Moment Of Inertia Of The Solid Cylinder

Moment of Inertia of the Solid Cylinder

Let I= Moment of inertia of the cylinder about an axis through its centre

To calculate I

Consider a cylinder of mass M, radius R and length L.

mass per unit volume of the cylinder $\rho=\frac{M}{V}=\frac{M}{\pi R^2 L}$

Imagine that the cylinder is made of a large number of coaxial cylindrical shells

Take a small elemental cylindrical shell of mass dm having internal radius x and external radius (x + dx).

So for that elemental cylindrical shell $d V=(2 \pi x d x) L$

And
$
dm=\rho d V=\frac{M}{\pi R^2 L}(2 \pi x d x) L
$

Now integrate this dI between the limits x=0 to x=R

$\begin{aligned}
& I=\int d I=\int x^2 * \rho d v \\
& =\int_0^R \frac{M}{\pi R^2 L}\left(2 \pi * L x^3 d x\right) \\
& =\frac{2 M}{R^2} \int_0^R x^3 d x=\frac{M R^2}{2}
\end{aligned}$

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Solved Example Based on the Moment of Inertia of the Solid Cylinder

Example 1: The moment of inertia of solid cylinder mass M, length L and radius R about its own axis is:

1) $M R^2$
2) $M L^2$
3) $\frac{M R^2}{2}$
4) $\frac{M L^2}{12}$

Solution:

Moment of inertia for solid cylinder -

$
I=\frac{1}{2} M R^2
$
wherein
About the axis passing through a central line.
The moment of inertia about its own axis is $\frac{M R^2}{2}$

Hence, the answer is option (3).

Example 2: Two solid cylinders are connected by a massless rod as shown in the figure. The distance b/w centre line of both cylinders is 3R. What is the moment (in terms of MR²) of inertia of the system about the axis shown in the below diagram?

1) 5.55

2) 6.25

3) 1.85

4) 1.6

Solution:

Moment of inertia for solid cylinder About axis passing through central line is given by:
$
I_c=\frac{1}{2} M R^2
$

For the figure

For cylinder 1

$\begin{gathered}
I_1=I_c+M d^2 \\
I_1=\frac{1}{2} M R^2+M\left[R+\frac{R}{3}\right]^2 \\
\frac{1}{2} M R^2+M\left[\frac{4 R}{3}\right]^2=\frac{41}{18} M R^2
\end{gathered}$

For cylinder 2

$\begin{aligned}
& I_2=I_c+M d^2 \\
I_2= & \frac{1}{2} M R^2+M\left[R+\frac{2 R}{3}\right]^2 \\
I_2= & \frac{59}{18} M R^2
\end{aligned}$

So,

$I=I_1+I_2=\frac{41}{18} M R^2+\frac{59}{18} M R^2=\frac{100}{18} M R^2=\frac{50}{9} M R^2$

Hence, the answer is option (1).

Example 3: The moment of inertia of a cylinder of mass $M$ and length $L$ and radius $\mathrm{R}$ about an axis passing through its centre and perpendicular to the axis of the cylinder is $I=M\left(\frac{R^2}{4}+\frac{L^2}{12}\right)$ If such a cylinder is to $\frac{L}{R}$ for it to have the be made for a given mass of a material, the ratio $\bar{R}$ for it to have the minimum possible I is:

1) $\frac{2}{3}$
2) $\frac{3}{2}$
3) $\sqrt{\frac{2}{3}}$
4) $\sqrt{\frac{3}{2}}$

Solution:

$
I=m\left[\frac{R^2}{4}+\frac{L^2}{12}\right]
$

Given,
$
M=\rho\left(\pi R^2 L\right)
$

So,
$
I=M\left[\frac{R^2}{4}+\frac{M^2}{\left.12\left(\rho \pi R^2\right)^2\right)}\right]
$

