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Here you see the topics that
will be covered by this exam--

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twice as much material
as last time.

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And, of course, the material is
also more difficult.

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I will touch upon most of
these topics today-- not all--

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and I can't go into great depth,
for obvious reasons.

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We don't have the time.

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And I want to emphasize that
what is not covered today

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does not mean at all that
it will not be on the exam.

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I want to ask your attention
for these two key concepts:

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the work-energy theorem,

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which tells you that if
an object moves from A to B

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that the work done
on that object

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is the kinetic energy
at point B

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minus the kinetic energy
at point A.

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This always applies
both for conservative forces,

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like gravitational and spring
forces, but it also holds

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for nonconservative forces
such as friction.

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Friction can remove
kinetic energy.

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It turns kinetic energy
into heat,

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and that is perfectly fine
in the work-energy theorem.

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However, the conservation
of mechanical energy only holds

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for conservative forces.

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There you have
the conservation

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of the sum of potential energy
and kinetic energy,

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and now, of course,

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you cannot afford to lose
kinetic energy through heat,

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because then the conservation
of mechanical energy

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would not hold,
and so that can only be used

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exclusively in the case
of conservative forces.

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Let's start
with a simple example

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of an incline at an angle theta.

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I have here an object, mass m,

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friction coefficient kinetic
is mu k,

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and the static friction
coefficient equals mu s.

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This point here,
let that be point A,

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and let the bottom
of the incline be B,

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and let the distance between
them along the slope be l.

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And the first thing
that you want to do

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with a problem like this,

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you want to make what we call
a free-body diagram.

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That means you want to draw
all the forces on that object.

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Clearly there is gravity,
which is mg.

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And then there is
the normal force

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perpendicular to the surface.

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The surface pushes up, and we
call this N, the normal force.

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The object wants to slide down.

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Friction holds it up.

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So there is also
a frictional force.

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And these are the only
three forces that are on it.

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So with a free-body diagram,
you won't know more.

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However, I want you
to appreciate the fact

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that the force from the surface
onto this object

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is, of course, the vectorial sum
between these two,

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and that is what's called
normally the contact force.

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And that contact force better be
exactly the same as mg,

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but in opposite direction,

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otherwise there could
never be equilibrium.

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Of course, we always split it
into perpendicular directions

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because of the way
that we analyze this.

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There is no acceleration
in the y direction,

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only in x direction
when it starts moving,

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and that's why we split it.

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But, of course,
the vectorial sum is really

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the force with which the surface
pushes onto that object.

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All right, let's suppose now
that we increase the angle theta

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until the object starts
to slide.

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So now I'm going to decompose
these forces.

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In the x direction
I have here mg sine theta,

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and the component of gravity
in the y direction

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equals mg cosine theta.

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Since there is no acceleration
in the y direction,

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the normal force must be
also mg cosine theta.

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I lifted up the incline

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to the point that it is
just about to start sliding.

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And so that means

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that the frictional force is
a maximum value possible,

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which is the static friction
coefficient times N,

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and therefore

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that is mu of s times mg
times the cosine of theta.

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And when will that happen?

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When the frictional force and
mg sine theta are exactly equal.

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If I go then one hair further,
it will start to slide,

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and so that's the case when
this equals mg sine theta,

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and so you lose your mg,

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and you find that
mu s equals tangent theta.

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That's when it will happen.

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This is a way
that you can measure

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the static friction coefficient.

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Alternatively, if you know
the static friction coefficient,

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you can predict at what angle
that will occur.

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Now, at that situation that
it's just hanging by its thumbs,

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so to speak, by its fingernails,

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we touch it very lightly,
we blow on it--

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(imitates blowing )--
and it starts to slide.

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And now I'm interested
in knowing

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what the acceleration
is downhill.

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The reason why
it will be accelerated,

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that this friction coefficient
goes down now to mu k,

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and so now there is a t force

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along the slope
in the plus x direction,

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and so I can write down now
Newton's Second Law:

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ma, which is the force...
the net force

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in the positive x direction,

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would be equal mg times
the sine of theta

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minus the frictional force,

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but now this becomes
mu k mg cosine theta.

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I lose my m,

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and so the acceleration
in the positive x direction--

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that's the way I defined
positive x direction--

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equals g times the sine of theta

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minus mu of k times
the cosine of theta.

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That's acceleration, and now
a natural question would be,

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what is the speed
at which it reaches point B?

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Well, what we would have done
early on in the course,

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we would have said, "Well,
if I know the acceleration,

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"and I know the distance,

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"and I know
the initial speed is zero,

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"then clearly that distance l
must be one-half a t squared,

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t being the time that it takes
to go from A to B."

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And so t would be the square
root of 2l divided by a.

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What speed would it have at B?

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Well, the speed at B would be
a t-- early on 801.

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And so that would become,
if we multiply this by a,

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the square root of 2al.

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And if you want to see that
in all glorious detail,

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then of course
you'll have to take that a

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and you have to multiply that
by 2l and take the square root.

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So we get 2gl times
the sine of theta

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minus mu of k times
the cosine of theta

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and the whole thing
to the power one-half.

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That is the speed at point B.

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I, however, would not do it
that way.

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I don't find it very elegant.

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I would say, "Hey, why not apply
the work-energy theorem?"

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Why not say

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that the work done when
that object moves from A to B

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must be the kinetic energy
at point B

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minus the kinetic energy
at point A?

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Now, I know that I release it
at zero speed, so this is zero.

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And so the kinetic energy at B

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is clearly
one-half m vB squared,

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so it comes down now to applying
the work-energy theorem

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and calculating
how much work was done

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by all the forces at stake--

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conservative
and nonconservative,

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it makes no difference--

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and that puts it equal
to one-half m vB squared.

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Well, first of all,
when it goes from point A to B,

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gravity is doing positive work.

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It's going down, and the amount
of work that gravity is doing

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is plus mgh, but h divided by l
is the sine of theta,

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so this is also
mgl times the sine of theta.

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That's positive work
done by gravity.

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So we have mgl times
the sine of theta--

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that's the positive
contribution by gravity--

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and now we get a negative
contribution by friction,

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because the frictional force
is uphill,

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but the motion is downhill.

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They're 180 degrees opposite,

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so I don't have to worry
about the dot product,

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because the cosine of the angle
is minus 1,

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and so I can ignore the dot
and simply take that force--

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the frictional force--

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and multiply it
by that distance l,

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but I have to, of course,
take the minus sign,

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because the cosine of that angle
equals minus 1.

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And so I let l times
the frictional force,

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which is mu k
mg cosine of theta.

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And that's the negative work
done by the friction,

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and this equals
one-half m vB squared.

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And I lose an m, and
if you look very carefully--

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if you're willing to do that
in your head--

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you bring the 2 to the left,
so you get 2gl,

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and then you get

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the g sine theta mu k cosine
theta to the power one-half.

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You get exactly the same result,

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and I find this
somehow more pleasing

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to use the work-energy theorem.

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So friction does negative work--
it's converted into heat.

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But that's no problem
for the work-energy theorem.

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We know in life that friction
takes kinetic energy out.

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Whenever something is moving,
it will come to a halt.

