Moment Of Inertia Of A Disc

Edited By Vishal kumar | Updated on Sep 25, 2024 05:41 PM IST

The moment of inertia of a disc is a measure of how hard it is to spin the disc around an axis. This concept is crucial in physics, as it helps us understand the rotational motion of objects. For example, the moment of inertia of a car's wheels affects how quickly the car can accelerate and stop. Understanding this property helps in designing efficient and safe vehicles.

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Moment Of Inertia Of A Disc

In this article, we will cover the concept of the moment of inertia of a disc. This topic falls under the broader category of rotational motion, which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), almost nine questions have been asked on this concept. And for NEET two questions were asked from this concept.

Let I=Moment of inertia of a DISC about an axis through its centre and perpendicular to its plane

To calculate I

Consider a circular disc of mass M, radius R and centre O.

$\text { And mass per unit area }=\sigma=\frac{M}{\pi R^2}$

Take an elementary ring of mass dm of radius x as shown in figure

$\begin{aligned}
& \text { So, } d m=\sigma *(2 \pi x d x)=\frac{M}{\pi R^2} *(2 \pi x d x) \\
& \Rightarrow d I=x^2 d m \\
& I=\int d I=\int_0^R x^2 *\left(\frac{M}{\pi R^2} *(2 \pi x d x)\right)=\frac{2 M}{R^2} \int_0^R x^3 d x=\frac{M R^2}{2}
\end{aligned}$

Also, we can also find the moment of inertia of a circular disc with respect to different situations. They are as follows:

Solid Disc

Here, the axis of rotation is the central axis of the disc. It is expressed as:
$$
\frac{1}{2} M R^2
$$

Axis at Rim

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In this case, the axis of rotation of a solid disc is at the rim. It is given as:
$$
\frac{3}{2} M R^2
$$

Disc with a Hole

Here, the axis will be at the centre. It is expressed as:
$$
\frac{1}{2} M\left(a^2+b^2\right)
$$
where,$a$ is the inner radius and $b$ is the outer radius.

Solved Examples Based on the Moment of Inertia of A Disc

Example 1: The moment of inertia of a uniform semicircular disc of mass M and radius $r$ about a line perpendicular to the plane of the disc through the centre is :

1) $M r^2$
2) $\frac{1}{2} M r^2$
3) $\frac{1}{4} M r^2$
4) $\frac{2}{5} M r^2$

Solution:

As we learnt in

Moment of inertia for disc -

$I=\frac{1}{2} M R^2$

wherein

About an axis perpendicular to the plane of the disc & passing through its centre.

We know that the moment of inertia of semicircular disc through the $centre I=\frac{M r^2}{2}$

Hence, the answer is the option (2).

Example 2: A circular disc $X$ of radius $\mathrm{R}$ is made from an iron plate of thickness $t$ and another disc $Y$ of radius $4 R$ is made from an iron plate of thickness $t / 4$. Then the relation between the moment of inertia. $I_X$ and $I_Y$ is :

1) $I_Y=32 I_X$
2) $I_Y=16 I_X$
3) $I_Y=I_X$
4) $I_Y=64 I_X$

Solution:

Mass of Disc
$
X=\left(\pi R^2 t \sigma\right) \Rightarrow I_X=\frac{M R^2}{2}=\frac{\left(\pi R^2 t \sigma\right) R^2}{2}=\frac{\pi R^4 \sigma t}{2}
$

Mass of Disc
$
\begin{aligned}
& Y=I_Y=\frac{M(4 R)^2}{2}=\frac{\pi(4 R)^2}{2} \frac{t}{4} \sigma 16 R^2=32 \pi R^4 t \sigma \\
& \frac{I_X}{I_Y}=\frac{\pi R^4 \sigma t}{2} \times \frac{1}{32 \pi R^4 \sigma t}=\frac{1}{64} \\
& I_Y=64 I_X
\end{aligned}
$

Hence, the answer is option(4).

