Topic 2.2 ExtendedG1 – 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 ExtendedG1 – Calculating rotational inertia

Different continuous mass distributions have different rotational inertias.

HoopI = MR2

axis

R

axisDisk/CylinderI = MR21

2

R

Hoop about diameterI = MR2

axis

R

12

axis

R

L

Disk/CylinderI = MR2 + ML21

4112

Furthermore, the rotational inertia depends on the location of the axis of rotation.

Why is Ihoop > Idisk for same M and R?

Topic 2.2 ExtendedG1 – Calculating rotational inertia

Spheres can be solid or hollow.

Solid sphereI = MR22

5

R

axis

axis

R

Spherical shellI = MR22

3

axis

R1

Annular cylinder/RingI = M(R1

2+R22)1

2

R2

So can disks.

Why is Isphere > Ishell for same M and R?

Which is greater - Isphere

or Ishell?

If R1 = R2 what does the ring become?

If R1 = 0 what does the ring become?

Topic 2.2 ExtendedG1 – Calculating rotational inertia

A thin rod can be rotated about a variety of perpendicular axes. Here are two:

Thin rod about center, ┴

I = ML2112

axis

L

Thin rod about end, ┴

I = ML213

axis

L

Why is Irod,cent < Irod,end?

Topic 2.2 ExtendedG1 – Calculating rotational inertia

Finally, we can look at a slab rotated through a perpendicular axis:

Slab about center, ┴

I = M(a2+ b2)112

axis

a

b

Why doesn’t the thickness of the slab matter?

Topic 2.2 ExtendedG1 – 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)112

axis

a

bM = 30, a = 2 and b = 4 so thatI = M(a2+ b2)1

12

= ·30(22 + 42)112

= 50 kg·m2

K = Iω212

= ·50·20212

= 10000 J

Topic 2.2 ExtendedG1 – 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 center, ┴

I = ML2112

axis

L

Thin rod about end, ┴

I = ML213

axis

L

Topic 2.2 ExtendedG1 – 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

xL2

-L2dm is shown.

ℓ

dm

Since this is a 1D problem, dm = λdℓ.

I = ∫r2dm rotational inertia(continuous masses)

I = ∫ℓ2 λdℓ-L/2

L/2

= λ∫ℓ2 dℓ-L/2

L/213ℓ

3

-L/2

L/2

= λ· λ3= -L3

8

-L3

8 =

λL3

12

Since λ = M/L, Irod,cent =

ML3

12L = ML2 112

Topic 2.2 ExtendedG1 – 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

xL0 ℓ

dm

I = ∫r2dm

= λ∫ℓ2 dℓ0

L

Since this is a 1D problem, dm = λdℓ.

I = ∫ℓ2 λdℓ0

L13ℓ

3

0

L

= λ· λ3

= L3 - 03 =

λL3

3

Since λ = M/L,

Irod,end = ML3

3L = ML2 13

Topic 2.2 ExtendedG1 – Calculating rotational inertia

Note that the axes of rotation all pass through either the center or the end.What if we have a rotating object like a sphere on the end of a rod, as shown.

axis

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 ExtendedG1 – Calculating rotational inertia

THE PARALLEL AXIS THEOREMSuppose 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

where h is the distance between the new axis and the axis passing through the cm.

parallel axis theorem

Topic 2.2 ExtendedG1 – Calculating rotational inertia

THE PARALLEL AXIS THEOREMSuppose 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.

8 m.1 m

12 kg200 kg

Since the rod already has a formula for Irod,end, we’ll find its rotational inertia first:

Irod,end = ML213 = ·12·821

3 = 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 = ·200·0.12 + 200·8.122

5= MR2 + Mh22

5

= 13122.8 kg·m2

Then

Itot = 256 + 13122.8 = 13378.8 kg·m2

cm

Topic 2.2 ExtendedG1 – Calculating rotational inertia

cm

PROOF OF THE PARALLEL AXIS THEOREMConsider 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 distance h from the cm, we can find I with respect to the new point.

Icm

I

The question is, how are I, Icm and h related.

h

Topic 2.2 ExtendedG1 – Calculating rotational inertia

cm

PROOF OF THE PARALLEL AXIS THEOREMWe’ll center our coordinates on the cm:h is the distance between the old axis and the new.

x

y

h

O

P

O is the old axis and P is the new parallel axis.Let dm be an arbitrary mass having coordinates (x,y).

(x, y)

Let P have coordinates (a,b).

a

b

dm

Let r be the distance from P to dm.

r

Then r forms the hypotenuse of the triangle shown:The legs of that triangle are shown:

x - ay - b

Topic 2.2 ExtendedG1 – Calculating rotational inertia

PROOF OF THE PARALLEL AXIS THEOREMThe distance from the original origin O to dm is x2 + y2 so that

cm

x

y

h

O

P

(x, y)

a

b

dm

r

x - ay - b

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

= ∫[(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

I = Icm + Mh2

00How do you know that xcm is 0?How do you know that ycm is 0?