MOMENT OF INERTIA
. The velocity of a particle in a rotating object is given by
v = r w
. This equation is used to derive the kinetic energy of a rotating object:
K= 1/2 mv2 = 1/2 m(r
)2 = 1/2 mr2
2
. Since the total energy is the sum of the energies of all the particles and as w
is the same for all particles, we have:
KTotal = 1/2 (
mr2)
2
. Comparing the given expression with the kinetic energy in case of linear
motion we get mr 2 analogus to mass in th linear motion
. It turns out that the value of the summation is the rotational equivalent of mass, and is referred to as rotational inertia or moment of inertia ( I): KRotational = 1/2 I
2 . For every particle on the rotating body, there is a specific moment of inertia
for that object and for that specific axis of rotation. As the axis changes I
changes and as the shape changes I changes.
. Thus Moment of inertia is dependent on,
1. Shape of the rotating body(i.e distribution of mass)
2. Axis of rotation
. It is independent of
1. Velocity of rotation
2. Duration of rotation
Derivation
. Before we start the derivation, we should clarify how to find I
. Since mr2 is different in different parts of a solid object, we will be integrating
by splitting up an object into pieces of mass each with the same mass of d m .
. Now, each piece of mass is probably at a different distance (r) from the axis
of rotation.
. Thus, I is given by
. Assuming that the object we are dealing with has a uniform density of r,
we will be concerning ourselves with the infinitely small pieces of volume (dV)
that the infinitely small pieces of masses ( dm) take up:
I =
r2
dV =
r2 dV Since density is constant, it can remain outside the integral.
. Here is a table of the ten most common moments of inertia that are used
| Shape and Axis | | Rotational Inertia
|
| Hoop About Central Axis | | |
| Hoop About Any Diameter | | I = 1/2 MR2 |
| Annular Cylinder (Ring) About Central Axis | | | I = M(R12 +R22) |
| Solid Cylinder (Disk) About Central Axis | |
| I = 1/2 MR2 |
| Solid Cylinder (Disk) About Central Diameter | | | I = 1/4 MR2 +1/12 ML2 |
| Thin Rod About Axis Through Center Perpendicular to Length | |
| I = 1/12 ML2 |
| | Thin Rod About Axis Through One End Perpendicular to Length | |
| I = 1/3 ML2 |
| | Solid Sphere About Any Diameter | |
| I = 2/5 ML2 |
| Thin Spherical Shell About Any Diameter | |
| I = 2/3 ML2 |
| Slab About Perpendicular Axis Through Center | |
| I = M(a2 +b2) |
. The Parallel and perpendicular axis theorems help in
determining the moment of inertia about a given axis.
Parallel-Axis Theorem
. Given the moment of inertia ICM of a body about an axis through its center of mass ,
then the moment of inertia about a new axis parallel to the first but displaced a
distance h can be found through a relation called the parallel-axis theorem.
. The relation is stated as:
I = ICM + Md2 . The expression added to the center of mass moment of inertia will be recognized as the moment of inertia of a point mass - the moment of inertia about a parallel axis is the center of mass moment plus the moment of inertia of the entire object treated as a point mass at the center of mass. Perpendicular Axis Theorem . For a planar object, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia of two perpendicular axes through the same point in the plane of the object. . The utility of this theorem goes beyond that of calculating moments of strictly planar objects. . It is a valuable tool in the building up of the moments of inertia of three dimensional objects such as cylinders by breaking them up into planar disks and summing the moments of inertia of the composite disks.  | |