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If you spin a top, that we know

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that the top will not last
very long because of friction.

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And I have here a top.

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I hope you can see it shortly--
there it is.

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It's a teeny-weeny little top.

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It's very cute--
I play with it in my office.

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I'm rotating it,

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and you see that friction
is taking out kinetic energy

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and it's converting it to heat.

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Please stay on the desk,
will you?

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Will you please not fall
on the floor?

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Ah, it does-- great.

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So you see,
that's what friction does.

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Kinetic energy has been removed,

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and now it has been
converted to heat.

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This surface is a little
smoother than the desk,

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and so I will give it
a spin here,

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so it may last a little longer,

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but obviously
it can't last very long,

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so very shortly, it, too, will
fall over and come to rest,

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and that kinetic energy will
have been converted into heat.

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All right, so let's continue now
on another subject,

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and the one that I want to talk
to you about now is pendulum.

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I have a pendulum, and the
pendulum at time t equals zero.

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This angle is theta zero,
and I know what that angle is.

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It's five degrees, which is
approximately 0.09 radians.

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At the time t equals zero,

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I give this pendulum
a tangential speed.

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I call it v of B,
because I call this point B.

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It's going to arc.

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I call this A.

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So this is a circle,

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and it comes to a halt,
let's say here at point C.

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And then this angle is
the maximum angle possible,

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is theta maximum.

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Let the length of the pendulum,
for simplicity,

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simply be one meter.

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It is small-angle approximation.

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The angles will never be
very large,

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as you will see later when
we calculate theta maximum,

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and so we know
that we're going to get

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a simple harmonic oscillation
to a very good approximation.

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So theta is going to be

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that angle theta maximum
times the cosine--

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or if you want to, be my guest,
you can make this a sine;

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I always work with cosines--
omega t plus phi.

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And omega equals
the square root of g over l.

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I will give you some equations
during your exam,

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but this one
I will not give you--

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I just assume
that you remember this--

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and that the period
of the pendulum equals

234
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2 pi times
the square root of l over g.

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So we know omega, because
we know g and we know l,

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and now a reasonable
question is,

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what would be theta maximum?

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Well, if we knew this,

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then we would know
what theta maximum is,

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because this part here equals
l times the cosine of theta max,

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and so this is l minus
l cosine theta max.

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So it comes down to calculating

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how high this object comes
above point A.

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Well, I will split this
into two heights,

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this one, which I call h1,
and this one, which I call h2.

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00:15:10 --> 00:15:16
And so the one that
we really want to know

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is what is h1 plus h2?

248
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Because h1 plus h2 will be

249
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l times 1 minus
the cosine of theta max.

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And so the moment
we have h1 plus h2,

251
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we immediately have
the maximum angle.

252
00:15:31 --> 00:15:37
Now, h1 is a piece of cake,
because we know that point B...

253
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that this object is at B when
theta zero is five degrees.

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00:15:41 --> 00:15:47
So h1 equals l times 1 minus
the cosine of theta zero.

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And you know theta zero.

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And so you'll find, then, if
you were interested in numbers--

257
00:15:54 --> 00:16:00
but don't bother
if you don't like numbers--

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for the numbers that I gave you,
this is 0.0038 meters.

259
00:16:01 --> 00:16:07
It's only 3.8 millimeters.

260
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It's a teeny-weeny little bit.

261
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It's a large displacement
in this direction

262
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but very little
in this direction.

263
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So we know already what h1 is.

264
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So now what is h2?

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Well, we could now use

266
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the conservation
of mechanical energy,

267
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and we could say,
let us call arbitrarily

268
00:16:26 --> 00:16:32
the gravitational potential
energy at point A,

269
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let's call that zero.

270
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And going to apply

271
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the conservation of mechanical
energy, which means

272
00:16:35 --> 00:16:41
that the kinetic energy at A
plus the potential energy at A

273
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must be the kinetic energy at B
plus the potential energy at B

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00:16:46 --> 00:16:52
equals the kinetic energy at C
plus the potential energy at C.

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Now, the potential energy
at A is zero

276
00:16:55 --> 00:17:01
because we define it that way.

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The kinetic energy at C is zero
because it comes to a halt.

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So now we can write down

279
00:17:03 --> 00:17:09
that one-half m vA squared
equals one-half m vB squared

280
00:17:16 --> 00:17:22
plus mg times h1-- that is the
potential energy at point B--

281
00:17:24 --> 00:17:30
it's higher than this point A--
vertical separation h1--

282
00:17:30 --> 00:17:36
and that equals the
potential energy at point C,

283
00:17:34 --> 00:17:40
which is mg times h1 plus h2.

284
00:17:41 --> 00:17:47
And if you compare
this part of the equation,

285
00:17:43 --> 00:17:49
you see that you lose mgh1,

286
00:17:45 --> 00:17:51
and so you'll find
that m vB squared--

287
00:17:53 --> 00:17:59
want to carry the one-half
for now, we can still do that--

288
00:17:57 --> 00:18:03
equals mg times h2.

289
00:18:02 --> 00:18:08
And so you will find
immediately h2.

290
00:18:06 --> 00:18:12
h2 equals v of B squared,
which is known.

291
00:18:14 --> 00:18:20
In our case it must be
a known number--

292
00:18:17 --> 00:18:23
I will give you in a minute
what that is--

293
00:18:19 --> 00:18:25
and that is divided by 2g.

294
00:18:22 --> 00:18:28
And this v of B
that I had in mind--

295
00:18:24 --> 00:18:30
numbers are really
not that important;

296
00:18:27 --> 00:18:33
you don't have a calculator
anyhow at your exam--

297
00:18:30 --> 00:18:36
but I was going to give this a
speed of 0.3 meters per second

298
00:18:34 --> 00:18:40
in this direction

299
00:18:35 --> 00:18:41
in order to keep the
maximum angle quite modest.

300
00:18:38 --> 00:18:44
So I know h2, and now
that I know h2 and I know h1,

301
00:18:42 --> 00:18:48
I can go to this equation

302
00:18:45 --> 00:18:51
and I can ask, what is
the cosine of theta maximum?

303
00:18:50 --> 00:18:56
h2, by the way, in case
you're interested in numbers,

304
00:18:53 --> 00:18:59
is about 0.0045.

305
00:18:57 --> 00:19:03
So it is 4.5 millimeters.

306
00:19:00 --> 00:19:06
So it was originally
3.8 millimeters, h1,

307
00:19:04 --> 00:19:10
and now h2 is
4.5 millimeters higher.

308
00:19:09 --> 00:19:15
So you can calculate
the cosine of theta max.

309
00:19:14 --> 00:19:20
However, I said to myself,

310
00:19:16 --> 00:19:22
if we use the conservation
of mechanical energy,

311
00:19:20 --> 00:19:26
can we not also use
the work-energy theorem?

312
00:19:24 --> 00:19:30
Why not?

313
00:19:25 --> 00:19:31
That should work as well.

314
00:19:28 --> 00:19:34
The work-energy theorem tells me

315
00:19:30 --> 00:19:36
that the work done
on that object

316
00:19:33 --> 00:19:39
when the object moves
from A to B

317
00:19:36 --> 00:19:42
equals the kinetic energy at B
minus the kinetic energy at A.

