### Parallel Axis Theorem- Dynamics

• To understand the parallel-axis theorem and its applications.

• Use the parallel axis theorem to calculate moment of inertia of a body or system about various axes.

### Solved Example:

#### 35-5-01

Which of the following defines the parallel axis theorem?

### Solved Example:

#### 35-5-02

The moment of inertia around a parallel axis will always be __________ the moment of inertia around the center of mass axis.

### Solved Example:

#### 35-5-03

What is the unit of mass moment of inertia? (RRB JE ME CBT 2 Aug 2019)

Solution:
$I = mk^2$

Hence units will be kg.m^2. This is mass moment of inertia and should NOT be confused with area moment of inertia.

### Solved Example:

#### 35-5-04

The mass moment of inertia of a solid sphere about its diameter is: (SSC Scientific Asst. Nov 2017-Shift 2)

### Solved Example:

#### 35-5-05

The mass moment of inertia of a thin spherical shell about its diameter is: (UPPSC AE Mech- Paper Dec 2020)

### Solved Example:

#### 35-5-06

It is the material's ability to resist twisting:

### Solved Example:

#### 35-5-07

The angular momentum of a rotating object can be calculated by the formula

### Solved Example:

#### 35-5-08

A flywheel connected to a punching machine has to supply energy of 400 Nm while running at a mean angular speed of 20 rad/s. If the total fluctuation of speed is not to exceed $\pm$2%, the mass moment of inertia of the flywheel in $kg-m^2$ is: (GATE ME 2013)

Solution:

Coefficient of fluctuation of speed $C_s$ is given by

$C_s = \dfrac{\omega_{max} - \omega_{min}}{\omega} = 0.04$

because it is 2% of either side

$\Delta E = I \omega^2 C_s$ \begin{align*} I &= \dfrac{\Delta E}{\omega^2 C_s}\\ &= \dfrac{400}{20^2\times 0.04}\\ &= 25 kg.m^2 \end{align*}

### Solved Example:

#### 35-5-09

A ring of mass 10 kg and diameter 0.4 m is rotated about its axis. If it makes 2100 revolutions per minute, then its angular momentum will be:

Solution:

For a ring, $I =mr^{2} =\left( 10\right) \left( 0.2\right) ^{2} =0.4kg\cdot m^{2}$

Angular momentum, $= I \cdot \omega = \left( 0.4\right) \left( \dfrac {2\pi \times 2100}{60}\right) = 87.96kg.m^{2}/s$

### Solved Example:

#### 35-5-10

A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc:

Solution:

I $\omega$ = constant

As the insect moves towards the centre, I decreases $\omega$ increases.

When the insect moves from centre to the rim, I increases $\omega$ decreases.