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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
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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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And so the one that
we really want to know
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is what is h1 plus h2?
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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,
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we immediately have
the maximum angle.
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Now, h1 is a piece of cake,
because we know that point B...
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that this object is at B when
theta zero is five degrees.
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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--
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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
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It's only 3.8 millimeters.
260
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It's a teeny-weeny little bit.
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It's a large displacement
in this direction
262
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but very little
in this direction.
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So we know already what h1 is.
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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
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the gravitational potential
energy at point A,
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let's call that zero.
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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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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
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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