318
00:19:41 --> 00:19:47
That is one-half m vB squared
minus one-half m vA squared.

319
00:19:53 --> 00:19:59
And how much work was done
by gravity in going from A to B?

320
00:19:56 --> 00:20:02
Because gravity is the
only force that does work.

321
00:20:00 --> 00:20:06
The tension is not doing
any work,

322
00:20:02 --> 00:20:08
because the tension,
which is in this direction,

323
00:20:05 --> 00:20:11
is always perpendicular
to the direction of motion.

324
00:20:08 --> 00:20:14
And work is a dot product

325
00:20:10 --> 00:20:16
between force and
the direction of motion,

326
00:20:12 --> 00:20:18
perpendicular to each other.

327
00:20:14 --> 00:20:20
So that's zero.

328
00:20:15 --> 00:20:21
So it's only mg that matters--
the gravitational force.

329
00:20:20 --> 00:20:26
Well, when it goes from A to B,

330
00:20:23 --> 00:20:29
the work done by gravity is
minus mgh1.

331
00:20:28 --> 00:20:34
Object goes up, it does negative
work, equals minus mgh1.

332
00:20:35 --> 00:20:41
Now, look at that equation,
and look, if you want to, here.

333
00:20:43 --> 00:20:49
Compare this one with this one.

334
00:20:48 --> 00:20:54
Completely identical.

335
00:20:50 --> 00:20:56
So the work-energy theorem is,
of course, in a way

336
00:20:53 --> 00:20:59
the same as the conservation
of mechanical energy.

337
00:20:57 --> 00:21:03
The work done when
the object goes from B to C

338
00:21:00 --> 00:21:06
is the kinetic energy at C
minus the kinetic energy at B.

339
00:21:05 --> 00:21:11
That equals
one-half m vC squared

340
00:21:09 --> 00:21:15
minus one-half m vB squared,

341
00:21:13 --> 00:21:19
and that is the work
that gravity is doing

342
00:21:16 --> 00:21:22
when the object goes
from B to C.

343
00:21:19 --> 00:21:25
And that work is minus mgh2.

344
00:21:23 --> 00:21:29
So this equals minus mgh2.

345
00:21:28 --> 00:21:34
This is zero.

346
00:21:30 --> 00:21:36
And so you see,

347
00:21:31 --> 00:21:37
what you see here is
exactly what you see here.

348
00:21:35 --> 00:21:41
The two are identical.

349
00:21:38 --> 00:21:44
And so you could have used
the work-energy theorem,

350
00:21:41 --> 00:21:47
or you can use the conservation
of mechanical energy.

351
00:21:43 --> 00:21:49
That makes no difference.

352
00:21:45 --> 00:21:51
And so you now can calculate
theta maximum,

353
00:21:48 --> 00:21:54
and for those of you
who want some numbers,

354
00:21:50 --> 00:21:56
I found that theta maximum was
plus or minus 7.4 degrees.

355
00:21:57 --> 00:22:03
You always get two angles.

356
00:22:02 --> 00:22:08
And in radians, that would be
plus or minus 0.13 radians.

357
00:22:07 --> 00:22:13
You know what the cosine
of the angle is,

358
00:22:09 --> 00:22:15
so you always get two angles.

359
00:22:11 --> 00:22:17
There's nothing
you can do about it.

360
00:22:13 --> 00:22:19
And so now you can ask yourself
the question, what is phi?

361
00:22:16 --> 00:22:22
Phi is always
a bit of a pain in the neck,

362
00:22:18 --> 00:22:24
and there is really not
all that much physics in phi.

363
00:22:21 --> 00:22:27
But I was sort of curious
for these initial conditions

364
00:22:24 --> 00:22:30
what phi would be.

365
00:22:26 --> 00:22:32
And so let's just take
a quick look at that.

366
00:22:29 --> 00:22:35
And so if we want
to know what phi is,

367
00:22:31 --> 00:22:37
we have to look at the initial
condition that at t equals zero

368
00:22:35 --> 00:22:41
the velocity at point B is
plus 0.3 in this direction,

369
00:22:42 --> 00:22:48
and we know that
theta equals theta zero,

370
00:22:46 --> 00:22:52
and we know what theta zero is--

371
00:22:48 --> 00:22:54
that was the five degrees,
that is the .09 radians.

372
00:22:52 --> 00:22:58
And so I'm going to substitute
that in my solution.

373
00:22:56 --> 00:23:02
So I know when t equals zero,

374
00:22:58 --> 00:23:04
I know that
theta equals theta zero.

375
00:23:01 --> 00:23:07
So theta zero--
I know what theta zero is;

376
00:23:04 --> 00:23:10
we just calculated
what theta max is;

377
00:23:07 --> 00:23:13
times the cosine of phi
at t equals zero.

378
00:23:10 --> 00:23:16
And so out pop
two angles of phi,

379
00:23:14 --> 00:23:20
plus and minus phi 1.

380
00:23:17 --> 00:23:23
You always find
a plus and a minus sign

381
00:23:19 --> 00:23:25
because the cosine
of plus the angle

382
00:23:21 --> 00:23:27
is the same as the cosine
of minus the angle.

383
00:23:23 --> 00:23:29
So now we have to find out
which of the two it is.

384
00:23:27 --> 00:23:33
By the way, when you find

385
00:23:29 --> 00:23:35
that theta maximum equals
plus or minus 0.13 radians,

386
00:23:34 --> 00:23:40
you could have picked
either plus or minus.

387
00:23:36 --> 00:23:42
I picked the plus.

388
00:23:38 --> 00:23:44
If you would have picked
the minus,

389
00:23:40 --> 00:23:46
you would have found
a different phase angle,

390
00:23:42 --> 00:23:48
but you can't pick minus.

391
00:23:44 --> 00:23:50
I just want to remind you

392
00:23:45 --> 00:23:51
that I picked the plus
in whatever follows.

393
00:23:48 --> 00:23:54
And so now I have
to take into account

394
00:23:50 --> 00:23:56
the fact that at t equals zero
that I know the velocity vB.

395
00:23:58 --> 00:24:04
And how does that come in?

396
00:24:00 --> 00:24:06
Well, I take the derivative
of that equation there,

397
00:24:03 --> 00:24:09
and so I get that d theta/dt--

398
00:24:05 --> 00:24:11
which is the angular velocity
in radians per second

399
00:24:09 --> 00:24:15
at any moment in time,

400
00:24:10 --> 00:24:16
but we're going to evaluate it
at t equals zero--

401
00:24:14 --> 00:24:20
equals minus omega theta max

402
00:24:18 --> 00:24:24
times the sine of omega t
plus phi,

403
00:24:24 --> 00:24:30
but we will evaluate it
at t equals zero.

404
00:24:28 --> 00:24:34
What is d theta/dt?

405
00:24:29 --> 00:24:35
Well, d theta/dt is
v divided by l.

406
00:24:34 --> 00:24:40
This is v divided by l
and is therefore plus 0.3--

407
00:24:38 --> 00:24:44
this plus because at t equals
zero, the angle is increasing.

