3/27/2012
MOMENT OF INERTIA
In rotational motion, the moment of inertia is the equivalent to the
mass in translational motion. It measures the resistance of a body
to an Angular Acceleration.
The equivalent equation to F = ma for a rotating
body is M = Iα, where I is the moment of inertia.
V
Second moment about the z-axis
For constant ρ , we have
I = ρ ∫ r 2 dV
y′
r
yG
xG =
r′
G
d
xG
O
x′
The units of I are kg ⋅ m 2 or slug ⋅ ft 2
∫ x dm
m
∫ dm
m
yG =
∫ y dm
m
∫ dm
2
Radius of Gyration
Sometimes the moment of inertia about an axis is given as
the Radius of Gyration, k in the form
I = mk2
m
x
I
m
k=
or
Composite bodies
I O = ∫ r dm = ∫ ⎡⎣( xG + x′) 2 + ( yG + y′) 2 ⎦⎤ dm
G1 • m1
m
I O = ∫ ( xG2 + 2 xG x′ + x′2 + yG2 + 2 yG y′ + y′2 ) dm
G2
•
m
I O = ( xG2 + yG2 ) ∫ dm + 2 xG ∫ x′dm + 2 yG ∫ y′dm + ∫ ( x′2 + y′2 ) dm
m
m
m
0
0 m
IO = d 2 m + IG
x
THE PARALLEL AXES THEOREM
2
m
•
G
1
V
y
dm
r
If I G is known, the moment of inertia
I O about any other parallel axis O is
given by
IO = IG + m d 2
Where m is the total mass of the body
and d is the distance between the
parallel axes.
I = ∫ r 2 dm = ∫ r 2 ρ dV
m
y
If the axis goes through the center of mass G
and is perpendicular to the plane of motion,
the moment of inertia is denoted by I G
IG
3
m2d 2
G3 •
3
EXAMPLES
SOLUTION
In this type of problem we create what is
called a shell element of volume dV as
shown in the figure below
Hence
r2
1
r1
m3
(
I G = ∑ I Gi + mi di2
i =1
1. Determine the moment of inertia of the cylinder shown
about the z - axis. The density ρ of the material is constant
d1
•G
d3
)
R
0
m
G
I G = I G1 − I G2
dI z = r 2 dm = 2πρ hr 3 dr
I z = ∫ r 2 dm = 2πρ h ∫ r 3 dr =
•
2
πρ h
2
4
and integrating
R4
The mass of the cylinder is m = ∫ dm = 2πρ h ∫ r dr = πρ hR 2
R
0
m
The volume of the element is
dV = (2π r )( h)( dr )
and the mass of the element is
dm = ρ dV = ρ (2π rh) dr
So in terms of the mass
5
1
Iz = m R 2
2
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2. The sphere is formed by revolving the shaded area around
the x - axis. Determine the moment of inertia I x and express
the result in terms of the total mass m of the sphere. The
maerial has a constant density ρ .
Using the fact that we can look at the ring as a thin cylinder:
1
dI x = y 2 dm
y
2
dx
SOLUTION
dI x =
Define a differential disk with volume
y
dV = π y 2 dx
dx
Now y = r − x
2
iy
x
z
2
Ix =
2
So dV = π (r 2 − x 2 ) dx
So the mass m is
r
m = 2 ρπ ∫ (r 2 − x 2 ) dx = 2 ρπ (r 3 −
0
r3
)
3
2
4
m = ρ π r3
3
1) OUTER RING:
I G(1) =
I G(1) =
x
z
8
ρ π r5
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8
ro = 2 ft , ri = 1 ft , d I = 0.25 ft
1
1
1
1
= mo ro2 − mi ri 2 = ( ρπ ro2 d I ) ro2 − ( ρπ ri 2 d I ) ri 2
2
2
2
2
2) INNER RING:
I G(2)
I G(2) =
SOLUTION
iy
2
Ix = mr2
5
7
3. Determine the moment of inertia of the assembly about
an axis which is perpendicular to the plane and passes
through the point O. The material has a specific weight
of γ = 90 lb / ft 3
RO = 2.5 fft , RI = 2.0 ft
f , dO = 1 ft
f
1
1
(1)
2
2
I G = mO RO − mI RI
2
2
r
2
1
ρπ ( r 2 − x 2 ) dx
ρ
2 ∫0
Ix =
and dm = ρπ (r − x )dx
2
1 2
( r − x 2 ) × ρπ ( r 2 − x 2 ) dx
2
1
1
ρπ d I ( ro4 − ri 4 ) = × 2.795 × 3.14159 × 0.25 × (24 − 14 )
2
2
I G(2) = 16.464 slug / ft 2
I G = I G(1) + I G(2) = 117.717
ρ = 2.795 slug / ft 3
1
( ρπ RO2 dO ) RO2 − 12 ( ρπ RI2d I ) RI2 = 12 ρπ dO ( RO4 − RI4 )
2
1
× 2.795 × 3.14159 × 1 × (2.54 − 24 )
2
I G(1) = 101.253 slug / 9ft 2
KINETIC EQUATIONS OF MOTION
The geometry and loading are
assumed to be such that we can
assume that the motion is planar.
