Topic 2.2 Extended
G1 – Calculating rotational inertia
If a body is made of discrete masses use
I = Σmiri2
rotational inertia
(discrete masses)
If a body is made of a continuous distribution of
masses, we have
I =
∫r2dm
rotational inertia
(continuous masses)
This equation is derived in a way analogous to that
for the cm.
Note: The important thing to remember is that you
choose mass elements dm that are located a distance
r from the axis of rotation.
Topic 2.2 Extended
G1 – Calculating rotational inertia
Why is Ihoop > Idisk
for same M and R?
Furthermore, the
rotational inertia
depends on the
location of the axis
of rotation.
R
Hoop
I = MR2
R
Disk/Cylinder
I = 12 MR2
R
axis
mass distributions
have different
rotational inertias.
axis
Different continuous
R
Hoop about diameter
Disk/Cylinder
1
I = 12 MR2
I = 14 MR2 + 12 ML2
Topic 2.2 Extended
G1 – Calculating rotational inertia
or hollow.
Which is greater Isphere or Ishell?
Why is Isphere > Ishell for
same M and R?
R
R
Solid sphere
I = 25 MR2
So can disks.
If R1 = R2 what does
the ring become?
If R1 = 0 what does
the ring become?
axis
axis
Spheres can be solid
R2
R1
Annular cylinder/Ring
1
I = 2 M(R12+R22)
Spherical shell
I = 23 MR2
Topic 2.2 Extended
G1 – Calculating rotational inertia
axis
rotated about a variety
of perpendicular axes.
Here are two:
axis
A thin rod can be
L
L
Thin rod about center, ┴
1
I = 12
ML2
Thin rod about end, ┴
I = 13 ML2
Why is Irod,cent < Irod,end?
Topic 2.2 Extended
G1 – Calculating rotational inertia
Finally, we can look at
axis
a slab rotated through
a perpendicular axis:
Why doesn’t the
thickness of the
slab matter?
b
a
Slab about center, ┴
1
I = 12 M(a2+ b2)
Topic 2.2 Extended
G1 – Calculating rotational inertia
axis
What is the rotational
kinetic energy of a 2-m
by 4 m, 30-kg slab
rotating at 20 rad/s?
M = 30, a = 2 and b = 4
so that
1
I = 12 M(a2+ b2)
=
1
2
12 ·30(2
a
+ 42)
= 50 kg·m2
K =
1
2
Iω2
=
1
2
·50·202
= 10000 J
b
Slab about center, ┴
1
I = 12 M(a2+b2)
Topic 2.2 Extended
G1 – Calculating rotational inertia
So how are all of the rotational inertias calculated?
We won’t derive all of them, but we will derive I for
the rod (because it is 1D, and therefore easiest).
axis
axis
L
Thin rod about center, ┴
1
I = 12
ML2
L
Thin rod about end, ┴
I = 13 ML2
Topic 2.2 Extended
G1 – Calculating rotational inertia
Consider a thin rod of mass M and length L with a
rotational axis perpendicular to its center.
We superimpose a
y
Cartesian cs over
the rod, centering
dm
it on the axis of
-L
rotation.
ℓ
2
dm is shown.
Since this is a 1D problem, dm = λdℓ.
I =
I =
∫ℓ
2
∫ℓ
λdℓ = λ
2
dℓ = λ·
-L/2
-L/2
Since λ = M/L,
Irod,cent =
x
rotational inertia
(continuous masses)
∫r2dm
L/2
L/2
L
2
ML3
12L
1 3
ℓ
3
=
L/2
=
-L/2
1
2
12ML
λ L3
3
8
-
-L3
8
=
λL3
12
Topic 2.2 Extended
G1 – Calculating rotational inertia
Now consider the same thin rod of mass M and length L
with a rotational axis perpendicular to its end.
We superimpose a
y
Cartesian cs over
the rod, centering
dm
it on the axis of
rotation.
