1. Topic 2.2 Extended G1 – Calculating rotational inertia
If a body is made of discrete masses use
rotational inertia
(discrete masses)
I = Σmiri2
If a body is made of a continuous distribution of masses, we have
rotational inertia
(continuous masses)
I = ∫r2dm
This equation is derived in a way analogous to that for the cm.
© 2006 By Timothy K. Lund
Note: The important thing to remember is that you choose mass elements dm that are located a distance r from the axis of rotation.

2. Topic 2.2 Extended G1 – Calculating rotational inertia
Hoop
I = MR2
axis R
Different continuous mass distributions have different rotational inertias.
axis
Disk/Cylinder
I = MR2 1 2 R
Why is Ihoop > Idisk for same M and R?
Hoop about diameter
I = MR2
axis R 1 2
axis R L
Disk/Cylinder
I = MR2 + ML2 1 4 12
Furthermore, the rotational inertia depends on the location of the axis of rotation.
© 2006 By Timothy K. Lund

3. Topic 2.2 Extended G1 – Calculating rotational inertia
Solid sphere
I = MR2 2 5 R
axis
axis R
Spherical shell
I = MR2 2 3
Spheres can be solid or hollow.
Which is greater - Isphere or Ishell?
Why is Isphere > Ishell for same M and R?
So can disks.
© 2006 By Timothy K. Lund
axis R1
Annular cylinder/Ring
I = M(R12+R22) 1 2 R2
If R1 = R2 what does the ring become?
If R1 = 0 what does the ring become?

4. Topic 2.2 Extended G1 – Calculating rotational inertia
A thin rod can be rotated about a variety of perpendicular axes. Here are two:
Thin rod about center, ┴
I = ML2 1 12
axis L
Thin rod about end, ┴
I = ML2 1 3
axis L
© 2006 By Timothy K. Lund
Why is Irod,cent < Irod,end?

5. Topic 2.2 Extended G1 – Calculating rotational inertia
Finally, we can look at a slab rotated through a perpendicular axis:
Slab about center, ┴
I = M(a2+ b2) 1 12
axis a b
Why doesn’t the thickness of the slab matter?
© 2006 By Timothy K. Lund

6. Topic 2.2 Extended G1 – Calculating rotational inertia
What is the rotational kinetic energy of a 2-m by 4 m, 30-kg slab rotating at 20 rad/s?
Slab about center, ┴
I = M(a2+b2) 1 12
axis a b
M = 30, a = 2 and b = 4 so that
I = M(a2+ b2) 1 12
= ·30( ) 1 12
© 2006 By Timothy K. Lund
= 50 kg·m2
K = Iω2 1 2
= ·50·202 1 2
= J

7. 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).
Thin rod about end, ┴
I = ML2 1 3
axis L
Thin rod about center, ┴
I = ML2 1 12
axis L
© 2006 By Timothy K. Lund

8. 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 Cartesian cs over the rod, centering it on the axis of rotation. y dm -L 2 L 2 x ℓ
dm is shown.
Since this is a 1D problem, dm = λdℓ.
rotational inertia
(continuous masses)
I = ∫r2dm
© 2006 By Timothy K. Lund
L/2
= λ∫ℓ2 dℓ
-L/2
L/2
I = ∫ℓ2 λdℓ 1 3 ℓ3
-L/2
L/2
= λ· λ 3 = L3 8
-L3 =
λL3 12
-L/2
Since λ = M/L,
Irod,cent =
ML3
12L
= ML2 1 12

9. 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 Cartesian cs over the rod, centering it on the axis of rotation. y dm x ℓ L
Since this is a 1D problem, dm = λdℓ.
I = ∫r2dm
© 2006 By Timothy K. Lund L
= λ∫ℓ2 dℓ L
I = ∫ℓ2 λdℓ 1 3 ℓ3 L
= λ· λ 3
= L3 - 03 =
λL3 3
Since λ = M/L,
Irod,end =
ML3 3L
= ML2 1 3

10. Topic 2.2 Extended G1 – Calculating rotational inertia
Note that the axes of rotation all pass through either the center or the end.
axis
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.
© 2006 By Timothy K. Lund

11. 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
parallel axis theorem
I = Icm + Mh2
where h is the distance between the new axis and the axis passing through the cm.
© 2006 By Timothy K. Lund

12. 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
8 m
.1 m
Since the rod already has a formula for Irod,end, we’ll find its rotational inertia first:
Irod,end = ML2 1 3
= ·12·82 1 3
= 256 kg·m2
© 2006 By Timothy K. Lund
Since the sphere is not rotating about its cm, we must use the parallel axis theorem, with h = 8.1 m:
I = Icm + Mh2
= MR2 + Mh2 2 5
= ·200· ·8.12 2 5
= kg·m2
Then
Itot =
= kg·m2

13. 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: cm
If we choose a parallel axis passing through some other point located a distance h from the cm, we can find I with respect to the new point. cm I
Icm
© 2006 By Timothy K. Lund
The question is, how are I, Icm and h related. h

14. 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). y
(x, y) cm dm
Let P have coordinates (a,b). r
y - b
Let r be the distance from P to dm. P
x - a
Then r forms the hypotenuse of the triangle shown: h b
© 2006 By Timothy K. Lund O a
The legs of that triangle are shown: x

15. Topic 2.2 Extended G1 – Calculating rotational inertia
PROOF OF THE PARALLEL AXIS THEOREM cm x y h O P
(x, y) a b dm r
x - a
y - b
The distance from the 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
© 2006 By Timothy K. Lund
= ∫[(x-a)2 + (y-b)2]dm
= ∫[(x2 - 2xa + a2) + (y2 - 2yb + b2)]dm
= ∫[(x2 + y2) + (a2 + b2) – 2xa - 2yb]dm
= ∫[(x2 + y2)dm + ∫h2dm – 2a∫xdm – 2b∫ydm
= Icm + h2∫dm – 2axcm – 2bycm
How do you know that xcm is 0?
How do you know that ycm is 0?
I = Icm + Mh2