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You see here the topics that
will be on your plate on Monday.

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It's clearly not possible for me

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in one exam
to cover all these topics,

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so I will have to make
a choice Monday.

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Today I also have
to make a choice.

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I cannot do justice to all
these topics in any depth,

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so I will butterfly over them

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and some of them
I won't even touch at all.

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However, what is not covered
today may well be on the exam.

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You will get loads of equations.

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Almost every equation
that I could think of

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will be on your exam.

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There's also special tutoring
this weekend--

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you can check that on the Web--
Saturday and Sunday.

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Let's start with a completely
inelastic collision.

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Completely inelastic collision.

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We have mass m1,
we have mass m2.

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It's a one-dimensional problem.

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This one has velocity v1,
and this has velocity v2.

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They collide, and after the
collision they are together--

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because that's what it means

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when the collision
is completely inelastic--

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and they have a velocity,
v prime.

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If there is no net
external force on the system,

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momentum must be conserved.

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So I now have
that m1 v1 plus m2 v2

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must be m1 plus m2
times v prime.

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One equation with one unknown--
v prime follows immediately.

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You may say,

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"Gee, you should really
have put arrows over here."

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Well, in the case that it is
a one-dimensional collision,

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you can leave the arrows off

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because the signs automatically
take care of that.

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Kinetic energy is not conserved.

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Before the collision,
your kinetic energy

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equals one-half m1 v1 squared
plus one-half m2 v2 squared.

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After the collision,
kinetic energy

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equals one-half m1 plus m2
times v prime squared.

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And you can easily prove that
this is always less than that

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in case of a completely
inelastic collision.

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There's always
kinetic energy destroyed,

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which then comes out
in the form of heat.

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Let's do now
an elastic collision.

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And I add the word
"completely" elastic,

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but "elastic" itself is enough

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because that means that
kinetic energy is conserved.

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I start with the same initial
condition: m1 v1, m2 v2,

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but now, after the collision,

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m1 could either go this way
or this way, I don't know.

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So this could be v1 prime,
this could be v1 prime.

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m2, however, will always go
into this direction.

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That's clear, because if you get
hit from behind by object one,

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after the collision, you
obviously go in this direction.

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Again, we don't have
to put the arrows over it,

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because the signs take care of
it in a one-dimensional case.

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This will be plus, then.

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That could be... you could adopt
that as your convention,

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and if it goes
in this direction,

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then it will be negative.

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So if you find, for v1 prime,
minus something,

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it means it's bounced back.

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So now we can apply
the conservation of momentum

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if there is no external force
on the system.

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Internal force is fine.

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All right, so now we have

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m1 v1 plus m2 v2 equals
m1 v1 prime plus m2 v2 prime--

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conservation of momentum.

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Now we get the conservation
of kinetic energy,

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because we know it's
an elastic collision.

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One-half m1 v1 squared,
one-half m2 v2 squared--

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that's before the collision.

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After the collision,
one-half m1 v1 prime squared

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plus one-half m2 v2
prime squared.

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Two equations with two unknowns

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and in principle, you can solve
for v1 prime and for v2 prime,

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except that it could be
time-consuming.

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And so on exams,
what is normally done

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when you get a problem
like that...

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Normally you get a problem
whereby either m1 is made m2

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or m1 is much,
much larger than m2

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or m1 is much,
much smaller than m2,

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like banging a basketball
onto a Ping-Pong ball

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or a Ping-Pong ball
onto a basketball.

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I will do a very simple example

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whereby I will take now
for you m1 equals m2,

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and I will call that m

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and I will even simplify
the problem by making v2 zero,

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so the second object
is standing still.

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One-dimensional collision,
one hits that object.

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Very special case.

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So the conservation of momentum
now becomes much simpler.

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m times v1-- there is no v2--

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equals m v1 prime
plus m v2 prime.

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Conservation of kinetic energy

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is one-half m v1 squared equals
one-half m v1 prime squared

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plus one-half
m v2 prime squared.

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Notice that now I lose my "m"s,
which is very convenient.

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Here I lose
my one-half "m"s, even

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and this is very easy to solve.

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If you square this equation

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you get something that will look
very similar to this.

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If you square it,

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you get v1 squared
equals v1 prime squared

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plus v2 prime squared
plus two v1 prime v2 prime.

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And compare this equation
with this equation--

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they're almost identical,
except for this term,

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so this term must be zero.

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But we know that v2 prime
is not zero.

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It got kicked
and so it'll go forward.

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So what that means, then,
is that v1 prime equals zero

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and if v1 prime equals zero,
you see that v2 prime equals v1.

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And this is that classic case

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whereby a ball hits
another ball.

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This one has no speed.

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It hits it with
a certain velocity, v1.

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They have the same mass.

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After the collision,
this one stands still

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and this one takes over
the speed.

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Famous Newton's Cradle.

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You see it often with pendulums.

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It is the logo
on the 8.01 home page,

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and I showed you
a demonstration here

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when we discussed that
in lectures.

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All right, let's move on,
and let's do something now

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on torques, angular momentum,
rotation

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and let's discuss
the Atwood machine.

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The Atwood machine
is a clever device

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that allows you to measure, to
a reasonable degree of accuracy,

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the gravitational acceleration.

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Here's a pulley, and the pulley
has mass m, has radius R.

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It's solid,
so it's a solid disk--

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rotates, frictionless,
about point P, radius R.

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And there is a rope here, near
massless-- we ignore the mass.

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Mass m2 is here
and mass m1 is here,

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and let's assume that m2
is larger than m1.

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So this will be accelerated
in this direction,

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this will be accelerated
in this direction,

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and this will start to rotate
with angular velocity omega,

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which will be
a function of time.

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And now the first thing
we want to do

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is to make up
"free-body" diagrams.

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"Free-body" diagrams,
for this one, is easy.

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m1 g down... and T1 up.

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For this one, we have m2 g down
and we have T2 up.

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For the pulley, it's
a little bit more complicated.

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This is that point P.

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If here is a tension T1 that's
pulling down on the pulley--

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so this is T1--

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and this T2 is pulling down
on the pulley-- so there's T2--

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it has a mass,
so it has weight mg.

