FYI: I is the rotational analog for m. In translation, the mass distribution
is not of importance. But in rotation, it is. Hence I = mr2.
Topic 2.2 Extended
We call the G
quantity
I the ROTATIONALdynamics
INERTIA of the mass.
– Rotational
FYI:
Sometimes it is called the MOMENT OF INERTIA.
MOMENT OF INERTIA
F
Consider a particle of
Ft F
mass m constrained to move
Fr
in a circle of radius r by
a force F.
r
How does the particle’s
acceleration depend on F?
Only Ft will cause the
mass to accelerate. Thus Ft = ma.
But since the mass is constrained to move in a
circle, the acceleration is tangential so that
Ft = mat
Recalling that at = rα we have
Ft = mrα
Finally, since τ = rFt we have rFt = mr2α or
τ = mr2α which we can write as
τ = Iα
Newton’s 2nd for rotation by single
(I = mr2)
torque
Topic 2.2 Extended
G – Rotational dynamics
MOMENT OF INERTIA
The moment of inertia for a group of point masses
about a given axis is
I = mr2
Rotational Inertia
Point Masses
Calculate the rotational inertia of the masses about
the axis shown.
4 kg
I = Σmiri2
2.0 m
= 1·12 + 2·2.52 + 4·32
= 49.5 kg·m2
2 kg
If a torque of 15 nm was
1.0 m
1 kg
applied to the axis, what
would the rotational
acceleration be?
2.5 m
τ = Iα
15 = 49.5α
 α = 0.303 rad/s2
Topic 2.2 Extended
G – Rotational dynamics
MOMENT OF 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
G – Rotational dynamics
MOMENT OF INERTIA
or hollow.
Which is greater Isphere or Ishell?
Why is Ishell > Isphere 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
G – Rotational dynamics
MOMENT OF 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
G – Rotational dynamics
MOMENT OF 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
G – Rotational dynamics
MOMENT OF INERTIA
=
1
2
12 ·30(2
+ 42)
axis
What is the rotational
inertia of a 2-m by 4 m,
30-kg slab?
M = 30, a = 2 and b = 4
so that
1
I = 12 M(a2+ b2)
b
a
= 50 kg·m2
What torque applied to the
axis of rotation will
accelerate the slab at 2.5
rad/s2?
τ = Iα
 τ = 125 n·m
τ = 50(2.5)
Slab about center, ┴
1
I = 12 M(a2+b2)
Topic 2.2 Extended
G – Rotational dynamics
THE PARALLEL AXIS THEOREM
Notice that most of the axes for the previous
rotational inertias were through the center of
mass.
Engineers would like to know the rotational
inertia through other axes, too.
Fortunately it is not necessary to list an
infinite number of I’s for each extended mass.
Instead, the parallel axis theorem is used:
If you know Icm then you can find I about any
parallel axis using
I = Icm + Md2
Parallel Axis Theorem
Where M is the total mass of the extended object,
and d is the distance from the center of mass to an
axis parallel to the cm axis.
FYI: We just added the rotational inertias of the constituent parts to get
the total rotational inertia.
Topic 2.2 Extended
G – Rotational dynamics
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 d = 8.1 m:
I = Icm + Md2 =
Then
2
MR2
5
+ Md2 =
=
2
·200·0.12
5
+ 200·8.12
13122.8 kg·m2
Itot = 256 + 13122.8 = 13378.8 kg·m2
Topic 2.2 Extended
G – Rotational dynamics
APPLICATIONS OF ROTATIONAL DYNAMICS
Consider a disk-like pulley of
α
mass m and radius R. A string is
connected to a block of mass M,
and wrapped around the pulley.
What is the acceleration of the
block as it falls?
We can insert the forces into our
diagrams, important dimensions, and
accelerations.
m
R
M
T
Note: The acceleration of the pulley is angular, α.
T
Note: The acceleration of the block is linear, a.
a
Mg
Note: At this point we have 3 unknowns, but only 2 equations. WE
Topic 2.2 Extended
NEED ANOTHER EQUATION…
G – That
Rotational
dynamics
equation is a = Rα.
Question:
If the mass
the pulley isDzero
what is the expected
A
PPLICATIONS
OF Rof
OTATIONAL
YNAMICS
acceleration
of the
mass M?
Now we get
ourfalling
equations:
For the pulley, τ = Iα.
m
Since T is tangent to the
R
pulley, at a distance R from
the pivot point, τ = RT.
1
Since for a pulley I = mR2
2
ΣF = Ma
τ = Iα
T =
1
2α
mR
2
1
mRα
2
T - Mg = -Ma
ma = Mg - Ma
T
T = Mg - Ma
a = Rα → α = a/R
1
2
T
Block analysis
Pulley analysis
RT =
α
→ T =
→ a =
1
2
2Mg
2M + m
ma
a
M
Mg