This requires that the geometry and
loading be symmetrical with respect
to a fixed reference plane, p. 409.
The reference
Th
f
frame
f
x - y - z mus be
b
INERTIAL with the origin at some
arbitrary point P not necessarily in the body.
The axes do not rotate, but can translate with a constant velocity
Translational Motion:
In this case the previous analysis for systems of particles applies,
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and we have:
m = mO + mI = ρπ dO ( RO2 − RI2 ) + ρπ d I (ro2 − ri 2 )
m = 2.795 × 3.14159 × ⎡⎣1× (2.52 − 22 ) + 0.25 × (22 − 12 ) ⎤⎦
I O = I G + m d 2 , m = 26.34 slug , d = 2.5 ft
I O = 117.717 + 26.34 × 2.52 ⇒
I O = 282 slug / ft 2
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The sum of all the external forces acting on the body is equal to
the mass of the body times the acceleration of its center of mass.
∑F = ma
G
In the case of translational motion in the x - y plane we have
∑F
∑F
x
= m (aG ) x
y
= m (aG ) y
Rotational Motion:
Let the z-axis is at point P and F is the
resultant ( ∑ Fi ) of all forces acting
on the body (ignoring internal forces).
Then consider an element of mass dm
at position r in the body:
dm
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d (r × F) = r × dm a
d M P = r × dm a
If α is the angular acceleration
and ω the angular velocity of the
body, using a = a P + α × r − ω 2r
dM P = dm r × ( a P + α × r − ω 2r )
And integrating over the whole body we get
M P ≡ ∑ M P = aPy ∫ x dm − aPx ∫ y dm + α ∫ r 2 dm
dm
m
Because x =
Thus
= dm ⎡⎣r × a P + r × (α × r ) − ω 2 (r × r ) ⎤⎦
In Cartesian coordinates:
dM P k = dm {( x i + y j) × (aPx i + aPy j) + ( x i + y j) × [α k × ( x i + y j)]}
= dm( − yaPx + xaPy + α x 2 + α y 2 )k
dM P = dm(− yaPx + xaPy + α r 2 )
∑M
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Using the parallel axis theorem I P = I G + m( x + y ) we get
2
∑M
P
2
= y m(− aPx + y 2α ) + x m(aPy + x 2α ) + α I G (∗)
aGx i + aGyy j = aPx i + aPyy j + α k × ( x i + y j) − ω 2 ( x i + y j)
and we get aGx = aPx − yα − xω 2
aGy = aPy + xα − yω 2
so that − aPx + yα = −aGx − xα
Substituting these last expressions into (‫ )٭‬we have
In scalar form,there are three independent equations of
planar motion:
y
= m aGy
G
= α I G or
)
= − y m aPx + x m aPy + I Pα
= I Gα
G
P
= − y m aGx + x m aGy + α I G
or
∑ M P = ∑ (Mk
)
P
)
P
are called the
Kinetic Moments.
The sum of the moments of the external forces about a
point P are equivalent to the sum of the Kinetic Moments
of the components of maG about P plus the Kinetic
Moment of I G α.
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We must remember that maG and I G α are not force or couple
moments. These are the Effects of forces and couples acting
on the body
= m aGx
m
The components maGx i and maGy are treated as Sliding Vectors
acting anywhere along their line of action.
α I G k is treated as a Free Vector and can act at any point.
aPy + xα = aGy + yω 2
x
(∫
IP
y dm m
The sum of the moments about the center of mass G of the body
due to all external forces is equal to the product of the moment of
inertia of the body about the center of mass G times the angular
acceleration α of the body.
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Where ( Mk
Now aG = a P + α × r − ω 2 r so
∑F
∑F
∑M
P
)
x dm m and y =
∑M
∑M
= − y m aPx + x m aPy + α I P
P
m
m
my
If P = G , the equation reduces to
= dm [r × a P + r × (α × r )]
Lets go back to
∑M
(∫
m
mx
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Equations of Motion in Pure Translation
Under translation only all points in a body have the same
acceleration a = aG and the angular acceleration α = 0.
Rectilinear Translation:
In this case all points in the body travel along parallel staright
line paths. The equations of motion become:
∑F = ma
∑F = ma
∑M = 0
x
∑ M =∑ ( M )
P
k
Gx
y
P
This underscores the need to always draw the free body
diagram to account for ∑ Fx , ∑ Fy , ∑ M G or ∑ M P .
We should also always draw the kinetic diagram to
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accont for the terms m aGx , m aGy and α I G .
Gy
G
If the sum of the moments is taken about another point
different from G , then the moment maG must be taken
into account
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3/27/2012
For example, if a point A as shown in the figure is chosen
∑M
A
= ∑ ( Mk
)
A
EXAMPLES
= ( m aG ) ⋅ d
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