Moment Of Inertia Of Hollow Cylinder

The moment of inertia is a crucial concept in rotational dynamics, representing an object's resistance to changes in its rotational motion. For a hollow cylinder, such as a metal pipe or a rolling hoop, the moment of inertia plays a significant role in how it behaves when spun or rolled. Unlike solid cylinders, where mass is distributed throughout, a hollow cylinder concentrates its mass along the outer edge. This unique distribution impacts how easily the cylinder can rotate and is pivotal in applications like flywheels in engines, where controlling rotational inertia is vital. Understanding the moment of inertia for a hollow cylinder helps in designing efficient mechanical systems, optimizing energy use, and ensuring stability in various engineering applications

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This Story also Contains

1. Moment of Inertia of the Hollow Cylinder
2. Solved Examples Based on the Moment of Inertia of the Hollow cylinder
3. Summary

Moment of Inertia of the Hollow Cylinder

The moment of inertia is a fundamental concept in rotational dynamics, representing an object's resistance to changes in its rotational motion. For a hollow cylinder, this property becomes particularly interesting due to its unique mass distribution. Unlike a solid cylinder, the mass of a hollow cylinder is concentrated further from the axis, resulting in a higher moment of inertia for the same mass.

Let I= Moment of inertia of the hollow cylinder about its axis passing through its C.O.M

To calculate I

Consider a cylinder of mass M, radius R and length L as shown in the figure

Now take an elemental ring of radius R and mass dm which is coaxial to hollow cylinder.

The moment of inertia of the elemental ring about the axis of the cylinder and $\text{ring is}dI=dm{R}^2$

So integrating the Moment of inertia of such elemental rings will give I

So,

$I=\int dI=\int dm{R}^2=M{R}^2$

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Solved Examples Based on the Moment of Inertia of the Hollow cylinder

Example 1: Three hollow cylinders each of mass M and radius R are arranged as shown in the figure. If the moment of inertia of the system about an axis passing through the central line is nMR² then find n?

1) 5

2) 7

3) 11

4) 8

Solution:

Moment of inertia for hollow cylinder
$I=M{R}^2$

wherein
About the axis passing through the central line.
moment of inertia passing through the central line of one cylinder A

$I_1=M{R}^2$

moment of inertia of cylinder B about that axis

$\begin{array}{rl}& I_2=I_c+M{d}^2\Rightarrow M{R}^2+M(2R{)}^2\\ & I_2=5M{R}^2\\ & I_{\text{net}}=I_1+I_2+I_2\Rightarrow M{R}^2+5M{R}^2+5M{R}^2\\ & I_{\text{net}}=11M{R}^2\\ & \because I=nM{R}^2\\ & \therefore n=11\end{array}$

Hence, the answer is the option (3).

Example 2: Moment of inertia (in MR²)of hollow cylinder mass M, Length R and Radius R about C.G and perpendicular to its own axis is (Give your answer till 2 places after the decimal point)

1) 0.25

2) 0.5

3) 0.58

4) 0.4

Solution:

Consider a hollow cylinder of mass M, length 'l ' and radius 'r' capable of rotating about its geometrical axis. Let m be its mass per unit length.

$\mathrm{m}=\mathrm{M}/\mathrm{l}$ Hence $\mathrm{M}=\mathrm{m}\cdot l$
The mass of such a ring is given by
Mass, $\mathrm{dm}=\mathrm{m}\cdot \mathrm{dx}=(\mathrm{M}/l)\mathrm{dx}$
The M.I. of such a ring about a transverse axis (passing through C) i given by

$\mathrm{dI}=\mathrm{dm}\cdot {\mathrm{R}}^2$
The M.I. of a ring about diameter is given by

$\mathrm{d}\mathrm{I}=\frac{\mathrm{dm}\cdot {\mathrm{R}}^2}{2}=\left(\frac{M}{l}\right)dx\cdot \frac{{\mathrm{R}}^2}{2}$
By parallel axes theorem

$\begin{array}{c}{\mathrm{dI}}_{\mathrm{c}}={\mathrm{dI}}_{\mathrm{G}}+\mathrm{dm}\cdot {\mathrm{x}}^2\\ {\mathrm{d}}_{\mathrm{c}}=\left(\frac{M}{l}\right)dx\cdot \frac{{\mathrm{R}}^2}{2}+\left(\frac{M}{l}\right)dx\cdot {\mathrm{x}}^2\end{array}$

Integrating the above expression into limits

$\begin{array}{rl}& \int {\mathrm{d}}_{\mathrm{c}}={\int }_{-\frac{1}{2}}^{+\frac{1}{2}}\left(\frac{M}{l}\right)dx\cdot \frac{{\mathrm{R}}^2}{2}+{\int }_{-\frac{1}{2}}^{-\frac{1}{2}}\left(\frac{M}{l}\right)dx\cdot {\mathrm{x}}^2\\ & \therefore {\mathrm{I}}_{\mathrm{C}}=\frac{M{\mathrm{R}}^2}{2l}{\int }_{-\frac{1}{2}}^{+\frac{1}{2}}dx\cdot +\frac{M}{l}{\int }_{-\frac{l}{2}}^{+\frac{1}{2}}{\mathrm{x}}^2\cdot dx\\ & \therefore {\mathrm{I}}_{\mathrm{c}}=\frac{M{\mathrm{R}}^2}{2l}[x{]}_{-\frac{1}{2}}^{+\frac{1}{2}}+\frac{M}{l}{\left[\frac{{\mathrm{x}}^3}{3}\right]}_{-\frac{1}{2}}^{+\frac{l}{2}}\\ & \therefore {\mathrm{I}}_{\mathrm{c}}=\frac{M{\mathrm{R}}^2}{2l}[+\frac{l}{2}-(-\frac{l}{2})]+\frac{M}{3l}[{(+\frac{l}{2})}^3-{(-\frac{l}{2})}^3]\\ & \therefore {\mathrm{I}}_{\mathrm{c}}=\frac{M{\mathrm{R}}^2}{2l}[\frac{l}{2}+\frac{l}{2}]+\frac{M}{3l}[\frac{l^3}{8}+\frac{l^3}{8}]\\ & \therefore I_c=\frac{M{R}^2}{2l}\left[\frac{2l}{2}\right]+\frac{M}{3l}\left[\frac{2l^3}{8}\right]\\ & \therefore {\mathrm{I}}_{\mathrm{c}}=\frac{M{\mathrm{R}}^2}{2l}[l]+\frac{M}{3l}\left[\frac{l^3}{4}\right]\\ & \therefore I_c=\frac{M{R}^2}{2}+\frac{M{l}^2}{12}\end{array}$
Given, $I=R$
So, $\therefore I_c=\frac{7M{R}^2}{12}$

Hence, the answer is the option (3).

Summary

A moment of inertia is a revolving object; for this reason, a moment of inertia is also known as angular momentum or acceleration. However, as the name implies, moment of inertia is for angular or rotational motion, whereas inertia is for linear motion. This makes it very distinct from linear inertia. The size and shape of the object's cross-section, such as its I, circular, or rectangular cross-section, among others, are some of the factors that determine the moment of inertia. Another factor is the object's density, and the distribution of particles and objects about the rotational axis is the last and most important factor.