Mechanics Sub-Menu Mechanics Main Menu Problem #0002 Chapters Chapters
1.Rotational Inertia of Geometrical Bodies             Click For PDF Version

(a)

Annular cylinder about its central axis

Let be the outer radius of the annular cylinder and be its inner radius, and l be its length. Let be its density.
*Move the Mouse over the image for an Animated View

Animated

We will calculate expression for the rotational inertia by integrating with variable r, the radial distance measured from the axis.

Mass of annular cylinder is given by the integral

Its rotational inertia is given by the integral

(b)

Solid cylinder (or ring) about central axis

*Move the Mouse over the image for an Animated View
Animated

Let the radius of the cylinder be R and its mass M. We can obtain its rotational inertia I from the formula for the rotational inertia of an annular cylinder by substituting

We have

 

(c)

Solid disk of width
*Move the Mouse over the image for an Animated View

Animated

Let R be the radius, thickness and be the density of the disk. For calculating the rotational inertia about the axis as shown in the figure we choose angular variable measured from the vertical direction, and consider an infinitesimal box of length dx, height dy and width .

The moment of inertia can be found by integrating

As

Therefore,

But the mass of the disk is

Thus the moment of inertia of a thin disk of mass M is

We will use the parallel axis theorem for finding the rotational inertia of a thin disk about an axis parallel to the vertical axis passing through its centre.

This gives

Using the expression of rotational inertia of a thin disk, we have

(d)

Cylinder about axis through its CM

We will use this result for calculating the rotational inertia of a solid cylinder of length L, radius R, and mass M about a vertical axis passing through its centre of mass.

*Move the Mouse over the image for an Animated View
Animated

Mass of the cylinder M is

We thus find that the rotational inertia of a cylinder about axis as shown in the figure is

(e)

Thin rod about an axis through its centre

*Move the Mouse over the image for an Animated View
Animated

Rotational inertia of a thin rod of length L and mass M about an axis passing through its centre can be obtained from the above result by putting in it R =0. We get

(f)

Thin rod about axis at one of its ends

*Move the Mouse over the image for an Animated View
Animated

By applying parallel axis theorem and using the expression of rotational inertia of a thin rod about axis through its CM, we get

(g)

Thin spherical shell about any diameter

Let radius of the shell be r, its thickness and be its density. Using spherical polar coordinates and measuring distance from the polar axis, we have

Mass of a spherical shell of radius r, thickness and density is

Using this expression for M, the rotational inertia of a thin spherical shell of radius r can be expressed as

(h)

Rotational inertia of a solid sphere

With this result we obtain next the rotational inertia of a solid sphere of radius R about any diameter.

*Move the Mouse over the image for an Animated View
Animated

Integrating the thin spherical shell expression with respect to r from 0 to R, we get

Mass of a homogeneous sphere of radius R and density is

We thus find

(i)

Rotational Inertia of Thin Slab

*Move the Mouse over the image for an Animated View
Animated

Let the length of the slab be a, its width be b, its thickness be and its density be . Its rotational inertia about an axis perpendicular to its plane and passing through its centre of mass can be calculated by integrating the following expression;

Mass of the slab is

We thus find