408
00:24:43 --> 00:24:49
So it is plus.

409
00:24:45 --> 00:24:51
How do we know it's v over l?

410
00:24:47 --> 00:24:53
Well, remember,
we discussed that before.

411
00:24:50 --> 00:24:56
If the pendulum changes
the angle by an amount d theta,

412
00:24:55 --> 00:25:01
and if this arc here--
I call that ds--

413
00:24:59 --> 00:25:05
and if the length is l

414
00:25:00 --> 00:25:06
then the definition of d theta--
the angle in radians--

415
00:25:04 --> 00:25:10
is ds divided by l.

416
00:25:08 --> 00:25:14
If you divide both sides by dt,

417
00:25:10 --> 00:25:16
which mathematicians cannot do,
but physicists can,

418
00:25:13 --> 00:25:19
then you get
d theta/dt equals ds/dt,

419
00:25:16 --> 00:25:22
and that is
the velocity divided by l.

420
00:25:20 --> 00:25:26
So you see that
d theta/dt is v divided by l.

421
00:25:23 --> 00:25:29
And so we now have
a second equation.

422
00:25:26 --> 00:25:32
We now can solve for sine phi.

423
00:25:29 --> 00:25:35
That gives you again two angles.

424
00:25:32 --> 00:25:38
That gives you an angle phi 2

425
00:25:35 --> 00:25:41
and that gives you
180 degrees minus phi 2.

426
00:25:39 --> 00:25:45
They have the same value
for sine phi.

427
00:25:42 --> 00:25:48
But only one of this will be
the same as one of these,

428
00:25:47 --> 00:25:53
and that's the one
that you pick.

429
00:25:49 --> 00:25:55
And in my case,
where for my value of plus 0.13,

430
00:25:54 --> 00:26:00
I find then that phi equals
minus 0.82 radians.

431
00:25:59 --> 00:26:05
It is about minus 47 degrees.

432
00:26:03 --> 00:26:09
There's not much physics
in the phase angle.

433
00:26:05 --> 00:26:11
What is interesting perhaps is
to mention

434
00:26:07 --> 00:26:13
that if we had chosen the speed
at times t equals zero,

435
00:26:12 --> 00:26:18
if we had given the speed
in this direction

436
00:26:15 --> 00:26:21
of 0.3 meters per second,

437
00:26:17 --> 00:26:23
the theta maximum would not
have changed-- of course not--

438
00:26:20 --> 00:26:26
but the phi would have changed.

439
00:26:22 --> 00:26:28
In fact, the phi
that you would have found

440
00:26:24 --> 00:26:30
would have been plus 0.82.

441
00:26:26 --> 00:26:32
And for those of you
who had preferred

442
00:26:29 --> 00:26:35
to use the theta maximum
to take minus 0.13,

443
00:26:33 --> 00:26:39
they would have found

444
00:26:35 --> 00:26:41
a different phase angle
altogether.

445
00:26:37 --> 00:26:43

446
00:26:40 --> 00:26:46
All right.

447
00:26:42 --> 00:26:48
By now this top, of course,
must be dead like a doornail.

448
00:26:45 --> 00:26:51
So let's just take
a look at that.

449
00:26:47 --> 00:26:53
And I can't believe--
what's going on?

450
00:26:51 --> 00:26:57
Do you understand
why it's still rotating?

451
00:26:54 --> 00:27:00
Why is friction not...?

452
00:26:56 --> 00:27:02

453
00:26:57 --> 00:27:03
Holy smoke!

454
00:26:59 --> 00:27:05
The basic foundations of physics
are at stake again.

455
00:27:04 --> 00:27:10
And it seems that
every time at lectures,

456
00:27:06 --> 00:27:12
we seem to have a way
to overthrow them.

457
00:27:10 --> 00:27:16
Friction must take
kinetic energy out.

458
00:27:14 --> 00:27:20
I'm extremely puzzled.

459
00:27:16 --> 00:27:22
Let's not look at it.

460
00:27:17 --> 00:27:23
I hate this.

461
00:27:19 --> 00:27:25
Things that I cannot explain,
I hate them.

462
00:27:21 --> 00:27:27
Let's just not look at it.

463
00:27:23 --> 00:27:29
Let's go on.

464
00:27:25 --> 00:27:31
We can always take a look
at the end of the lecture

465
00:27:26 --> 00:27:32
and see what it's doing.

466
00:27:27 --> 00:27:33
By that time
it'sgot to be dead, right?

467
00:27:29 --> 00:27:35
It can't go on forever.

468
00:27:30 --> 00:27:36
So, let's now talk
about a spring.

469
00:27:36 --> 00:27:42
What I just did
with the pendulum,

470
00:27:39 --> 00:27:45
I can do something similar
with a spring,

471
00:27:43 --> 00:27:49
similar in the sense
that at time t equals zero

472
00:27:47 --> 00:27:53
I stretch the spring a little
and I give it a kick

473
00:27:51 --> 00:27:57
and then I let it oscillate.

474
00:27:54 --> 00:28:00
Very similar to what I just did,
but now it is a simpler problem,

475
00:28:00 --> 00:28:06
because a spring is
exactly one-dimensional.

476
00:28:05 --> 00:28:11
Let this be the relaxed length l
of the spring.

477
00:28:08 --> 00:28:14
This is point A.

478
00:28:10 --> 00:28:16
I stretch it to point B.

479
00:28:13 --> 00:28:19
I give it a kick.

480
00:28:15 --> 00:28:21
I will not put in
any numbers now.

481
00:28:18 --> 00:28:24
I'll give it a speed vB

482
00:28:20 --> 00:28:26
and it comes to a halt
here at point C.

483
00:28:23 --> 00:28:29

484
00:28:25 --> 00:28:31
Let the spring constant be k,
and we know

485
00:28:28 --> 00:28:34
we're going to get a...
so this is a t equals zero--

486
00:28:32 --> 00:28:38
we're going to get
a simple harmonic oscillation,

487
00:28:35 --> 00:28:41
x equals x max.

488
00:28:38 --> 00:28:44
This is going to be x max.

489
00:28:42 --> 00:28:48
This is going to be xB.

490
00:28:45 --> 00:28:51
And this is x of A,
which is zero.

491
00:28:49 --> 00:28:55
So we're going to get

492
00:28:51 --> 00:28:57
x max times the cosine
of omega t plus phi--

493
00:28:55 --> 00:29:01
same old story,
it's getting pretty boring.

494
00:28:58 --> 00:29:04
Omega equals
the square root of k over m,

495
00:29:01 --> 00:29:07
and T equals
two pi divided by omega.

496
00:29:04 --> 00:29:10
These equations
you will not see on your exam.

497
00:29:09 --> 00:29:15
What is now x max?

498
00:29:13 --> 00:29:19
How would you approach it?

499
00:29:14 --> 00:29:20
You've just seen how I did it
with the pendulum.

500
00:29:18 --> 00:29:24
I used first the conservation
of mechanical energy--

501
00:29:21 --> 00:29:27
there was no friction--
and then I showed you

502
00:29:23 --> 00:29:29
that it's completely consistent
with the work-energy theorem,

503
00:29:26 --> 00:29:32
so what would you do now?

