Mass Moment of Inertia: Definition, Calculation, and Importance for Rigid Body Rotation, Study notes of Physics

The concept of Mass Moment of Inertia (IG), its definition, units, and significance in rigid body rotation problems. It covers the F=ma analysis moment equation, rotational kinetic energy, angular momentum, and the parallel axis theorem for calculating IG about axes other than the mass center. Common shapes like disks, spheres, rods, and rings have their specific IG values provided.

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IGis the “mass moment of inertia” for a body about an
axis passing through the body’s mass center, G.
IGis defined as: IG= r2dm Units: kg-m2or slug-ft2
IGis used for several kinds of rigid body rotation
problems, including:
(a) F=ma analysis moment equation ( ΣMG= IGα).
(b) Rotational kinetic energy ( T = ½ IGω2)
(c) Angular momentum ( HG= IGω)
IGis the resistance of the body to angular acceleration.
That is, for a given net moment or torque on a body, the
larger a body’s IG, the lower will be its angular
acceleration, α.
IGalso affects a body’s angular momentum, and how a
body stores kinetic energy in rotation.
Mass Moment of Inertia, IG
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I^ G

is the “mass moment of inertia” for a body about an axis passing through the body’s mass center, G.I^ G

is defined as:

I^ G

∫^

(^2) r dm

Units:

kg-m

2

or

slug-ft

2

I^ G

is used for several kinds of rigid body rotation problems, including:

(a)

F=ma analysis moment equation (

M

G^

= I

αG

(b)

Rotational kinetic energy ( T = ½ I

ωG

2

(c)

Angular momentum ( H

G^

= I

ωG

I^ G

is the resistance of the body to angular acceleration. That is, for a given net moment or torque on a body, thelarger a body’s I

, the lower will be its angularG

acceleration,

I^ G

also affects a body’s angular momentum, and how a

Mass Moment of Inertia, body stores kinetic energy in rotation.

I

G

Mass

Moment

of Inertia,

I

G^

(cont’d)

I^ G

for a body depends on the body’s mass and the location of the mass.The greater the distance the mass is from the axis ofrotation, the larger I

G^

will be.

For example, flywheels have a heavy outer flange thatlocates as much mass as possible at a greater distancefrom the hub.If I is needed about an axis other than G, it may becalculated from the “parallel axis theorem.”

I G

’s for Common Shapes

Thin^ I

G^

2

=^

mR 1 2

y

x

R

Disk: G

y

x

R

G

IG

2

=^

mR 2 5

Sphere:

x

y G

P

L

L 2 Slender IP

2

=^

mL 1 3

IG

2

=^

mL (^112)

About P(end of rod)

About G(center of rod)

Rod:

IG

2

2

=^

m(a

  • b )

(^112)

x

y

G

a

b

Rectangular

Plate:

y

G

R

ThinRing:

IG

2

= mR

(All mass is atthe same radius, R)

x

Radius of Gyration, k

, for Complex ShapesG^

Some problems with a fairly complex shape, such as adrum or multi-flanged pulley, will give the body’s mass mand a radius of gyration, k

, that you use to calculate IG

.G

If given these, calculate I

G^

from:

I^ G

= mk

(^2) G

As illustrated below, using k

G^

in this way is effectively

modeling the complex shape as a thin ring.

G^

R

G

k^ G IG

= mk

(^2) G

Radius of Gyration, k

G

Some problems involving a complex shape with mass,m, and an outer radius, R, will give a “radius of gyration”,k^

, that can be used to determine I

for that shape.

The

equation, I

= mk , indicates that the complex shape

G is being modeled dynamically by a thin ring with mass, m,and a radius, k.

G

G^

G G

2

ComplexShape:

Thin Ring,k^

Model:G