Calculation of Moment of Inertia in different bodies

Table Of Contents

1] Moment of Inertia of the thin rod about an axis perpendicular to its plane and passing through a center.
2] Moment of Inertia of the thin rod about an axis perpendicular to its plane and passing through one end.
C] Moment of Inertia of a circular ring about a perpendicular axis passing through a center
D] The moment of inertia of a circular disk about an axis perpendicular to its plane and passing through a center.

1] Moment of Inertia of the thin rod about an axis perpendicular to its plane and passing through a center.

moment of inertia of thin rod about its center — fig: M.I of the thin rod about its center

Let us consider a thin uniform rod of mass ‘M’ and length ‘L’ rotating about a perpendicular axis and passing through a center. Consider a small element of thickness ‘dx’ of mass ‘dm’ at a distance ‘x’ from the axis of rotation.

Then, Mass of rod per unit length = $\frac{M}{L}$

mass of small element of thickness ‘ $dx$ ‘ is: $dm =\frac{M}{L}dx$

As we know the moment of inertia of a particle can be calculated by the product of mass and perpendicular distance from the axis of rotation. So,

Moment of inertia of small element is: $dI = dm x^2 \quad \longrightarrow \quad dI = \frac{M}{L} x^2 dx$

Since the total moment of inertia of a rod can be calculated by adding M.I of all elements, we can do so by integrating $dI$ with a limit from one end to another end.

The moment of inertia of a thin rod about the given axis is:

$I = \int_{-\frac{L}{2}}^{\frac{L}{2}} dI$

$I = \int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{M}{L} x^2 dx$

$I = \frac{M}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} x^2 dx$

$I = \frac{M}{L} \left [ \frac{x^3}{3} \right]_{-\frac{L}{2}}^{\frac{L}{2}}$

$I = \frac{M}{3L} \left [ \left( \frac{L}{2} \right)^3 - \left( \frac{-L}{2} \right)^3 \right]$

$I = \frac{M}{3L} \left [ \frac{L^3}{8} + \frac{L^3}{8} \right]$

$I = \frac{M}{3L} \times \frac{L^3}{4}$

$\boxed{I = \frac{1}{12}ML^2}$

If ‘K’ is the radius of gyration, then we can write:

$MK^2 = \frac{1}{12}ML^2}$

$\boxed{K = \frac{L}{\sqrt{12}}}}$

2] Moment of Inertia of the thin rod about an axis perpendicular to its plane and passing through one end.

moment of inertia of thin rod about one end — fig: M.I of the thin rod about its one end.

Let us consider a thin uniform rod of mass ‘M’ and length ‘L’ rotating about a perpendicular axis and passing through one end as shown in the figure above. Consider a small element of thickness ‘dx’ of mass ‘dm’ at a distance ‘x’ from the axis of rotation.

Then, Mass of rod per unit length = $\frac{M}{L}$

mass of small element of thickness ‘ $dx$ ‘ is: $dm =\frac{M}{L}dx$

As we know the moment of inertia of a particle can be calculated by the product of mass and perpendicular distance from the axis of rotation. So,

Moment of inertia of small element is: $dI = dm x^2 \quad \longrightarrow \quad dI = \frac{M}{L} x^2 dx$

Since the total moment of inertia of a rod can be calculated by adding M.I of all elements, we can do so by integrating $dI$ with a limit from one end to another end.

The moment of inertia of a thin rod about the given axis is:

$I = \int_0^L dI$

$I = \int_0^L \frac{M}{L} x^2 dx$

$I = \frac{M}{L} \int_0^L x^2 dx$

$I = \frac{M}{L} \left [ \frac{x^3}{3} \right]_0^L$

$I = \frac{M}{3L} \left [ L^3 - 0 \right]$

$I = \frac{M}{3L} \times L^3$

$\boxed{I = \frac{1}{3}ML^2}$

If ‘K’ is the radius of gyration then,

$MK^2 = \frac{1}{3}ML^2}$

$\boxed{K = \frac{L}{\sqrt{3}}}$

Alternative way

we can use theorem of parallel axis on the moment of inertia to find the moment of inertia of a thin rod about a perpendicular axis passing through at one end. According to the theorem of parallel axis:

$I = I_{CM} + MD^2$

where $I_{CM}$ = moment of inertia about an axis through a center of mass (i.e. center of rod) and ‘D’ is the perpendicular distance between the given axis and parallel axis.

Since we know the moment of inertia of a thin rod about perpendicular distance passing through the center is $\frac{1}{12}ML^2$ So, $I_{CM} = \frac{1}{12}ML^2$ and the distance between two parallel axes becomes $\frac{L}{2}$ .

Hence,

$I = I_{CM} + MD^2$

$I = \frac{1}{12}ML^2 + M\left( \frac{L}{2} \right)^2$

$I = \frac{1}{12}ML^2 + \frac{1}{4}ML^2$

$\boxed{I = \frac{1}{3}ML^2}$

C] Moment of Inertia of a circular ring about a perpendicular axis passing through a center

Consider a thin circular ring of mass ‘M’ and radius ‘R’ rotating around a perpendicular axis and passing through a center as shown in the figure. As shown in the figure, let’s consider a small element of mass ‘dm’.

Moment of inertia of small element $dI = dm R^2$

Then, the moment of inertia of the whole circular ring around that axis can be calculated by integrating $dI$ over the limit.

$I = \int_0^{2 \pi R} dI$

$I = \int_0^{2 \pi R} dm R^2$

$I = R^2 \int_0^{2 \pi R} dm$

$I = R^2M$

$\boxed{I =M R^2}$

If ‘K’ is the radius of gyration. then,

$MK^2 = MR^2$

$\boxed{K = R}$

The radius of gyration in a circular ring is equal to the radius of a circular ring.

D] The moment of inertia of a circular disk about an axis perpendicular to its plane and passing through a center.

moment of inertia of circular disk about a perpendicular disk about its center — fig: MI of the circular disk about the center

Consider a thin circular disk of Mass ‘M’ and radius ‘R’ about a perpendicular axis passing through a center. We can consider a circular disk made up of many concentric circular rings. Consider one circular ring of thickness ‘dx’ at a distance ‘x’ from the axis of rotation.

Mass per unit area of a circular disk is $\frac{M}{\pi R^2}$

The area of considered ring is

$\begin{align*} da &= \pi \left( x+dx \right)^2 - \pi x^2 \\ da &= \pi(x^2 + 2xdx + dx^2) - \pi x^2 \\ da &= \pi x^2 +2\pi xdx + \pi dx^2 - \pi x^2 \\ da &= 2\pi xdx + \pi dx^2 \end{align*}$