504
00:29:28 --> 00:29:34
How do we get x max?

505
00:29:31 --> 00:29:37
Any suggestions?

506
00:29:32 --> 00:29:38
Any proposals?

507
00:29:34 --> 00:29:40
Come on!

508
00:29:35 --> 00:29:41
You must have learned something
from the last ten minutes.

509
00:29:38 --> 00:29:44
Any idea?

510
00:29:39 --> 00:29:45
It's the same concept.

511
00:29:40 --> 00:29:46
We know the initial conditions,

512
00:29:42 --> 00:29:48
and we know it's a simple
harmonic oscillation,

513
00:29:45 --> 00:29:51
and we want to know

514
00:29:47 --> 00:29:53
what the maximum displacement
is going to be.

515
00:29:50 --> 00:29:56
Anyone with the courage
to speak out?

516
00:29:52 --> 00:29:58
In the worst case, you're wrong.

517
00:29:54 --> 00:30:00
Believe me, I am wrong so often,
if only you knew.

518
00:29:59 --> 00:30:05
(student responds )

519
00:30:00 --> 00:30:06
LEWIN:
So speak up.

520
00:30:01 --> 00:30:07
Shall we try the conservation
of mechanical energy?

521
00:30:03 --> 00:30:09
Is there anything wrong
with that?

522
00:30:05 --> 00:30:11
Nothing wrong with it.

523
00:30:07 --> 00:30:13
Let's try the conservation
of mechanical energy.

524
00:30:10 --> 00:30:16
The conservation
of mechanical energy tells me

525
00:30:14 --> 00:30:20
that the kinetic energy at A
plus the potential energy at A

526
00:30:19 --> 00:30:25
must be the kinetic energy at B
plus the potential energy at B,

527
00:30:24 --> 00:30:30
and that must be equal

528
00:30:26 --> 00:30:32
to the kinetic energy at C plus
the potential energy at C.

529
00:30:29 --> 00:30:35
Right?

530
00:30:31 --> 00:30:37
There is no friction, there are
no nonconservative forces,

531
00:30:34 --> 00:30:40
so this has got to work.

532
00:30:36 --> 00:30:42
Well, the potential energy at A
is zero, because remember,

533
00:30:41 --> 00:30:47
the potential energy of a spring
is one-half k x squared,

534
00:30:46 --> 00:30:52
if x is the displacement
from its relaxed position.

535
00:30:49 --> 00:30:55
So this is zero.

536
00:30:51 --> 00:30:57
When it comes to a halt here,
this is zero.

537
00:30:54 --> 00:31:00

538
00:30:56 --> 00:31:02
When I watched this lecture,
I noticed a slip of the tongue.

539
00:31:00 --> 00:31:06
Clearly the potential energy
at A is zero

540
00:31:03 --> 00:31:09
That's fine.

541
00:31:05 --> 00:31:11
But at C, the object
is standing still,

542
00:31:09 --> 00:31:15
so at C it is the kinetic energy
that is zero

543
00:31:12 --> 00:31:18
and not the potential energy.

544
00:31:15 --> 00:31:21
And I continued
the problem correctly,

545
00:31:18 --> 00:31:24
because I wrote down here
one-half k x max squared,

546
00:31:26 --> 00:31:32
which is exactly this term.

547
00:31:28 --> 00:31:34
That is the potential energy
at C, which reaches a maximum.

548
00:31:32 --> 00:31:38
It is the kinetic energy
that is zero here.

549
00:31:34 --> 00:31:40
And I accidentally mentioned

550
00:31:36 --> 00:31:42
that it was the potential energy
that is zero,

551
00:31:39 --> 00:31:45
which is obviously not so.

552
00:31:42 --> 00:31:48

553
00:31:44 --> 00:31:50
So we get

554
00:31:45 --> 00:31:51
that one-half m vA squared
equals one-half m vB squared

555
00:31:54 --> 00:32:00
minus one-half
k x of B squared--

556
00:32:00 --> 00:32:06
that is the potential energy
at this location--

557
00:32:05 --> 00:32:11
and that equals
one-half k x of C,

558
00:32:09 --> 00:32:15
which we have called x max,
squared,

559
00:32:12 --> 00:32:18
and you are in business.

560
00:32:15 --> 00:32:21
If I tell you
where the object is--

561
00:32:17 --> 00:32:23
with the pendulum, I gave you

562
00:32:19 --> 00:32:25
the initial angle,
my theta zero--

563
00:32:21 --> 00:32:27
I tell you where the object is,
so you know x of B,

564
00:32:24 --> 00:32:30
I tell you what speed I gave it,
so you know this one,

565
00:32:26 --> 00:32:32
you can calculate what
the maximum displacement is.

566
00:32:29 --> 00:32:35
Exactly the same idea,
and if you want to apply

567
00:32:32 --> 00:32:38
the work-energy theorem,
of course,

568
00:32:33 --> 00:32:39
you will get exactly
the same result.

569
00:32:35 --> 00:32:41
This is, in fact,
the same thing.

570
00:32:38 --> 00:32:44
So let's now turn to Newton's
universal law of gravity,

571
00:32:46 --> 00:32:52
and believe it or not, I think
I even give you on the exam

572
00:32:52 --> 00:32:58
the gravitational force

573
00:32:55 --> 00:33:01
of Newton's universal law
of gravity.

574
00:32:59 --> 00:33:05
I don't think you need it, but
you will find it on the exam.

575
00:33:04 --> 00:33:10
Let's take the Earth going
around the sun,

576
00:33:07 --> 00:33:13
and let's approximate
the orbit by a circle.

577
00:33:11 --> 00:33:17
So here is the Earth,
here is the sun, mass sun,

578
00:33:16 --> 00:33:22
and the orbital radius is
capital R.

579
00:33:21 --> 00:33:27
The Earth has an orbital
velocity, I call it v orbital,

580
00:33:29 --> 00:33:35
and there is on the object the
angular velocity, say, is omega.

581
00:33:35 --> 00:33:41
It goes around with
a certain angular velocity.

582
00:33:39 --> 00:33:45
And there has to be here
a force on the system,

583
00:33:42 --> 00:33:48
the gravitational force,

584
00:33:44 --> 00:33:50
which is the same
as the centripetal force.

585
00:33:47 --> 00:33:53
That's the only way the object
can go around in a circle.

586
00:33:51 --> 00:33:57
So this gravitational force
equals mass of the sun

587
00:33:57 --> 00:34:03
mass of the Earth times G
divided by R squared--

588
00:34:02 --> 00:34:08
the force of gravity.

589
00:34:05 --> 00:34:11
But it also must equal
to the centripetal acceleration,

590
00:34:09 --> 00:34:15
and this is here the mass
of the Earth, and so it also is

591
00:34:14 --> 00:34:20
the mass of the Earth times
the orbital speed v squared

592
00:34:19 --> 00:34:25
divided by the orbital radius,

593
00:34:21 --> 00:34:27
and if you prefer to write that
in terms of omega squared R,

594
00:34:26 --> 00:34:32
that's fine, too.

595
00:34:28 --> 00:34:34
Look at this here.

