To this point we have used a particle model

for our study, for the most part at least.

This chapter branches out a bit and we

start to study rigid bodies, and rotation.

A rigid body is one that doesn't bend,

think of things like a disk, a ball, a gear

merry-go-rounds.

We will also play with divers, ice skaters,

and other such people in motion.

Now they aren't strictly rigid, but for

some of their motion we'll pretend.

Rigid bodies have two basic types of motion.

Translational: We have used this to great extent.

Blocks slide down ramps, projectile motion, etc

Rotational motion: we also did this to some extent.

Disks spinning, objects turning.

In addition to these two types a combination can be made

as well.

This hammer has a translational motion

as well as a rotational motion.

maybe easier to see is this wrench. It is sliding across a desk

in a straight line, but it is also rotating.

For a bit of review we have rotational quantities

and their relationships

With sign convention as well

Further review, we have rotational kinematic equations

Finally the connection between rotational

and translational motion.

As we begin to talk about rigid body

rotation, we need to talk about what

the body rotates around.

this is the center of mass.

You can see this in the previous pictures.

Here is another example that illustrates this motion.

What then is the center of mass?

Here are three methods

1. Spin the object.

It will rotate around the center of mass.

think back to the youtube video.

2. Hang the object by 2 or 3 different

locations, and draw a line straight down

from the fixed point.

The intersection of the lines will be the

center of mass.

3. Use an equation

This equation assumes that you can add up every single

particle and its mass at its location. in practicality , not good.

Your book then goes into a bit of calculus and

takes the limit of this summation to an integral

and gives you the integral form of the sum.

This is still not a practical equation.

You would need to know an equation describing

the mass density of an object. As well as a perfect

size or dimension of an object.

This would work great in uniform rods, and circles,

but for any engineering purpose, no good.

Also note that in the current form this integral

is not helpful, you need a dx or a dy.

For a uniform rod you could consider

that the ratio of mass and distance over the total

mass and distance are the same.

This would give:

If I throw a hammer, or tennis racket, and asked you to find the velocity of any point on it at any time,

could you do it?

"What is torque exactly?"

"Why is the motion of the center of mass constrained to the tangential velocity at the edge?"

"Why doesn't torque take into account mass?"

"Can you please explain the moment of inertia?"

Main Ideas

Rotational Motion

Center of mass

Torque

Moment of Inertia

Dynamics

Statics

Energy

Work

Momentum

Rotational Energy

Yet if it is rotating the 'translational or tangential' velocity

is related to angular velocity.

In speaking of energy, there is both energy of position and energy of

motion.

When an object rotates it moves. Amazing.

Thus it has kinetic (moving) energy.

With these two together we have a

rotational kinetic energy.

Note this is specifically of only one mass

at one position.

this would be like a rock in a sling

However, most objects have mass and

distances that vary.

Thus you would have to add them all up.

This summation is an important term that

is specific for different types of objects.

It is called the moment of inertia.

Thus we can write the rotational

kinetic energy as:

This moment of inertia is analogous to

to the linear inertia that we have had before.

They both depend on the mass of the object

but when you rotate it also depends on how

far away that mass is from the axis of rotation.

Consider trying to rotate a 2x4 wooden board

by the end of the board, versus

by the middle of the board.

It has the same mass, but the mass is further away

from the axis of rotation, thus it has a larger moment of

inertia.

Demo

We could have much fun

with integration, finding the moments

of inertial for different objects.

Here is a list from your text.

Which has

a larger moment of

inertia

Solid sphere or

Solid Cylinder

A sphere

B Cylinder

C need more info

To find the moment of inertia of any object we

need to change our sum into an integral

In doing this we need to somehow compare

the r and the dm.

for example consider a disk.

As I go out radially the mass scales the same as the

area scales.

Here the area of the disk is

and a little small area can be

thought of by unwinding some

small ring having a height and length

multiplying them together gives:

Thus

Back to where we started :

Combining gives

Doing the integral gives:

Which is the same as the table.

With these comes the ability to also

find moments of inertia that are

not in the center of an object, or at

the edge.

This leads to the parallel-Axis theorem

The parallel - axis theorem

says that the moment of inertial

about an axis that is parallel to the

center of mass can be solved as:

Where M is the mass of the object

and d is the distance from this

parallel axis to the center of mass

**Torque**

1. The magnitude of the force.

2. The distance the force is from the pivot.

3. The angle of the force.

Consider opening a door.

How do these things affect the ability I have

to open this door?

As noted the torque is defined as:

The only tricky bit is the angle.

It is measured between the vectors.

With that, we have defined how

a force causes rotation.

Torque then takes the place of all

'forces' in our rotational dynamics.

Like other rotational variables it

has a sign that is positive when

causing CCW rotation, and negative

with CW rotation.

