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S18 PH 121 12 introduction

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Richard Datwyler

on 28 November 2018

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Transcript of S18 PH 121 12 introduction

Rotation
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.

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.

Initially I only have potential energy
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)

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.

Here is one such table showing these comparison
quantities.

moment of inertia rods
Full transcript