596
00:34:30 --> 00:34:36
You lose your Earth mass,

597
00:34:33 --> 00:34:39
if you want to know
what your orbital speed is,

598
00:34:36 --> 00:34:42
you lose one R,

599
00:34:38 --> 00:34:44
and so you find that the orbital
speed equals the mass of the sun

600
00:34:42 --> 00:34:48
times g divided by R
to the power one-half,

601
00:34:48 --> 00:34:54
and that is about
30 kilometers per second.

602
00:34:53 --> 00:34:59
I give it in kilometers
per second, but of course

603
00:34:56 --> 00:35:02
when you calculate it,
make sure you always stay in MKS

604
00:34:58 --> 00:35:04
before you make
a conversion to other units.

605
00:35:02 --> 00:35:08
So you know what the period is
to go around the sun.

606
00:35:07 --> 00:35:13
That is 2 pi R
divided by the orbital,

607
00:35:11 --> 00:35:17
and that is about 365½ days--
that's one year.

608
00:35:16 --> 00:35:22
That's how long it takes us
to go around the sun.

609
00:35:20 --> 00:35:26
Now I wonder, what is
the kinetic energy of the Earth

610
00:35:25 --> 00:35:31
in orbit around the sun?

611
00:35:27 --> 00:35:33
Well, that's completely trivial,
you would say,

612
00:35:31 --> 00:35:37
because it is one-half
of the mass of the Earth

613
00:35:35 --> 00:35:41
times the orbital speed squared.

614
00:35:38 --> 00:35:44
Indeed, that is
not very difficult.

615
00:35:44 --> 00:35:50
So the kinetic energy
of the Earth in orbit

616
00:35:49 --> 00:35:55
equals one-half mass of the
Earth times v orbit squared.

617
00:35:57 --> 00:36:03
So I take this equation
and I square it.

618
00:35:59 --> 00:36:05
I lose the half there,

619
00:36:00 --> 00:36:06
and so I get M earth M sun
times G divided by R.

620
00:36:10 --> 00:36:16
That is the kinetic energy
in orbit.

621
00:36:14 --> 00:36:20
But that...
oh, there's a half here.

622
00:36:18 --> 00:36:24
Whenever I make a slip,
you should always interrupt.

623
00:36:20 --> 00:36:26
So there is the half here.

624
00:36:22 --> 00:36:28
This is also minus one-half
the potential energy.

625
00:36:30 --> 00:36:36
And remember,
the potential energy equals

626
00:36:35 --> 00:36:41
minus M sun M earth G
divided by R.

627
00:36:44 --> 00:36:50
So here you see plus half,
here you see minus 1,

628
00:36:48 --> 00:36:54
so this is indeed minus
half the potential energy.

629
00:36:52 --> 00:36:58
What is the total energy
of the Earth around the sun?

630
00:36:58 --> 00:37:04
That means the sum of potential
energy and kinetic energy.

631
00:37:05 --> 00:37:11
Well, the kinetic energy is
minus one-half U,

632
00:37:11 --> 00:37:17
and the potential energy is U,

633
00:37:13 --> 00:37:19
and so this is...
this is plus one-half U.

634
00:37:20 --> 00:37:26
That means it's negative,
because U itself is negative.

635
00:37:26 --> 00:37:32
It is also equal to minus K.

636
00:37:30 --> 00:37:36
I just calculated for fun
how large that number is--

637
00:37:36 --> 00:37:42
it's a negative number--
how large that is,

638
00:37:38 --> 00:37:44
and it turns out to be

639
00:37:40 --> 00:37:46
approximately minus 2.7 times
ten to the 33 joules.

640
00:37:46 --> 00:37:52
Deeply negative.

641
00:37:48 --> 00:37:54
Now, if we get bored and we want
to leave the solar system,

642
00:37:54 --> 00:38:00
then we can ask ourselves
the question,

643
00:37:56 --> 00:38:02
if we mount a rocket to the
Earth and we fire that rocket,

644
00:38:00 --> 00:38:06
how much speed should we give it
to get away from the sun

645
00:38:04 --> 00:38:10
once and for all
and never come back?

646
00:38:06 --> 00:38:12
Well, all we would have to do

647
00:38:08 --> 00:38:14
is make the total energy
of the Earth zero,

648
00:38:11 --> 00:38:17
so it can cruise
all the way to infinity

649
00:38:14 --> 00:38:20
and have zero speed
when it gets there.

650
00:38:17 --> 00:38:23
How can I make this zero?

651
00:38:19 --> 00:38:25
Well, I have to add to the
kinetic energy exactly K,

652
00:38:24 --> 00:38:30
because minus K plus K is zero.

653
00:38:26 --> 00:38:32
If I have to add
the kinetic energy K,

654
00:38:31 --> 00:38:37
then the new kinetic energy
after the rocket burn

655
00:38:34 --> 00:38:40
would be exactly 2K.

656
00:38:35 --> 00:38:41
The potential energy
is not changing,

657
00:38:38 --> 00:38:44
because you fire the rocket
when you're there,

658
00:38:40 --> 00:38:46
and so you're still at the same
position after the rocket fire,

659
00:38:44 --> 00:38:50
and so your potential energy
hasn't changed,

660
00:38:46 --> 00:38:52
but your kinetic energy
must have doubled.

661
00:38:49 --> 00:38:55
If it hasn't doubled,
this doesn't become zero.

662
00:38:52 --> 00:38:58
Aha!

663
00:38:53 --> 00:38:59
But if the kinetic energy
has doubled,

664
00:38:55 --> 00:39:01
then the speed must have gone up
by the square root of two,

665
00:38:59 --> 00:39:05
because kinetic energy is
proportional to v squared,

666
00:39:02 --> 00:39:08
so the escape velocity
that you need

667
00:39:05 --> 00:39:11
must be the square root of two
times the orbital velocity,

668
00:39:09 --> 00:39:15
and that would be for the Earth
the square root of two times 30,

669
00:39:13 --> 00:39:19
which is about
42 kilometers per second.

670
00:39:16 --> 00:39:22
So by looking at the energy
considerations, you can find

671
00:39:22 --> 00:39:28
the connection between
total energy, escape velocity,

672
00:39:26 --> 00:39:32
and make the whole picture
internally consistent.

673
00:39:32 --> 00:39:38
We talked for one whole lecture
about resistive forces.

674
00:39:36 --> 00:39:42
I still remember the number
of that lecture--

675
00:39:39 --> 00:39:45
it was number 12--
because it's the one lecture

676
00:39:42 --> 00:39:48
that I worked on it
more than on any other

677
00:39:44 --> 00:39:50
with the help of my graduate
student Dave Pooley,

678
00:39:48 --> 00:39:54
who did these
wonderful calculations,

679
00:39:51 --> 00:39:57
these numerical solutions.

680
00:39:53 --> 00:39:59
And so let's talk a little bit
about the resistive forces.

681
00:39:59 --> 00:40:05
If an object moves through
a medium-- air or a liquid--

682
00:40:03 --> 00:40:09
there is a resistive force which
always opposes the velocity.

