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.

\[I_{new} = I_c + md^2\] \(I_{new}\) = the mass moment of inertia about any specified axis
\(I_c\) = the mass moment of inertia about an axis parallel to the above specified axis but passes through the body’s mass center.
m = mass of the body
d = normal distance from the body’s mass center to the above specified axis.

Solved Examples

Solved Example:

35-5-01

Which of the following defines the parallel axis theorem?

Correct Answer: A

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.

Correct Answer: A

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.

Correct Answer: D

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)

Correct Answer: B

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)

Correct Answer: C

Solved Example:

35-5-06

It is the material's ability to resist twisting:

Correct Answer: D

Solved Example:

35-5-07

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

Correct Answer: C

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*}

Correct Answer: A

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

Correct Answer: C

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.

Correct Answer: C