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The sum of all forces
on the pulley must be zero;

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otherwise it would accelerate
itself down, which it doesn't.

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And so there has to be
a force up-- I'll call it n.

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And that force n has to cancel
out all these three forces.

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So that must be
T1 plus T2 plus mg.

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We will not need it any further
in our calculations,

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but there has to be a force to
hold that in place, so to speak.

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So now we're going
to calculate the acceleration

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under the condition
that the rope does not slip.

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That means there
is friction with the pulley--

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not friction here, but here.

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Otherwise, the rope would slip,
the rope would slip.

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What it means,
if there is no slip,

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that if the rope moves
one centimeter

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that the wheel also turns

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at the circumference
one centimeter.

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That's what it means
when there is no slip.

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That means the velocity
of the rope--

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v of r, v of the rope--

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must be omega times
R of the pulley.

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That's what "no slip" means.

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So the acceleration--

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which is the derivative
of that velocity of the rope--

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A, is omega dot times R,
which is alpha times R.

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Omega is the angular velocity

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and alpha is
the angular acceleration.

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This is a condition... it is an
important condition for no slip.

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So let's now start
at object number one

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and write down
Newton's second law.

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I call this the positive
direction for object one

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and I call this the positive
direction for object two--

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just easier for me.

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So we get T1 minus m1 g
must be m1 a-- one equation.

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I don't know what T1 is,
I don't know what a is.

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Second equation, for this one.

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I call this
the positive direction.

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m2 g minus T2 must be m2 a--
second equation.

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One unknown has been added,
so I need more.

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Of course I need more.

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I also have to think
about the pulley.

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The pulley... the net force
on the pulley is zero.

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That's why this stays in place,
but it is going to rotate

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because this force T2
is larger than this T1.

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There is a torque
relative to that point P,

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and torque is defined
as r cross F.

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The torque relative
to point P,

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the magnitude is this position
vector times this force.

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That is a torque
in the blackboard.

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What is in the blackboard,
I will call positive.

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The torque, due to this force,
is out of the blackboard,

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and I will call that negative.

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Since this angle is 90 degrees,

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I simply get that the torque
relative to point P

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equals the radius
of the cylinder--

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the radius of that pulley--
times T2.

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That's the positive part,

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and the negative part
is the radius times T1.

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Notice that this force
and this force go through P,

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do not contribute to the torque.

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And that now equals

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the moment of inertia about
that point P, times alpha.

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But since we have no slip,
alpha is a divided by R.

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So it's the moment
of inertia about point P

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times a divided by R.

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But since it is a rotating disk

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which is rotating
about its center of mass...

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I know the moment of inertia,
I looked that up--

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if you need it during your exam,
you will find that in the exam--

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that is one-half MR squared...

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one-half MR squared
times a divided by R

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and I lose one R,
and so I find, then,

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that T2 minus T1
equals one-half Ma.

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Notice I also lose my second R.

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And so now I have
a third equation

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and I can solve for T2,
I can solve for T1

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and I can solve for a.

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And you can do that
as well as I can.

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If you find the result,

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you should always do
a little bit of testing

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to make sure
that your result makes sense.

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And what you should do
is you should say...

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You should make sure
that m2 g is larger than T2.

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That's a must-- otherwise,
it's not being accelerated down.

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You should also check that
T1 comes out larger than m1 g.

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That's also a must.

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And you should also check
that T2 be larger than T1.

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00:15:01 --> 00:15:07
Otherwise, the pulley wouldn't
rotate in clockwise direction.

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00:15:05 --> 00:15:11
It would also be useful,
which is a trivial check,

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00:15:09 --> 00:15:15
to stick in your results
m1 equals m2.

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That should give you that
the acceleration should be zero

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and it should give you
that T1 equals T2.

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00:15:20 --> 00:15:26
Those are obvious things and
that can be done very simply.

252
00:15:23 --> 00:15:29
It takes you no more
than ten seconds.

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And if it's... any one
of these is not met,

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00:15:27 --> 00:15:33
then somewhere you slipped up

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00:15:29 --> 00:15:35
and it will give you
an opportunity

256
00:15:30 --> 00:15:36
to go over the problem again.

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00:15:32 --> 00:15:38
 

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00:15:34 --> 00:15:40
All right, let's now do
another problem--

259
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simple harmonic oscillation
of a physical pendulum.

260
00:15:42 --> 00:15:48
 

261
00:15:44 --> 00:15:50
I have here a rod--
mass M, length l--

262
00:15:50 --> 00:15:56
rotating here about an axis
perpendicular to the blackboard

263
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without friction.

264
00:15:57 --> 00:16:03
Here is the center of the rod,
and let this angle be theta.

265
00:16:03 --> 00:16:09
 

266
00:16:07 --> 00:16:13
There is a torque
relative to point P.

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00:16:09 --> 00:16:15
There is also a force
that goes through point P.

268
00:16:13 --> 00:16:19
I'm not even interested
in that force.

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I know that it's here, mg.

270
00:16:17 --> 00:16:23
Since I'm going to take
the torque relative to point P,

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I don't worry about the force,

272
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but there has to be a force
through point P.

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Otherwise this ruler
would be accelerated down

274
00:16:29 --> 00:16:35
with acceleration g, if this
is the only force there were.

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But we know that's not the case,
it's going to swing.

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So there has to be a force to P.

277
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I don't want to know what it is,
but there has to be one.

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So the torque relative
to point P-- the magnitude--

279
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is the position vector, r of P,
from here to here

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00:16:51 --> 00:16:57
crossed with this force.

281
00:16:55 --> 00:17:01
And so that makes it
one-half l times mg

282
00:17:04 --> 00:17:10
times the sine of the angle,
and that's... the angle is theta

283
00:17:08 --> 00:17:14
so that is times
the sine of theta.

284
00:17:12 --> 00:17:18
The cross product has
the sine of the angle in it.

285
00:17:16 --> 00:17:22
So this is the torque
relative to point P

286
00:17:18 --> 00:17:24
which also must be

287
00:17:20 --> 00:17:26
the moment of inertia
about point P times alpha.