L
ℓ
0
Since this is a 1D problem, dm = λdℓ.
I =
∫r2dm
L
I =
x
∫
ℓ2
∫
L
ℓ2
λdℓ = λ
dℓ = λ·
0
0
Since λ = M/L,
Irod,end =
ML3
3L
=
1
3
ML2
1 3
ℓ
3
L
=
0
λ L3 - 03
3
=
λL3
3
Topic 2.2 Extended
G1 – Calculating rotational inertia
Note that the axes of rotation all pass through either
axis
the center or the end.
What if we have a rotating object like a
sphere on the end of a rod, as shown.
The rotational inertia of the rod is
Irod,end, but the sphere is not rotating
about its center (it is located the
length of the rod plus its radius from
the axis).
This section tells how to find
the rotational inertia through
any axis parallel to the one
passing through the cm of the
object.
Topic 2.2 Extended
G1 – Calculating rotational inertia
THE PARALLEL AXIS THEOREM
Suppose we know Icm, the rotational inertia of any
mass M through it’s center of mass.
Then the rotational inertia of that mass through
any parallel axis is given by
I = Icm + Mh2
parallel axis theorem
where h is the distance between the new axis and
the axis passing through the cm.
Topic 2.2 Extended
G1 – Calculating rotational inertia
THE PARALLEL AXIS THEOREM
Suppose a 200-kg solid sphere of radius 0.1-m is
placed on the end of a 12-kg thin rod of length 8 m.
12 kg
200 kg
8m
.1 m
Since the rod already has a formula for Irod,end,
we’ll find its rotational inertia first:
1
Irod,end = 13 ML2 = 3 ·12·82 = 256 kg·m2
Since the sphere is not rotating about its cm, we
must use the parallel axis theorem, with h = 8.1 m:
I = Icm + Mh2 =
Then
2
MR2
5
+ Mh2 =
=
2
·200·0.12
5
+ 200·8.12
13122.8 kg·m2
Itot = 256 + 13122.8 = 13378.8 kg·m2
Topic 2.2 Extended
G1 – Calculating rotational inertia
PROOF OF THE PARALLEL AXIS THEOREM
Consider the irregularly-shaped object of mass M
shown below:
If we choose an axis passing through the cm, we can
find Icm:
If we choose a parallel
axis passing through some
other point located a
I
distance h from the cm,
Icm
we can find I with
respect to the new point.
The question is, how
are I, Icm and h
related.
cm
h
Topic 2.2 Extended
G1 – Calculating rotational inertia
PROOF OF THE PARALLEL AXIS THEOREM
We’ll center our coordinates on the cm:
h is the distance between the old axis and the new.
O is the old axis and P is the new parallel axis.
Let dm be an arbitrary mass
having coordinates (x,y).
Let P have coordinates (a,b).
Let r be the distance from P
to dm.
Then r forms the
cm
hypotenuse of the
triangle shown:
O
The legs of that
triangle are shown:
y
dm
r
P
h
(x, y)
y-b
x-a
b
a
x
Topic 2.2 Extended
G1 – Calculating rotational inertia
PROOF OF THE PARALLEL AXIS THEOREM
The distance from the
y
dm
original origin O to dm is
x2 + y2 so that
Icm =
∫(x2
+ y2)dm
To find I with
respect to the new
axis P we have
I = ∫r2dm
= ∫[(x-a)2 + (y-b)2]dm
=
=
=
=
r
P
h
cm
O
(x, y)
y-b
x-a
b
a
x
∫[(x2 - 2xa + a2) + (y2 - 2yb + b2)]dm
∫[(x2 + y2) + (a2 + b2) – 2xa - 2yb]dm
∫[(x2 + y2)dm + ∫h2dm – 2a∫xdm – 2b∫ydm
0
0
Icm + h2∫dm – 2axcm – 2bycm
How
is 0?
How do
do you
you know
know that
that xycm
cm is 0?
I = Icm + Mh2