Example 3: Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is :

1) $\frac{181}{2} M R^2$
2) $\frac{19}{2} M R^2$
3) $\frac{55}{2} M R^2$
4) $\frac{73}{2} M R^2$

Solution:

Moment of inertia for the disc

$I=\frac{1}{2} M R^2$

wherein

About an axis perpendicular to the plane of the disc & passing through its centre.

Parallel Axis Theorem -

$I_{b b^{\prime}}=I_{a a^{\prime}}+m R^2$

wherein

$b b^{\prime}$ is an axis parallel to $a a^{\prime} \& a a^{\prime}$ an axis passing through the centre of mass.
$
\begin{aligned}
\mathrm{I}_{\mathrm{p}} & =\mathrm{I}_0+(7 \mathrm{~m}) \cdot(3 \mathrm{R})^2 \\
& =\left[\frac{m R^2}{2}+6\left[\frac{m R^2}{2}+m \cdot(2 R)^2\right]+(7 m)(3 R)^2\right] \\
& =\frac{181}{2} m R^2
\end{aligned}
$

Example 4: From a uniform circular disc of radius R and mass 9 M, a small disc of radius $\frac{R}{3}$ is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through the centre of the disc is :

1) $\frac{37}{9} M R^2$
2) $4 M R^2$
3) $\frac{40}{9} M R^2$
4) $10 \mathrm{MR}^2$

Solution:

Moment of inertia for disc -

$I=\frac{1}{2} M R^2$

wherein

About an axis perpendicular to the plane of the disc & passing through its centre.

Perpendicular Axis theorem -

$I_z=I_x+I_y$

(for a body in XY plane )

- wherein

$I_z=$ moment of inertia about the $z$-axis
$I_x . I_y$ : the moment of inertia about the $\mathrm{x} \& \mathrm{y}$ axis in the plane of the body respectively.

Mass of removed part

let the mass density be $\sigma$

$9 m=\sigma \pi r^2$
mass of removed part $=\sigma \frac{\pi r^2}{3^2}=m$
$
\begin{aligned}
I & =\frac{9}{2} M R^2-\left[\frac{M\left(\frac{R}{3}\right)^2}{2}+M\left(\frac{2 R}{3}\right)^2\right] \\
& =M R^2\left[\frac{9}{2}-\frac{1}{18}-\frac{4}{9}\right] \\
I & =4 M R^2
\end{aligned}
$

Example 5: What is the moment of inertia of a disc having an inner radius $R_1$ and outer radius $R_2$ about the axis passing through the centre and perpendicular to the plane as shown in diagram?

1) $\frac{M}{2}\left(R_2^2-R_1^2\right)$
2) $\frac{M}{2} \pi\left(R_2^2-R_1^2\right)$
3) $M\left(R_2^2-R_1^2\right)$
4) $\frac{M}{2}\left(R_2^2+R_1^2\right)$

Solution:

Moment of inertia for continuous body -

$I=\int r^2 d m$

- wherein

r is the perpendicular distance of a particle of mass dm of a rigid body from the axis of rotation

taking a strip of radius x and thickness dx

$\begin{aligned}
& \text { MI of ring } \mathrm{dl}=\mathrm{dm} x^2 \\
& \qquad I=\int d I=\int_{R_1}^{R_2}\left[\frac{M}{\pi\left(R_2^2-R_1^2\right)} 2 \pi x \times d x\right] x^2 \\
& I=\int d I=\frac{M}{\pi\left(R_2^2-R_1^2\right)} 2 \pi\left[\frac{x^4}{4}\right]_{R_1}^{R_2} \Rightarrow I=\frac{M}{2}\left(R_1^2+R_2^2\right)
\end{aligned}$

Summary

The moment of inertia for a rigid body is a physical quantity that combines mass and shape in Newton's equations of motion, momentum, and kinetic energy. The moment of inertia is applied in both linear and angular moments, although it manifests itself in planar and spatial movement in rather different ways. One scalar quantity defines the moment of inertia in planar motion.