288
00:17:24 --> 00:17:30
No different from what
we just had with the pulley.

289
00:17:26 --> 00:17:32
So alpha equals omega dot.

290
00:17:32 --> 00:17:38
It's also theta double dot.

291
00:17:35 --> 00:17:41
This omega is the angular
velocity-- it's d theta/dt

292
00:17:41 --> 00:17:47
and the derivative gives me
the angular acceleration.

293
00:17:47 --> 00:17:53
 

294
00:17:50 --> 00:17:56
There has to be a minus
sign here, that's important

295
00:17:53 --> 00:17:59
because the torque
is restoring--

296
00:17:55 --> 00:18:01
the same situation we had
when we had a spring.

297
00:18:00 --> 00:18:06
We were oscillating a spring,
the spring force is minus kx.

298
00:18:04 --> 00:18:10
The minus sign here plays
exactly the same role.

299
00:18:08 --> 00:18:14
So I can write down here

300
00:18:11 --> 00:18:17
minus I of p
theta double dot.

301
00:18:15 --> 00:18:21
Now, if we take
a small angle approximation--

302
00:18:18 --> 00:18:24
"small angle approximation"--

303
00:18:24 --> 00:18:30
then the sine of theta
is approximately theta,

304
00:18:30 --> 00:18:36
if theta is in radians.

305
00:18:32 --> 00:18:38
And so I can replace now
the sine of theta here by theta

306
00:18:37 --> 00:18:43
and so now I find, if I bring
this to the other side,

307
00:18:44 --> 00:18:50
I get theta double dot
plus one-half lMg

308
00:18:50 --> 00:18:56
divided by the moment
of inertia about point P

309
00:18:53 --> 00:18:59
times theta equals zero.

310
00:18:55 --> 00:19:01
Needless to say that I am happy
like a clam at high tide

311
00:19:00 --> 00:19:06
because I see here an equation
which clearly tells me

312
00:19:04 --> 00:19:10
that we have
a simple harmonic oscillation--

313
00:19:07 --> 00:19:13
theta double dot plus a constant
times theta is zero.

314
00:19:12 --> 00:19:18
And so we must get as a solution

315
00:19:15 --> 00:19:21
that theta must be
some maximum angle

316
00:19:19 --> 00:19:25
times the cosine of omega t
plus or minus phi.

317
00:19:24 --> 00:19:30
This omega hasnothing to do
with that omega.

318
00:19:29 --> 00:19:35
This is the angular frequency.

319
00:19:35 --> 00:19:41
This is related to
the period T of the oscillation,

320
00:19:38 --> 00:19:44
which is two pi
divided by omega.

321
00:19:40 --> 00:19:46
This is a constant.

322
00:19:41 --> 00:19:47
This omega is not a constant.

323
00:19:44 --> 00:19:50
It is unfortunate, in physics,
that we use the same symbol.

324
00:19:48 --> 00:19:54
The angular velocity is zero
when the object stands still;

325
00:19:52 --> 00:19:58
is a maximum
when the object is here.

326
00:19:54 --> 00:20:00
That omega is always the same.

327
00:19:56 --> 00:20:02
It's related to how long it
takes to make one oscillation.

328
00:20:01 --> 00:20:07
Both are called omega,
both are radians per second.

329
00:20:05 --> 00:20:11
Couldn't be more confusing.

330
00:20:09 --> 00:20:15
All right, if we find I of P

331
00:20:11 --> 00:20:17
then we can solve for
the frequency, angular frequency

332
00:20:15 --> 00:20:21
and we can solve for the period.

333
00:20:17 --> 00:20:23
So let's do the... calculate the
moment of inertia about point P.

334
00:20:23 --> 00:20:29
In order to do that, we have to
apply the Parallel-Axis Theorem

335
00:20:28 --> 00:20:34
because you will
probably be given

336
00:20:30 --> 00:20:36
the moment of inertia
for rotation about the axis

337
00:20:34 --> 00:20:40
through the center of mass
parallel to this axis.

338
00:20:37 --> 00:20:43
And then you will have to add

339
00:20:39 --> 00:20:45
the mass times the distance
between these two axes squared

340
00:20:44 --> 00:20:50
to apply
the Parallel Axis Theorem.

341
00:20:46 --> 00:20:52
So this is the moment of inertia
about the center of mass

342
00:20:50 --> 00:20:56
plus m times
the distance squared.

343
00:20:53 --> 00:20:59
And the distance between
these two axes is one-half l,

344
00:20:57 --> 00:21:03
so this is
one-quarter l squared.

345
00:21:00 --> 00:21:06
I look up in a table
what the moment of inertia is

346
00:21:04 --> 00:21:10
for rotation of a rod
about its center--

347
00:21:08 --> 00:21:14
a perpendicular rod
is perpendicular to the axis--

348
00:21:11 --> 00:21:17
and that is 1/12 Ml squared
plus one-quarter Ml squared

349
00:21:19 --> 00:21:25
give me one-third Ml squared.

350
00:21:24 --> 00:21:30
So I know now what I of P is,

351
00:21:27 --> 00:21:33
and so now I can solve
for the value for omega--

352
00:21:31 --> 00:21:37
the angular frequency.

353
00:21:33 --> 00:21:39
I'll do that here

354
00:21:35 --> 00:21:41
so that we keep everything
on one blackboard.

355
00:21:39 --> 00:21:45
So this term here,
which is omega squared...

356
00:21:44 --> 00:21:50
Omega squared equals

357
00:21:47 --> 00:21:53
one-half lMg divided
by one-third Ml squared.

358
00:22:00 --> 00:22:06
I lose one l, I lose one M--

359
00:22:03 --> 00:22:09
very common, always lose
your "M"s in gravity--

360
00:22:07 --> 00:22:13
and so this is three-halves
g divided by l.

361
00:22:11 --> 00:22:17
And so the period
of one oscillation

362
00:22:15 --> 00:22:21
is two pi divided by omega--

363
00:22:18 --> 00:22:24
omega is the square root
of this--

364
00:22:22 --> 00:22:28
so that becomes the square root
of two-thirds l over g.