$
\begin{aligned}
& \frac{d I}{d R}=0 \\
\Rightarrow & M\left[\frac{2 R}{4}+\frac{-4 M^2}{12 \rho^2 \pi^2 R^5}\right]=0 \\
R= & (2 / 3)^{1 / 6} \times\left(\frac{M}{\rho \pi}\right)^{1 / 3} \ldots( 1)\\
L= & \frac{M}{\rho \pi\left(\frac{2}{3}\right)^{1 / 3}\left(\frac{M}{\rho \pi}\right)^{2 / 3}} \ldots(2)
\end{aligned}
$

So, (2) divided by (1) will give
$
\frac{L}{R}=\sqrt{\frac{3}{2}}
$

Hence, the answer is the option (4).

Example 4: The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to AB as shown in the figure is 2.7 kg m².

1) $1.49 \times 10^2 \frac{\mathrm{kg}}{\mathrm{m}^3}$
2) $7.5 \times 10^1 \frac{\mathrm{kg}}{\mathrm{m}^3}$
3) $14 \cdot 9 \frac{\mathrm{kg}}{\mathrm{m}^3}$
4) $7.5 \times 10^2 \frac{\mathrm{kg}}{\mathrm{m}^3}$

Solution:

$\begin{aligned}
& I_{C D}=I_{A B}+M d^2 \\
& \quad=\left(\frac{M R^2}{2}\right)+M\left(\frac{L}{2}\right)^2 \\
& 2 \cdot 7=\left(\frac{M \times 0 \cdot 04}{2}\right)+\left(\frac{M \times 0 \cdot 64}{4}\right) \\
& 10 \cdot 8=0 \cdot 08 \mathrm{M}+0 \cdot 64 \mathrm{M} \\
& 10 \cdot 8=0 \cdot 72 \mathrm{M} \\
& M=\frac{10 \cdot 8}{0 \cdot 72}=15 \mathrm{~kg} \\
& M=\rho \times\left(\pi r^2 \mathrm{~L}\right) \\
& s=\frac{15}{\frac{22}{7} \times 0 \cdot 04 \times 0 \cdot 8} \\
& s=\frac{105}{22 \times 0 \cdot 04 \times 0 \cdot 8}=\frac{105 \times 10^{+3}}{22 \times 32} \\
& =0 \cdot 149 \times 10^3 \\
& s=1 \cdot 49 \times 10^2 \frac{\mathrm{kg}}{\mathrm{m}^3}
\end{aligned}$
Hence, the answer is option (1)

Example 5: A uniform solid cylinder with radius $\mathrm{R}$ and length $\mathrm{L}$ has a moment of inertia $\mathrm{I}_1$, about the axis of the cylinder. A concentric solid cylinder of radius $\mathrm{R}^{\prime}=\frac{\mathrm{R}}{2}$ and length $\mathrm{L}^{\prime}=\frac{\mathrm{L}}{2}$ is carved out of the original cylinder. If $\mathrm{I}_2$ is the moment of inertia of the carved-out portion of the cylinder then $\frac{\mathrm{I}_1}{\mathrm{I}_2}=$ $\qquad$ (Both $\mathrm{I}_1$ and $\mathrm{I}_2$ are about the axis of the cylinder)

1) 32

2) 34

3) 36

4) 38

Solution:

$\begin{aligned}
& \mathrm{I}_1=\frac{\mathrm{MR}^2}{2} \\
& \text { mass }=\rho \pi \frac{\mathrm{R}^2}{4} \cdot \frac{\mathrm{L}}{2} \\
& \mathrm{~m}_2=\frac{\mathrm{M}}{8} \\
& \mathrm{I}_2=\frac{\mathrm{m}_2 \mathrm{R}_2^2}{2}=\frac{\mathrm{MR}^2}{8 \times 4 \times 2} \\
& \frac{\mathrm{I}_1}{\mathrm{I}_2}=32
\end{aligned}$

Hence, the answer is option (1).

Summary

A moment of inertia is a revolving object; for this reason, a moment of inertia is also known as angular momentum or acceleration. But 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.