Transcript Rotational Motion and Equilibrium

-Calculation of Moments of
Inertia for Rigid Objects of
Different Geometries
-Parallel Axis Theorem
AP Physics C
Mrs. Coyle
Remember: Angular and Linear
Quantities
• Displacements
• Speeds
• Accelerations
s r
v  r
a  r
Remember: 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
1
K R   Ki   mi ri 
i
i 2
2
2
1
1 2
2 2
K R    mi ri   I
2 i
2

• I is called the moment of inertia
Remember: 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:
I   ri mi
2
i
Moment of Inertia of a Rigid Object
I
lim
mi 0
 r m   r dm
2
2
i
i
i
or in terms of density:
I    r dV
2
• 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 constant
I   r dm  R
2
I  MR
2
2
dm

If the Object has Uniform Density:
we will use the constant density
expression to evaluate the integral.
Density Expressions
• Volumetric Mass Density –> mass per unit
volume: 
=m/V
• Linear Mass Density –> mass per unit
length of a rod of uniform cross-sectional
area: l = m / L = A
• Face Mass Density –> mass per unit
thickness of a sheet of uniform thickness, t :
s  t
Ex: Moment of Inertia of a Uniform
Rigid Rod for Axis passing through
center of mass
l = m / L = A
dm = l dx
M
I   r dm   x
dx
L/ 2
L
1
2
I  ML
12
2
L/2
2
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
=m/V
I   r dm 
2
1
2
I z  MR
2
r
2

Lr
dr



2
Parallel-Axis Theorem
Is 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
I CM
1
2
 ML
12
I  I CM  MD
2
2
1
 L 1
2
2
I  ML  M    ML
12
3
2
• 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 # 27
The 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