365
00:22:27 --> 00:22:33
 

366
00:22:29 --> 00:22:35
And that is the period of
the oscillation of this ruler.

367
00:22:33 --> 00:22:39
We worked on something similar
in lectures

368
00:22:36 --> 00:22:42
and we even measured the period

369
00:22:38 --> 00:22:44
and we found
very, very good agreement

370
00:22:40 --> 00:22:46
with the theoretical prediction.

371
00:22:43 --> 00:22:49
 

372
00:22:46 --> 00:22:52
You can ask, now, what is
the kinetic energy of rotation

373
00:22:49 --> 00:22:55
of this rod,
which changes with time?

374
00:22:51 --> 00:22:57
Kinetic energy of rotation
is one-half I omega squared.

375
00:22:57 --> 00:23:03
Remember, the linear kinetic
energy is one-half mv squared.

376
00:23:03 --> 00:23:09
The m becomes I when you go to
rotation, and v becomes omega.

377
00:23:09 --> 00:23:15
So the kinetic energy
of rotation

378
00:23:12 --> 00:23:18
equals
one-half I omega squared--

379
00:23:17 --> 00:23:23
you'll find this equation
on your exam.

380
00:23:21 --> 00:23:27
This is I about that point P.

381
00:23:25 --> 00:23:31
You know what theta is,
as a function of time.

382
00:23:29 --> 00:23:35
So omega equals d theta/dt,
so you can find what omega is.

383
00:23:34 --> 00:23:40
You know what I of P is,
we just calculated it.

384
00:23:38 --> 00:23:44
And so you know what the kinetic
energy of rotation is

385
00:23:41 --> 00:23:47
at any moment in time.

386
00:23:43 --> 00:23:49
It will change.

387
00:23:44 --> 00:23:50
It will be zero
when this comes to a halt,

388
00:23:48 --> 00:23:54
and it will be a maximum
when it is here.

389
00:23:51 --> 00:23:57
 

390
00:23:53 --> 00:23:59
It's a continuous conversion

391
00:23:55 --> 00:24:01
from gravitational potential
energy, which is a maximum here,

392
00:23:58 --> 00:24:04
to kinetic energy,
which is a maximum here.

393
00:24:05 --> 00:24:11
And so this will obviously
change with time--

394
00:24:08 --> 00:24:14
the kinetic energy of rotation.

395
00:24:10 --> 00:24:16
 

396
00:24:12 --> 00:24:18
I thought problem 8-1
was a nice example

397
00:24:17 --> 00:24:23
of how to apply Kepler's laws,
Newton's laws of gravity

398
00:24:24 --> 00:24:30
in the case that we have
an elliptical orbit.

399
00:24:29 --> 00:24:35
Here we have a planet,
mass M, radius R.

400
00:24:33 --> 00:24:39
It's not rotating
and it has no atmosphere.

401
00:24:36 --> 00:24:42
Grant you,
a little bit artificial.

402
00:24:42 --> 00:24:48
We launch a satellite

403
00:24:44 --> 00:24:50
and the satellite
gets a velocity here

404
00:24:48 --> 00:24:54
which was called v zero

405
00:24:51 --> 00:24:57
as it is right here
at the surface

406
00:24:54 --> 00:25:00
and this angle, 20 degrees.

407
00:24:57 --> 00:25:03
And we were told
that the point farthest away

408
00:25:02 --> 00:25:08
was five planets' radii away.

409
00:25:05 --> 00:25:11
So if I try to sketch
the elliptical orbit--

410
00:25:08 --> 00:25:14
assuming that the velocity
is instantaneously reached;

411
00:25:12 --> 00:25:18
that means, this point must
also be a point of the orbit.

412
00:25:16 --> 00:25:22
If I try to make a sketch,

413
00:25:18 --> 00:25:24
then that elliptical orbit
would look something like this.

414
00:25:24 --> 00:25:30
 

415
00:25:31 --> 00:25:37
Something like this.

416
00:25:33 --> 00:25:39
And then, this distance
is five R.

417
00:25:36 --> 00:25:42
I call this point A.

418
00:25:40 --> 00:25:46
It's the point that is farthest
away from the center.

419
00:25:43 --> 00:25:49
It's clear that the satellite
will crash back onto the planet,

420
00:25:48 --> 00:25:54
but that's no concern to us.

421
00:25:50 --> 00:25:56
 

422
00:25:52 --> 00:25:58
When I draw this ellipse,
I made the assumption

423
00:25:55 --> 00:26:01
that all the mass of the planet
was inside the ellipse--

424
00:26:00 --> 00:26:06
for instance, somewhere here.

425
00:26:02 --> 00:26:08
Now, that is not the case,

426
00:26:04 --> 00:26:10
so only this part
of the ellipse is realistic

427
00:26:07 --> 00:26:13
and this part, of course,
is not.

428
00:26:10 --> 00:26:16
 

429
00:26:12 --> 00:26:18
What you're being asked
is to calculate what v zero was

430
00:26:16 --> 00:26:22
only based on the information

431
00:26:18 --> 00:26:24
of the 5R and the 20 degrees
and on the mass.

432
00:26:22 --> 00:26:28
And I can add to it

433
00:26:23 --> 00:26:29
what is the velocity here at A,
when it is farthest away,

434
00:26:27 --> 00:26:33
and I can even add to it

435
00:26:29 --> 00:26:35
what is the semimajor axis
where this is 2a?

436
00:26:35 --> 00:26:41
All of that comes for free
with that initial condition

437
00:26:40 --> 00:26:46
and the knowledge that it goes
out five planets' radii.

438
00:26:44 --> 00:26:50
The angular momentum, l,
equals R cross P.

439
00:26:51 --> 00:26:57
Angular momentum is conserved

440
00:26:54 --> 00:27:00
for this object in orbit--
mass m-- but only if you take

441
00:27:00 --> 00:27:06
the angular momentum
relative to this point.

442
00:27:04 --> 00:27:10
I don't know
what I called this point.

443
00:27:07 --> 00:27:13
It looks like I called it P.

444
00:27:09 --> 00:27:15
I can hardly read it,
I'll call it C for now.