As we consider this equation of torque

we can view it in two different ways.

The 'sine' can go with the Force

or the distance.

If with the force then it implies the

tangential component of the force.

If with the radial distance then it is

called the moment arm.

Both ideas are just fine.

F

r

d

Just as with forces acting on a particle

we need to add up all the torques

acting on a rigid body.

a few good steps to consider

1. Identify the 'forces'

2. Cleverly pick a pivot or axis of rotation

3. Note the radial distance from axis to force

4. Note the angle between them.

5. Assign positive or negative values.

As an example here is a beam with two

masses on the ends.

A

B

C

D

Which of these forces

will produce the most

Torque

A

B

C

D

At which location

would the axis be

best located

A

B

C

D

Which of these forces

will produce a positive

value of torque?

A A & B

B C & D

C B

D A

As you may have noticed there was a bit of

a random force acting on this rigid body at

point C.

This is a special torque of gravity.

Gravity is acting at all locations on this beam

And we could add up all of these individual

torques and find a resultant

This says that the Torque due to gravity

can be considered by taking the force of gravity

acting on the objects Center of mass only

Dynamics

F

F

F

t

r

Consider the force

shown here acting on

the rigid body about this

pivot

Only the tangential force

will cause it to move

and the only acceleration

will be in the

tangential direction

If we look at this force and use Newton's

2nd law we have

Using the definition of alpha we have:

Multiply both sides by r

But that is one definition of Torque

Finally if we summed up all of these

we have a moment of inertial

This then is Newton's

2nd law for rotations

Before we jump into solving

some problems

Let us consider a constraint

to the system.

If a disk is to be rotated, by having a

rope pulled over the outside edge

'think rope and pulley'

we assume they don't slip, and thus

they are constrained to move together

The rope produces a Torque, and the

angular velocity/acceleration of the pulley

and the

tangential velocity/acceleration of the

rope are fixed

**Equilibrium**

Recall that we now have a version of

Newton's law in rotational as well as

translational reference frames.

When we have a rigid body in

equilibrium BOTH sets work to our

advantage.

In solving these problems we:

1. Note all the forces (FBD)

2. Pick any ANY axis for rotation

3. Use equations:

**Rolling motion**

**Rolling motion**

is caused by torque, in most cases it is

from friction.

Here we will note the motion and constraints.

is caused by torque, in most cases it is

from friction.

Here we will note the motion and constraints.

With rolling motion we need

both the translational and

rotational kinetic energy.

All rotating objects have a

moment of inertia.

And they all come in the

generic form of mr^2

If we choose to use a constant

'c' we can write any Moment as:

With that, and also relation of angular

to linear velocity, as shown here:

We can rewrite our Kinetic energy.

Substitute in for moment of inertia (I) and

angular velcity (omega), gives:

Or...

This is our simplified version of kinetic energy of an

object that is translating and rotating.

Two objects with the same mass and radius both begin with the same initial velocity at the bottom of an inclined plane. If one is a solid sphere, and other is a hollow shell sphere. Which one travels further up the hill

A. solid (c=2/5)

B. shell (c=2/3)

C. neither, g is fixed.

**Down hill race.**

**Note we did this before as an example of**

moment of inertia.

Let us now predict a final velocity of a

solid sphere, disk, and hoop.

moment of inertia.

Let us now predict a final velocity of a

solid sphere, disk, and hoop.

**Initially I only have potential energy**

due to gravity.

Finally (if we ignore drags) we only

have kinetic energy, So...

due to gravity.

Finally (if we ignore drags) we only

have kinetic energy, So...

**Solving for Velocity**

**Note there is no**

Mass or radius

Rather only the shape (c)

Mass or radius

Rather only the shape (c)

Vectors in rotational motion

Right hand rule:

We use the right hand rule to give the direction

of Omega and Alpha.

We said that they were '+' CCW, and '-' CW

In truth the direction points along the axis

following the right hand rule.

We can now revisit the idea of Torque

How to calculate a Cross product...

Also we can define the Angular Momentum

of a particle as it moves, measured from the

origin

Doing some calculus, or maybe review of previous

material we have a relation between the change of

momentum and force.

This is true in rotational frames, just with Rotational

variables.

Angular momentum

Finally we can relate radial distance

and classic momentum to get

an angular momentum.

In addition we can compare to the classical derivation. (p=mv)

This angular momentum is a

conserved quantity here as

well, just like linear momentum.

Follow the same steps as before

find initial momentum

Find final momentum

equate the two and solve for unknown.

**Also as we have been going we have made**

conversions between rotational and translational

here is a bit of a list.

conversions between rotational and translational

here is a bit of a list.

**Here is one such table showing these comparison**

quantities.

quantities.

moment of inertia rods