Calculation of Moments of Inertia for Rigid Objects of

DifferentGeometries

Parallel Axis Theorem

Relations Angular and Linear Quantities

• Displacements

• Speeds

• Accelerations

s r

a r

v r

Rotational Kinetic Energy and Moment of Inertia

• The total rotational kinetic energy of the rigid object is the sum of the energies of all its particles

• I is called the moment of inertia

2 2

2 2 2

1

2

1 1

2 2

R i i ii i

R i ii

K K m r

K m r I

Moment of Inertia

• Moment of Inertia, I, is a measure of the resistance of an object to changes in its rotational motion.

• Moment of Inertia is analogous to mass in translational motion.

• When the object is made up of point masses you calculate moment of inertia using:

2i i

i

I r m

Moment of Inertia of a Rigid Object

lim 2 20im i ii

I r m r dm

2I r dVor in terms of density:

• When objects of different geometric shapes are involved, (ie. cylinders, hoops, rods, spheres) we will use geometric relationships to evaluate the integral.

• Note that for the following examples that the axis of rotation will coincide with the axis of symmetry of the objects.

Ex: Moment of Inertia of a Uniform Thin Hoop – for Axis perpendicular to the plane at the Center of the Hoop

• Assume r is constant22

2

I r dm R dm

I MR

:

we will use the constant density expression to evaluate the integral.

If the Object has Uniform Density than Density Expressions:

• Linear Mass Density –> mass per unit length: l = m / L atau m = l L

• Face Mass Density –> mass per unit area = sm/A atau m = s A

• Volumetric Mass Density –> mass per unit volume: r = m / V atau m = r V

Ex: Moment of Inertia of a Uniform Rigid Rod for Axis passing through

center of mass

l = m / L dm = l dx

/ 22 2

/ 2

21

12

L

L

MI r dm x dx

L

I ML

Ex: Moment of Inertia of a Uniform Solid Cylinder for axis along z axis

• For concentric shells with radius r, thickness dr and length L

r = m / V

2 2

2

2

1

2z

I r dm r Lr dr

I MR

Moment of inertia of objects

Parallel-Axis TheoremIs used to find I ,

if the axis of rotation does not coincide with the axis of symmetry but is parallel to the axis through the center

of mass of the object.

Parallel Axis Theorem

I = ICM + MD 2

-ICM is the moment of inertia about the axis through the center of mass

-D is the distance from the center of mass axis to the arbitrary axis

Moment of Inertia for a Rod Rotating Around One End

D= ½ L

21

12CMI ML

2CM

22 21 1

12 2 3

I I MD

LI ML M ML

• For a uniform density object made up of various shapes the total moment of inertia is the sum of the moments of inertia for the individual objects.

Example #25 • A uniform thin solid door has height 2.20m,

width 0.870m and mass 23.0kg. Find its moment of inertia fro rotation on one hinge across its bottom. Is any piece of data unnecessary? (Hint: Assume that all the mass is concentrated across the bottom of the door along the same level as the hinge)

• Ans: 5.80 kg m2 , h is unnecessary data.

Example # 27The density of the Earth at any distance r from its center, is approximately:

ρ= [14.2 – 11.6 (r/R)] x 10 3 km/m3 where R is the radius of the Earth.

Show that this density leads to a moment of inertia I= 0.330 MR2 about an axis through the center, where M is the mass of the Earth.

Hint: start with thin spherical shell equation: dI= (2/3) r2 dm

Example #23

Example #23

• Three identical thin rods, each of length L and mass m, are welded perpendicular to one another as shown. The assembly is rotated about an axis that passes through the end of one rod and is parallel to another. Determine the moment of inertia of this structure.

• Ans: (11/12)mL2