445
00:27:13 --> 00:27:19
So angular momentum
of this object in this orbit

446
00:27:16 --> 00:27:22
is not conserved relative
to any point,

447
00:27:18 --> 00:27:24
but it's only conserved
relative to this point.

448
00:27:21 --> 00:27:27
So the angular momentum

449
00:27:24 --> 00:27:30
relative to point C
at this moment in time

450
00:27:28 --> 00:27:34
is the position vector--
which is R-- times the velocity,

451
00:27:33 --> 00:27:39
but I have to multiply it
by the sine of 20 degrees.

452
00:27:37 --> 00:27:43
So it is m... it is mv, right?

453
00:27:41 --> 00:27:47
so we get a v zero, here.

454
00:27:44 --> 00:27:50
We get a capital R here, and we
get the sine of 20 degrees.

455
00:27:49 --> 00:27:55
R times... indeed.

456
00:27:51 --> 00:27:57
That must also be the angular
momentum right here.

457
00:27:56 --> 00:28:02
The fact that the angular
momentum is in the blackboard

458
00:27:59 --> 00:28:05
is no concern to me;

459
00:28:00 --> 00:28:06
I just want to know
the magnitude.

460
00:28:02 --> 00:28:08
When the object is here,

461
00:28:03 --> 00:28:09
the angular momentum
is the position vector

462
00:28:07 --> 00:28:13
which has length 5R.

463
00:28:08 --> 00:28:14
The velocity is v of A,

464
00:28:10 --> 00:28:16
but the angle is now 90 degrees,
so the sine of the angle is one.

465
00:28:15 --> 00:28:21
So I get m times vA
times five R.

466
00:28:20 --> 00:28:26
I lose my little m, and I have
one equation with two unknowns.

467
00:28:24 --> 00:28:30
I don't know what v zero is
and I don't know what v of A is.

468
00:28:28 --> 00:28:34
I need another equation.

469
00:28:30 --> 00:28:36
Well, these
are conservative forces--

470
00:28:32 --> 00:28:38
we're dealing with gravity--

471
00:28:34 --> 00:28:40
so mechanical energy
must be conserved

472
00:28:36 --> 00:28:42
or, what we used to say--

473
00:28:40 --> 00:28:46
the total energy
of the system is conserved.

474
00:28:43 --> 00:28:49
The total energy of the system

475
00:28:45 --> 00:28:51
is kinetic energy
plus potential energy...

476
00:28:47 --> 00:28:53
must be conserved, must be
the same here as it is there.

477
00:28:53 --> 00:28:59
What is the total energy here?

478
00:28:56 --> 00:29:02
The kinetic energy
is one-half m v zero squared.

479
00:29:00 --> 00:29:06
What is the potential energy
here?

480
00:29:03 --> 00:29:09
It is minus mMG divided
by its distance to the center,

481
00:29:08 --> 00:29:14
which is capital R.

482
00:29:10 --> 00:29:16
That is the total energy
right here.

483
00:29:13 --> 00:29:19
The total energy here
could not have changed.

484
00:29:16 --> 00:29:22
It must therefore be
one-half m vA squared

485
00:29:22 --> 00:29:28
minus mMG, but now
the distance equals 5R.

486
00:29:28 --> 00:29:34
Apart from the fact
that I lose my little m,

487
00:29:32 --> 00:29:38
notice I have
a second equation.

488
00:29:34 --> 00:29:40
I didn't add any unknowns,

489
00:29:36 --> 00:29:42
so I have two equations with
two unknowns, v zero and v of A.

490
00:29:41 --> 00:29:47
And you can solve for that,
you can find both.

491
00:29:44 --> 00:29:50
Interestingly enough, you can
also find the semimajor axis

492
00:29:50 --> 00:29:56
because the total energy
of the system

493
00:29:54 --> 00:30:00
is also minus mMG
divided by 2A.

494
00:29:57 --> 00:30:03
This equation is
also on your exam.

495
00:30:01 --> 00:30:07
And so if you know v zero,
or you know v A, then you can...

496
00:30:05 --> 00:30:11
The m cancels and you can also
calculate the semimajor axis.

497
00:30:10 --> 00:30:16
 

498
00:30:12 --> 00:30:18
Okay, Doppler shift.

499
00:30:14 --> 00:30:20
Let's do something
about Doppler shift.

500
00:30:18 --> 00:30:24
First, Doppler shift
of electromagnetic radiation:

501
00:30:22 --> 00:30:28
light, radio waves, X rays.

502
00:30:26 --> 00:30:32
There is a star moving relative
to us with a velocity v.

503
00:30:33 --> 00:30:39
Since we're dealing with
electromagnetic radiation,

504
00:30:35 --> 00:30:41
we don't have to ask whether the
star is moving relative to us

505
00:30:39 --> 00:30:45
or we are moving relative
to the star.

506
00:30:41 --> 00:30:47
That question is a meaningless
question in special relativity.

507
00:30:45 --> 00:30:51
You are here.

508
00:30:47 --> 00:30:53
You are receiving
the frequency F prime,

509
00:30:50 --> 00:30:56
and here is that star

510
00:30:52 --> 00:30:58
and that star is moving
with a certain velocity--

511
00:30:56 --> 00:31:02
let's say this velocity, v,
and this angle is theta.

512
00:31:03 --> 00:31:09
So this component of
the velocity in our direction

513
00:31:07 --> 00:31:13
is v cosine theta.

514
00:31:08 --> 00:31:14
We call that
the radial component.

515
00:31:10 --> 00:31:16
And whether the star
is moving to us

516
00:31:12 --> 00:31:18
or we to the star
makes no difference.

517
00:31:15 --> 00:31:21
 

518
00:31:16 --> 00:31:22
If the velocity of the star
is much, much smaller than C,

519
00:31:21 --> 00:31:27
then F prime, the one
that you will receive,

520
00:31:25 --> 00:31:31
equals F times one plus v over C
times the cosine of theta.

521
00:31:30 --> 00:31:36
This equation is on your exam.

522
00:31:32 --> 00:31:38
C is three times ten to the
fifth kilometers per second.