683
00:40:11 --> 00:40:17
It has two terms: k1 times v--

684
00:40:14 --> 00:40:20
this is the speed,
this is always positive--

685
00:40:17 --> 00:40:23
plus k2 times v squared,

686
00:40:22 --> 00:40:28
and this is the unit vector
of the velocity.

687
00:40:25 --> 00:40:31
It's always opposing
the velocity,

688
00:40:27 --> 00:40:33
so this is the magnitude
of the resistive force.

689
00:40:32 --> 00:40:38
We dealt exclusively
with spheres-- remember?

690
00:40:36 --> 00:40:42
And we did some
huge experiments.

691
00:40:39 --> 00:40:45
For spheres,

692
00:40:40 --> 00:40:46
k1 equals c1 times r and
k2 equals c2 times r squared.

693
00:40:51 --> 00:40:57
And we did some experiments
with corn syrup.

694
00:40:56 --> 00:41:02
And this was a syrup for which
c1 was approximately 160

695
00:41:02 --> 00:41:08
and c2 was

696
00:41:04 --> 00:41:10
about 1.2 times ten to the third
kilograms per cubic meters.

697
00:41:10 --> 00:41:16
I always remember
the units of this

698
00:41:12 --> 00:41:18
because that's always density.

699
00:41:14 --> 00:41:20
In fact, it's always
a little smaller

700
00:41:17 --> 00:41:23
than the actual density of the
liquid, of the syrup itself.

701
00:41:21 --> 00:41:27
This is a more complicated unit,

702
00:41:22 --> 00:41:28
but you can figure that one out
for yourself.

703
00:41:25 --> 00:41:31
And so we had ball bearings

704
00:41:27 --> 00:41:33
from an eighth-of-an-inch
diameter

705
00:41:29 --> 00:41:35
to a quarter of an inch,

706
00:41:31 --> 00:41:37
and we dropped them
in the Karo syrup.

707
00:41:34 --> 00:41:40
If we take the one for now,

708
00:41:36 --> 00:41:42
which is
a quarter-of-an-inch diameter,

709
00:41:38 --> 00:41:44
then the mass of that one is
about 0.1 grams.

710
00:41:43 --> 00:41:49
We knew the density of steel,

711
00:41:46 --> 00:41:52
and so you can figure out the
mass, just take my word for it.

712
00:41:49 --> 00:41:55
And so the question now is,

713
00:41:51 --> 00:41:57
if this ball bearing
starts to fall,

714
00:41:53 --> 00:41:59
what is the speed
that it will achieve?

715
00:41:56 --> 00:42:02
Well, in the beginning,
there is only mg on that ball

716
00:42:01 --> 00:42:07
and a teeny-weeny
little resistive force.

717
00:42:05 --> 00:42:11
As the speed builds up,
the resistive force will grow

718
00:42:09 --> 00:42:15
and there comes a time
that it's equal to mg,

719
00:42:12 --> 00:42:18
and then the object will
not be accelerated anymore

720
00:42:15 --> 00:42:21
and it will have what we call
the terminal velocity.

721
00:42:19 --> 00:42:25
And so what are the conditions
for terminal velocity?

722
00:42:23 --> 00:42:29
That this term equals mg.

723
00:42:25 --> 00:42:31
In other words, that c1 r v
plus c2 r squared v squared

724
00:42:35 --> 00:42:41
becomes equal to mg,
and when that's the case,

725
00:42:38 --> 00:42:44
then we have reached
the terminal velocity.

726
00:42:44 --> 00:42:50
Well, this is a quadratic
equation in v of t.

727
00:42:52 --> 00:42:58
You know c1, you know c2,

728
00:42:55 --> 00:43:01
you know the radius
of these ball bearings,

729
00:42:57 --> 00:43:03
you know the mass
of the ball bearings,

730
00:42:59 --> 00:43:05
so you can solve
for the terminal velocity.

731
00:43:01 --> 00:43:07
You will get two solutions;
it's a quadratic equation.

732
00:43:04 --> 00:43:10
One will be nonphysical,
you will see,

733
00:43:06 --> 00:43:12
so you keep
the one that is physical.

734
00:43:09 --> 00:43:15
But if you did that, that
would be a silly thing to do.

735
00:43:12 --> 00:43:18
And the reason why
that would be silly is

736
00:43:15 --> 00:43:21
that I claim that this term...

737
00:43:17 --> 00:43:23
which by the way we call
the pressure term,

738
00:43:20 --> 00:43:26
that this term can be completely
ignored compared to this term,

739
00:43:25 --> 00:43:31
which we call the viscous term.

740
00:43:28 --> 00:43:34
How do we know that that
pressure term can be ignored?

741
00:43:32 --> 00:43:38
Well, let's first ask
the question,

742
00:43:34 --> 00:43:40
when is the viscous term
and the pressure term the same

743
00:43:38 --> 00:43:44
for what velocity?

744
00:43:40 --> 00:43:46
So forget for now
that here are t's.

745
00:43:41 --> 00:43:47
We just want to know that this
term is the same as this term.

746
00:43:45 --> 00:43:51
That will be the case
when the velocity,

747
00:43:47 --> 00:43:53
which we call
the critical velocity,

748
00:43:50 --> 00:43:56
equals c1 divided by c2 times r.

749
00:43:53 --> 00:43:59
It's very easy to show

750
00:43:55 --> 00:44:01
that this term is then
exactly identical to that one.

751
00:43:58 --> 00:44:04
There's nothing critical
about that number.

752
00:44:01 --> 00:44:07
Really, it's a wrong name,
critical velocity,

753
00:44:04 --> 00:44:10
but we give it that name.

754
00:44:05 --> 00:44:11
It simply means
that the two terms are equal--

755
00:44:07 --> 00:44:13
that's all it means,
nothing else.

756
00:44:09 --> 00:44:15
If you calculate

757
00:44:11 --> 00:44:17
for a quarter-inch-diameter
steel ball bearing,

758
00:44:16 --> 00:44:22
you can calculate r,
you will find

759
00:44:18 --> 00:44:24
that the critical velocity is
about 100 miles per hour,

760
00:44:21 --> 00:44:27
and we know very well
there is no way

761
00:44:24 --> 00:44:30
when you drop a quarter-inch
ball bearing into this syrup

762
00:44:27 --> 00:44:33
that it's going to come
anywhere near that,

763
00:44:31 --> 00:44:37
so the speed will remain
way below 100 miles per hour,

764
00:44:35 --> 00:44:41
and therefore this is a term
that will dominate

765
00:44:38 --> 00:44:44
and this term can be
completely ignored.

766
00:44:41 --> 00:44:47
And if that term can be ignored,
you can see that immediately

767
00:44:45 --> 00:44:51
the terminal velocity is going
to be mg divided by c1r,

768
00:44:50 --> 00:44:56
and that, by the way, for
our quarter-inch ball bearing,

769
00:44:54 --> 00:45:00
turns out to be something like
two centimeters per second--

770
00:44:58 --> 00:45:04
many, many orders
of magnitude lower

771
00:45:00 --> 00:45:06
than the 100 miles per hour,
which was the critical velocity.

772
00:45:04 --> 00:45:10
What we did is we measured the
times for these ball bearings

773
00:45:09 --> 00:45:15
to fall over a distance
of four centimeters.

