kg m^{2}