523
00:31:39 --> 00:31:45
If cosine theta is positive,
the object is approaching

524
00:31:42 --> 00:31:48
and you observe
a higher frequency.

525
00:31:45 --> 00:31:51
If cosine theta is negative,
the object is going away.

526
00:31:48 --> 00:31:54
The radial component
is away from us

527
00:31:50 --> 00:31:56
and the frequency is lower.

528
00:31:52 --> 00:31:58
In optical astronomy,
we cannot measure frequencies.

529
00:31:55 --> 00:32:01
We can only measure wavelength.

530
00:31:57 --> 00:32:03
And the connection between
wavelength and frequency...

531
00:32:01 --> 00:32:07
Lambda equals the speed of light
divided by the frequency.

532
00:32:06 --> 00:32:12
So we can substitute
in that equation

533
00:32:09 --> 00:32:15
F equals C divided by lambda

534
00:32:12 --> 00:32:18
and F prime equals C
divided by lambda prime.

535
00:32:16 --> 00:32:22
When we do that and we also make
use of this approximation--

536
00:32:21 --> 00:32:27
that one divided by one plus x
is approximately one minus x

537
00:32:25 --> 00:32:31
as long as x is much,
much smaller than one--

538
00:32:29 --> 00:32:35
this is the first order term
in the Taylor Expansion.

539
00:32:32 --> 00:32:38
Then you can find
that now lambda prime

540
00:32:35 --> 00:32:41
becomes lambda times one minus--
it becomes a minus sign;

541
00:32:39 --> 00:32:45
the plus changes to a minus
sign, for that reason--

542
00:32:42 --> 00:32:48
times v divided by C
times cosine theta.

543
00:32:46 --> 00:32:52
Notice that here,
you have that radial velocity.

544
00:32:50 --> 00:32:56
And you have that here.

545
00:32:54 --> 00:33:00
If lambda prime
is larger than lambda,

546
00:32:57 --> 00:33:03
the object is going away
from us.

547
00:33:00 --> 00:33:06
We call that "red shift."

548
00:33:03 --> 00:33:09
 

549
00:33:04 --> 00:33:10
If the object is receding--

550
00:33:07 --> 00:33:13
if lambda prime is less
than lambda--

551
00:33:10 --> 00:33:16
we call that "blue shift,"

552
00:33:13 --> 00:33:19
because the wavelengths
become shorter in this time--

553
00:33:16 --> 00:33:22
move towards the blue end
of the spectrum.

554
00:33:18 --> 00:33:24
Here it becomes larger, shifts
to the red part of the spectrum.

555
00:33:22 --> 00:33:28
The blue shift means
the object is approaching you.

556
00:33:26 --> 00:33:32
During the lecture
that we discussed this,

557
00:33:28 --> 00:33:34
we took an example

558
00:33:29 --> 00:33:35
where lambda prime
over lambda was 1.00333.

559
00:33:35 --> 00:33:41
If you substitute that
in this equation,

560
00:33:39 --> 00:33:45
you'll find that v cosine theta

561
00:33:43 --> 00:33:49
equals minus 100 kilometers
per second.

562
00:33:48 --> 00:33:54
So what it means-- all you know
is the radial velocity;

563
00:33:51 --> 00:33:57
you never get any information
about the angle--

564
00:33:53 --> 00:33:59
that the radial velocity
is 100 kilometers per second

565
00:33:56 --> 00:34:02
away from us,

566
00:33:58 --> 00:34:04
or we are moving 100 kilometers
per second away from the star.

567
00:34:02 --> 00:34:08
That's the same thing
in special relativity.

568
00:34:05 --> 00:34:11
 

569
00:34:10 --> 00:34:16
If we apply this to sound, then
we get a very similar equation.

570
00:34:17 --> 00:34:23
Suppose you sit still in 26.100

571
00:34:19 --> 00:34:25
and I move a sound source
to you, but you don't move.

572
00:34:24 --> 00:34:30
Then you get the similar
equation that f prime--

573
00:34:28 --> 00:34:34
that's the frequency
that you will hear--

574
00:34:32 --> 00:34:38
equals f times one plus v over v
sound times the cosine of theta,

575
00:34:38 --> 00:34:44
and the sound's speed is
about 340 meters per second

576
00:34:43 --> 00:34:49
at room temperature.

577
00:34:44 --> 00:34:50
 

578
00:34:46 --> 00:34:52
In other words, if I twirl
something around in a circle

579
00:34:50 --> 00:34:56
and you are in the plane of
that circle-- you are here--

580
00:34:55 --> 00:35:01
then when the object
comes to you,

581
00:34:58 --> 00:35:04
you hear a maximum frequency--
f prime is larger than f.

582
00:35:03 --> 00:35:09
When it goes away from you, you
hear a decrease in frequency--

583
00:35:08 --> 00:35:14
smaller than f.

584
00:35:09 --> 00:35:15
And when the object is here,
and when it is here--

585
00:35:13 --> 00:35:19
when the angle is 90 degrees
and the cosine theta is zero--

586
00:35:17 --> 00:35:23
you'll find f prime equals f,
f prime equals f.

587
00:35:22 --> 00:35:28
So when I twirl
something around...

588
00:35:25 --> 00:35:31
let's suppose I twirl it
around with an orbital velocity.

589
00:35:29 --> 00:35:35
If I call that orbital velocity
of 3.4 meters per second--

590
00:35:33 --> 00:35:39
just to get some nice numbers,
then this is .01...

591
00:35:37 --> 00:35:43
And so if it comes to you,

592
00:35:38 --> 00:35:44
you hear an increase
of one percent in the frequency.

593
00:35:42 --> 00:35:48
When it goes away from you, you
hear a decrease of one percent.

594
00:35:46 --> 00:35:52
I have here a sound source which
produces roughly 1,500 hertz.

595
00:35:54 --> 00:36:00
(machine emits
high-pitched tone )

596
00:35:57 --> 00:36:03
I could twirl it around.

597
00:35:59 --> 00:36:05
I can do it
once per second around,

598
00:36:01 --> 00:36:07
which gives you
even twice this speed.

