PES 1110 Fall 2013, Spendier Lecture 28/Page 1

Today:

- Rotational Kinetic Energy and Rotational Inertia (10.6-10.7)

(Exam 2 will include the material covered in today’s lecture)

- %ext Friday, %ov 8th

: Exam 2

Rotational Kinetic Energy

A spinning wheel has kinetic energy. Any rotating object has a kinetic energy due to its

motion. The equation for kinetic energy is

21

2K mv=

How many different values of linear speed does a rotating wheel have? It has infinitely

many different linear speeds. So does this mean that the wheel has an infinite amount of

kinetic energy? NO! We need to use calculus to compute rotational kinetic energy.

We have to imagine splitting the rotating object up into many small pieces. Each of these

small pieces will go around a circle as the object rotates. We first need to index the

pieces, for example we can look at the i-th piece and follow it around a circle as it rotates.

The kinetic energy of the i-th piece is

21

2i i iK m v=

mi = mass of grey dot

vi = linear speed of grey dot

we can relate linear speed to angular speed

i iv rω=

by using ri, the distance from the axis of rotation to the grey dot. There is no subscript on

ω, since all pieces have the same angular speed. A rotating wheel has only ONE angular

speed!

( )2 2 21 1

2 2i i i i iK m r m rω ω= =

To obtain the total rotational kinetic energy for the entire wheel we need to add up all the

kinetic energies for each piece:

PES 1110 Fall 2013, Spendier Lecture 28/Page 2

2 2 2 21 1

2 2i i i i i

i i i

K K m r m rω ω ≈ = = ∑ ∑ ∑

This is an approximation for the total kinetic energy. It is not precise because the pieces

are too big. (There are white spaces in between.) This expression becomes exact in the

limit as the number of pieces approaches infinity (and so the size of each piece

approaches zero). We need to take some sort of a continuum limit of the inside sum. We

need to take an integral.

Rotational Inertia or Moment of Inertia

The inside sum 2

i i

i

m r∑ , is a sum over all the masses at distances ri from the axis of

rotation. This quantity depends on the shape and density of the object. It is given a special

name “moment of inertia” or “rotational inertia”

2

i i

i

I m r=∑ (this is correct for point particles)

But for continuous objects we need to take the limit and the sum becomes

2 2 2lim

i ii

i

I m r r dm r dVρ→∞

= = =∑ ∫ ∫

ρ = density (mass divided by volume: M/V)

V = volume

The end result is that the kinetic energy of a rotating object is simply

21

2K Iω= (for rotating object)

What is this moment of inertia physically?

The equation for the kinetic energy of a rotating object

21

2K Iω= (for rotating object)

looks just like the equation we had before for kinetic energy

21

2K mv= (for object moving in straight line)

but with different symbols in it.

For rotating objects we replace linear speed v with angular speed ω and we replaced mass

m with the moment of inertia I.

So the physical idea is that the moment of inertia, I, is the rotational counterpart to

mass m. It plays the same role in rotation as mass does in linear motion. We started to

think about mass when we did Newton’s 2nd

law: F = ma. The more mass an object has

PES 1110 Fall 2013, Spendier Lecture 28/Page 3

the more force you need to make it move. Hence, the moment of inertia tells us how

“hard" it is to make an object rotate (spin).

After the exam, we will talk about what plays the role of force in rotation. We will learn

that this is called the torque. The more moment of inertia an object has the more torque

you need to apply to make it spin.

The moment of inertia depends on:

(a) The object’s shape.

(b) The axis of rotation

(c) The total mass of the object and how its mass is distributed.

(a) The object’s shape.

For a fixed mass (uniformly distributed) and radius, round objects naturally rotate easier,

so they have a smaller moment of inertia.

In our calculations for I, the location of the mass determines how we split it into pieces:

2lim

i ii

i

I m r→∞

= ∑

We can split a round object into many pieces and a lemon shaped object into many

pieces.

The pieces are at different locations because of the different shape, i.e. different shapes

have different ri’s. The pieces are at different distances from the center of rotation.

Example 1:

A solid circle made out of wood and a hoop made out of metal have the same mass.

Assuming that both have the same radii and that both are rotating about their center,

which object will have a smaller moment of inertia?

Answer: For the hoop the mass is only around the edge. Basically, it has only one value

of ri from the center out. The solid circle has ri’s anywhere from zero all the way out to

the outer edge of the hoop. This means that the solid circle will have smaller moment of

inertia since it has smaller values of ri’s.

(see table on page 5)

PES 1110 Fall 2013, Spendier Lecture 28/Page 4

(b) The axis of rotation. The moment of inertia is different depending how I spin the

object. It depends on how ri is defined. One single object has more than one moment of

inertia. You have to specify around which axis you wan to spin the object.

Rotation of lemon about its the Center: Rotation of lemon about One End:

(smaller ri � smaller I) (bigger ri � larger I)

If we look at the same point, the moment of inertia will be bigger for rotation about One

End.

Example 2:

Two particles, each with mass m, are joined by a massless rod of length L.

a) Calculate the moment of inertia I1 for an axis of rotation located at the center of mass.

b) Calculate the moment of inertia I2 for an axis of rotation about one of the masses.

c) If in both cases the masses rotate at the same angular speed, which system has a larger

rotational kinetic energy?

PES 1110 Fall 2013, Spendier Lecture 28/Page 5

(c) The total mass of the object. A heavier wheel is harder to spin than a lighter wheel. It

also depends on how the mass is distributed:

Demo: two identical sticks have their mass differently distributed

Moment of Inertia for standard shapes

As we know for continuous objects: 2 2

I r dm r dVρ= =∫ ∫

We can use this equation to calculate the moment of inertia for standard shapes and

axes of rotation. Look in book for examples on page 256.

The book has Table 10-2 (page 255) for different shapes and axes of rotation obtained by

this integration:

PES 1110 Fall 2013, Spendier Lecture 28/Page 6

Parallel-axis theorem:

We just learned from in class example 2 that a body doesn't have just one moment of

inertia. In fact, it has infinitely many, because there are infinitely many axes about which

it might rotate. But there is a simple relationship between the moment of inertia Icom of a

body of mass M about an axis through its center of mass and the moment of inertia I

about any other axis parallel to the original one but displaced from it by a distance h. This

relationship is called parallel-axis theorem:

2

comI I Mh= −

h = distance between the 2 axes of rotation.

Example 4:

A part of a mechanical linkage has a mass of 3.6 kg. We measure its moment of inertia

about an axis 0.15 m from its center of mass to be I = 0.132 kg m2. What is the moment

of inertia Icom about a parallel axis through the center of mass?

PES 1110 Fall 2013, Spendier Lecture 28/Page 7

O%LY IF TIME permits

Example 5:

The pulley has a radius R and a moment of inertia I. The rope does not slip over the

pulley, and the pulley spins on a frictionless axle. The coefficient of kinetic friction

between block A and the tabletop is µk. The system is released from rest, and block B

descends. Bloch A has a mass mA and block B has a mass mB. Use energy methods to

calculate the speed vf of block B as a function of the distance d that it has descended.

PES 1110 Fall 2013, Spendier Lecture 28/Page 8

Example 6: (for our calculus II students in class - not on exam)

A uniform rod with mass M and length L rotates about an axis through point O, at an

arbitrary distance h from one end. Compute the moment of inertia

a) for an arbitrary distance h from one end,

b) for h = 0 (axis of rotation is at one end),

c) for h = L (axis of rotation is at the other end), and

d) for h = L/2 (axis of rotation passes through center).