774
00:45:12 --> 00:45:18
And if you know the time to go
a distance of four centimeters,

775
00:45:16 --> 00:45:22
you know the terminal velocity.

776
00:45:18 --> 00:45:24
And we know the mass
of the ball bearing,

777
00:45:20 --> 00:45:26
we know the radii of
the ball bearing, we know c1...

778
00:45:23 --> 00:45:29
no, we didn't know c1.

779
00:45:25 --> 00:45:31
I just wanted
to point out to you

780
00:45:26 --> 00:45:32
that you can measure c1
that way.

781
00:45:28 --> 00:45:34
You measure the speed,
you know r, you know m,

782
00:45:31 --> 00:45:37
you measure c1,

783
00:45:32 --> 00:45:38
and that's what I did before
the lecture that morning.

784
00:45:34 --> 00:45:40
That's how I came up
with that number 160,

785
00:45:37 --> 00:45:43
otherwise I wouldn't have known.

786
00:45:39 --> 00:45:45
c1 is extremely
temperature-dependent.

787
00:45:40 --> 00:45:46
In fact, it even changed
between classes.

788
00:45:43 --> 00:45:49
If the temperature goes up,
the value for c1 goes down.

789
00:45:48 --> 00:45:54
So that's the way you can
measure the value for c1.

790
00:45:53 --> 00:45:59
Okay, conservation of momentum.

791
00:45:59 --> 00:46:05
We dealt with the conservation
of momentum in one lecture--

792
00:46:02 --> 00:46:08
your parents were here--
and we only dealt exclusively

793
00:46:05 --> 00:46:11
with completely
inelastic collisions.

794
00:46:10 --> 00:46:16
Whenever we have

795
00:46:12 --> 00:46:18
a system of objects
with zero net external force,

796
00:46:18 --> 00:46:24
the momentum must be conserved.

797
00:46:20 --> 00:46:26
They can collide,
they can cause fireworks,

798
00:46:23 --> 00:46:29
they can explode,
they can break up in pieces,

799
00:46:27 --> 00:46:33
but momentum will be conserved

800
00:46:30 --> 00:46:36
in the absence
of any net external force.

801
00:46:34 --> 00:46:40
It's a very
nonintuitive concept.

802
00:46:37 --> 00:46:43
Kinetic energy can be destroyed,
but momentum cannot be destroyed

803
00:46:41 --> 00:46:47
in the absence of a net external
force on the system as a whole.

804
00:46:46 --> 00:46:52
We have an object m1 here,
speed v1, object m2, speed v2.

805
00:46:53 --> 00:46:59
Let's make it very simple--

806
00:46:55 --> 00:47:01
m1 is one kilogram,
v1 is three meters per second,

807
00:47:00 --> 00:47:06
m2 is two kilograms,
and v2 equals...

808
00:47:03 --> 00:47:09
I make this
five meters per second,

809
00:47:06 --> 00:47:12
and I make this
three meters per second.

810
00:47:10 --> 00:47:16
Momentum is conserved,

811
00:47:11 --> 00:47:17
so I can write down that
m1 times v1 plus m2 times v2

812
00:47:19 --> 00:47:25
must be m1 plus m2-- it is a
completely inelastic collision--

813
00:47:22 --> 00:47:28
times v prime.

814
00:47:24 --> 00:47:30
They come together,
they stick together,

815
00:47:27 --> 00:47:33
the total mass is going
to be m1 plus m2,

816
00:47:30 --> 00:47:36
and they're going to fly

817
00:47:31 --> 00:47:37
either in this direction
or in this direction,

818
00:47:33 --> 00:47:39
but whatever it is,
I call it v prime--

819
00:47:35 --> 00:47:41
it's after the collision.

820
00:47:38 --> 00:47:44
You may say, "Hey,
you forgot your arrows here,"

821
00:47:41 --> 00:47:47
because this is
a vector equation.

822
00:47:43 --> 00:47:49
Yeah, I purposely forgot them,

823
00:47:45 --> 00:47:51
because this is
a one-dimensional case.

824
00:47:47 --> 00:47:53
You can always leave
the arrows off,

825
00:47:49 --> 00:47:55
because the signs
take care of it.

826
00:47:50 --> 00:47:56
And you, of course,
have to be generous,

827
00:47:52 --> 00:47:58
namely you have to put
the minus signs in when needed.

828
00:47:56 --> 00:48:02
mv is one times five,
that is plus five.

829
00:48:00 --> 00:48:06
m2 v2 is going to be negative,

830
00:48:03 --> 00:48:09
because I must now observe
this as a minus three,

831
00:48:06 --> 00:48:12
so I get minus six,

832
00:48:07 --> 00:48:13
so the momentum before and after
the collision must be minus one

833
00:48:13 --> 00:48:19
and that must be equal
to three times v prime.

834
00:48:18 --> 00:48:24
And v prime is therefore minus
one-third meters per second.

835
00:48:22 --> 00:48:28
And so this object will stick,
and after the collision,

836
00:48:26 --> 00:48:32
they will move with a speed of
one-third of a meter per second.

837
00:48:30 --> 00:48:36
Kinetic energy was
almost exclusively lost.

838
00:48:33 --> 00:48:39
There was almost
nothing left, remember?

839
00:48:35 --> 00:48:41
Clearly that's
because of friction--

840
00:48:37 --> 00:48:43
not external friction;
there is no external friction,

841
00:48:40 --> 00:48:46
because if there were
any external friction,

842
00:48:42 --> 00:48:48
momentum would not be conserved.

843
00:48:44 --> 00:48:50
Think of them
as the two car wrecks--

844
00:48:46 --> 00:48:52
they go into each other.

845
00:48:47 --> 00:48:53
Fireworks!

846
00:48:48 --> 00:48:54
Man, you hear them scratch,
metal over metal.

847
00:48:51 --> 00:48:57
A lot of heat is produced,
kinetic energy is removed.

848
00:48:55 --> 00:49:01
But momentum is conserved,

849
00:48:57 --> 00:49:03
because those frictional forces
are all internal.

850
00:49:00 --> 00:49:06
And that's fine.

851
00:49:02 --> 00:49:08
So that is why it is such
a very nonintuitive concept.

852
00:49:08 --> 00:49:14
Let's now make sure that this
crazy object has come to a halt.

853
00:49:15 --> 00:49:21
If not, I... oh, I better not
say what I was going to do.

854
00:49:19 --> 00:49:25
My goodness!
It's still rotating!

855
00:49:23 --> 00:49:29
Now, I realize that it may cause
you sleepless nights,

856
00:49:29 --> 00:49:35
but on the other hand,
with an exam coming up,

857
00:49:31 --> 00:49:37
I would suggest
you just forget about this.

858
00:49:33 --> 00:49:39
Pretend you haven't seen it.

859
00:49:35 --> 00:49:41
Pretend this never happened.

860
00:49:37 --> 00:49:43
That may make you happier.

861
00:49:39 --> 00:49:45
Okay, see you Monday,
and good luck.

862
00:49:41 --> 00:49:47

863
00:49:50 --> 00:49:56.000