599
00:36:04 --> 00:36:10
And so you would hear
a two percent increase

600
00:36:06 --> 00:36:12
when it comes to you;

601
00:36:07 --> 00:36:13
two percent decrease
when it goes away from you.

602
00:36:10 --> 00:36:16
(tone wavers between higher
and lower pitch )

603
00:36:13 --> 00:36:19
Can you hear the Doppler shift?

604
00:36:16 --> 00:36:22
It's always difficult
in a lecture hall

605
00:36:18 --> 00:36:24
because you get reflections
from the wall.

606
00:36:20 --> 00:36:26
Can you hear it, when it comes
to you, that the pitch is higher

607
00:36:24 --> 00:36:30
than when it goes away from you?

608
00:36:26 --> 00:36:32
 

609
00:36:30 --> 00:36:36
Okay, rolling objects.

610
00:36:33 --> 00:36:39
Let's roll something
down the hill.

611
00:36:38 --> 00:36:44
Classic problem.

612
00:36:41 --> 00:36:47
Very often you see that
on exams.

613
00:36:45 --> 00:36:51
 

614
00:36:49 --> 00:36:55
We take a ball
and we roll it down an incline.

615
00:36:55 --> 00:37:01
This is the incline
and the angle is beta.

616
00:37:01 --> 00:37:07
Here is the ball, it's a sphere.

617
00:37:04 --> 00:37:10
It's a solid sphere,
it has mass m with radius R

618
00:37:07 --> 00:37:13
and it is solid.

619
00:37:10 --> 00:37:16
This is the center of that ball.

620
00:37:13 --> 00:37:19
I call this point P.

621
00:37:17 --> 00:37:23
I put all the forces on
that I can think of.

622
00:37:21 --> 00:37:27
There is mg here...

623
00:37:26 --> 00:37:32
at this point P, the normal
force from the surface...

624
00:37:31 --> 00:37:37
and there is friction.

625
00:37:34 --> 00:37:40
This is the frictional force.

626
00:37:35 --> 00:37:41
The sum of Ff
and the normal force

627
00:37:37 --> 00:37:43
is what we call
the contact force.

628
00:37:39 --> 00:37:45
That is the force
with which the incline

629
00:37:42 --> 00:37:48
pushes back onto the ball,
but we normally decompose that

630
00:37:46 --> 00:37:52
in a direction along the slope
and perpendicular to the slope.

631
00:37:50 --> 00:37:56
So these are
the only three forces

632
00:37:52 --> 00:37:58
that this object is experiencing

633
00:37:54 --> 00:38:00
and I can ask you now
what is "a"--

634
00:37:57 --> 00:38:03
the acceleration at which
it goes down the slope--

635
00:38:00 --> 00:38:06
and I can ask you
what is the frictional force.

636
00:38:05 --> 00:38:11
I will ask you that under
the conditions of "pure roll."

637
00:38:11 --> 00:38:17
What does pure roll mean?

638
00:38:13 --> 00:38:19
It means that if this--
the circumference here--

639
00:38:16 --> 00:38:22
moves one centimeter

640
00:38:17 --> 00:38:23
that the center of mass
has also moved one centimeter.

641
00:38:21 --> 00:38:27
So it means that the velocity
of the center--

642
00:38:26 --> 00:38:32
so this velocity, v of c--

643
00:38:29 --> 00:38:35
at all moments in time
will be omega R.

644
00:38:34 --> 00:38:40
And omega is
the angular velocity

645
00:38:36 --> 00:38:42
which is changing with time.

646
00:38:39 --> 00:38:45
That is the condition
of pure roll.

647
00:38:42 --> 00:38:48
Atall moments in time

648
00:38:44 --> 00:38:50
if I have a rotating sphere
or cylinder

649
00:38:48 --> 00:38:54
always will the circumferential
speed be omega R.

650
00:38:52 --> 00:38:58
It could be standing still
and just rotating like this,

651
00:38:56 --> 00:39:02
slipping like hell
and not moving forward--

652
00:38:58 --> 00:39:04
then the circumferential speed
is always omega R.

653
00:39:02 --> 00:39:08
But if the center also moves
with that same speed,

654
00:39:06 --> 00:39:12
only then do we call it
pure roll.

655
00:39:08 --> 00:39:14
If you take
the derivative of this,

656
00:39:11 --> 00:39:17
you get a equals omega dot
times R.

657
00:39:14 --> 00:39:20
That equals alpha R, alpha being
the angular acceleration.

658
00:39:20 --> 00:39:26
All right, we are going
to write down now

659
00:39:23 --> 00:39:29
the torque relative
to that point C.

660
00:39:26 --> 00:39:32
If I take point C,

661
00:39:27 --> 00:39:33
then I lose the force mg
and I lose the force N.

662
00:39:31 --> 00:39:37
There is only one force
that I have to deal with.

663
00:39:34 --> 00:39:40
For the torque,
relative to point C...

664
00:39:37 --> 00:39:43
The fact that the torque
is in the blackboard

665
00:39:40 --> 00:39:46
is of no concern to me.

666
00:39:42 --> 00:39:48
I just want to know
the magnitude--

667
00:39:44 --> 00:39:50
that is R times Ff,
R times the frictional force.

668
00:39:48 --> 00:39:54
That must be the moment
of inertia

669
00:39:51 --> 00:39:57
about that point C
times alpha.

670
00:39:54 --> 00:40:00
We've seen this
now twice already.

671
00:39:57 --> 00:40:03
And so, since alpha
is a divided by R,

672
00:40:00 --> 00:40:06
this is I of c
times a divided by R.

673
00:40:03 --> 00:40:09
This is one equation

674
00:40:05 --> 00:40:11
and I have a as an unknown
and I have Ff as an unknown.

675
00:40:10 --> 00:40:16
There is not enough--
I need more.

676
00:40:14 --> 00:40:20
Newton's second law always
holds-- the center of mass...

677
00:40:18 --> 00:40:24
For the center of mass, it must
always hold that F equals Ma.

678
00:40:23 --> 00:40:29
So all I do now is I go
to the center of mass and I say,

679
00:40:26 --> 00:40:32
What are the forces which
are acting down the slope?

