Moment Of Inertia Of A Disc

Moment of Inertia of A Disc

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

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.

Frequently Asked Questions (FAQs):

Q 1: What is the formula for the moment of inertia of a disc?

Ans: The formula for the moment of inertia of a rod when the axis is through the centre is $I =\frac{1}{2} MR^2.$

Q 2: Is there any difference between the moment of inertia and rotational inertia?

Ans: No

Q 3: Is the moment of inertia a scalar or a vector quantity?

Ans: Scalar quantity

Q 4: Does the moment of inertia change with the change of the axis of rotation?

Ans: Yes