680
00:40:30 --> 00:40:36
I am not interested in the ones
perpendicular to the slope--

681
00:40:33 --> 00:40:39
only the ones down the slope.

682
00:40:35 --> 00:40:41
That must be Ma--
it's nonnegotiable,

683
00:40:37 --> 00:40:43
because that is
the quality, characteristic

684
00:40:39 --> 00:40:45
of the center of mass.

685
00:40:41 --> 00:40:47
Well, this one is uphill

686
00:40:43 --> 00:40:49
and the only one
that is downhill

687
00:40:45 --> 00:40:51
is the component of mg along the
slope, which is Mg sine beta.

688
00:40:50 --> 00:40:56
I call that the plus direction,

689
00:40:54 --> 00:41:00
so I get Mg sine beta
minus that frictional force

690
00:40:59 --> 00:41:05
equals M times a.

691
00:41:06 --> 00:41:12
It's my second equation.

692
00:41:07 --> 00:41:13
I have two equations with
two unknowns, and I can solve.

693
00:41:12 --> 00:41:18
a is an unknown, and the
frictional force is an unknown.

694
00:41:15 --> 00:41:21
You can do that
as well as I can--

695
00:41:17 --> 00:41:23
I will just give you
the results.

696
00:41:19 --> 00:41:25
I don't want to waste
your time on the algebra.

697
00:41:21 --> 00:41:27
a, the acceleration, under
the conditions of pure roll--

698
00:41:25 --> 00:41:31
butonly under the conditions
of pure roll--

699
00:41:28 --> 00:41:34
equals MR squared times g
times the sine of beta

700
00:41:36 --> 00:41:42
divided by MR squared

701
00:41:38 --> 00:41:44
plus the moment of inertia
about that point C.

702
00:41:44 --> 00:41:50
And the frictional force--

703
00:41:48 --> 00:41:54
you can see that
from this equation--

704
00:41:50 --> 00:41:56
is the moment of inertia
times a divided by R squared.

705
00:41:53 --> 00:41:59
And I will leave "a" as it is

706
00:41:56 --> 00:42:02
because it becomes
too complicated otherwise.

707
00:41:59 --> 00:42:05
And so
the frictional force becomes

708
00:42:02 --> 00:42:08
the moment of inertia about C
times a divided by R squared.

709
00:42:06 --> 00:42:12
So this a has
to be substituted in here.

710
00:42:11 --> 00:42:17
The acceleration is
independent of the mass

711
00:42:13 --> 00:42:19
and independent of the radius--
very nonintuitive.

712
00:42:16 --> 00:42:22
It doesn't matter

713
00:42:18 --> 00:42:24
whether you take
a ball this big or this big.

714
00:42:20 --> 00:42:26
It doesn't matter
what the mass is.

715
00:42:22 --> 00:42:28
The reason for that is
that, as you will see shortly,
 

716
00:42:25 --> 00:42:31
that the moment of inertia

717
00:42:26 --> 00:42:32
also contains
the term MR squared,

718
00:42:28 --> 00:42:34
and so all the MR squares
will disappear.

719
00:42:30 --> 00:42:36
 

720
00:42:32 --> 00:42:38
Let's first take a look
at beta related to a.

721
00:42:36 --> 00:42:42
Notice when beta equals zero,
that a equals zero.

722
00:42:41 --> 00:42:47
Itbetter be.

723
00:42:42 --> 00:42:48
If there is no slope, there
cannot be any acceleration.

724
00:42:46 --> 00:42:52
Notice if a equals zero

725
00:42:47 --> 00:42:53
that the frictional force
is also zero.

726
00:42:50 --> 00:42:56
It better be.

727
00:42:51 --> 00:42:57
So that is an easy check
that you can do--

728
00:42:54 --> 00:43:00
internal consistency check.

729
00:42:55 --> 00:43:01
So now let's look at I of C.

730
00:42:58 --> 00:43:04
What is I of C?

731
00:42:59 --> 00:43:05
I look it up in the table.

732
00:43:02 --> 00:43:08
A ball, sphere rotating about an
axis through the center of mass.

733
00:43:07 --> 00:43:13
It's one of the few
that I happen to remember--

734
00:43:10 --> 00:43:16
is two-fifths MR squared.

735
00:43:12 --> 00:43:18
If I substitute that in here,
I'll find that the acceleration

736
00:43:17 --> 00:43:23
under the conditions
of pure roll

737
00:43:20 --> 00:43:26
is five-sevenths g
times the sine of beta.

738
00:43:27 --> 00:43:33
And I find that
the frictional force, F of f,

739
00:43:32 --> 00:43:38
equals two-sevenths, I believe,
times Mg sine beta.

740
00:43:38 --> 00:43:44
Indeed.

741
00:43:40 --> 00:43:46
 

742
00:43:42 --> 00:43:48
These are the results

743
00:43:44 --> 00:43:50
by substituting the moment
of inertia in there.

744
00:43:47 --> 00:43:53
 

745
00:43:49 --> 00:43:55
Let's look at these results
a little closer.

746
00:43:53 --> 00:43:59
If there were no friction,
then we know

747
00:43:56 --> 00:44:02
that the acceleration
would have been g sine beta.

748
00:44:00 --> 00:44:06
We remember that--

749
00:44:02 --> 00:44:08
a sliding object with no
friction would be g sine beta.

750
00:44:05 --> 00:44:11
Why is it now lower?

751
00:44:08 --> 00:44:14
Five-sevenths is lower than one.

752
00:44:10 --> 00:44:16
It's obvious.

753
00:44:11 --> 00:44:17
Because when this object
comes rolling down,

754
00:44:14 --> 00:44:20
the kinetic energy,
the total kinetic energy

755
00:44:17 --> 00:44:23
is kinetic energy of rotation

756
00:44:19 --> 00:44:25
plus the linear portion
of the kinetic energy--

757
00:44:22 --> 00:44:28
the forward motion due
to this one-half Mv squared.

758
00:44:25 --> 00:44:31
When you have a sliding object,

759
00:44:27 --> 00:44:33
there is
only the linear component.

760
00:44:30 --> 00:44:36
There is no rotational
kinetic energy.

761
00:44:32 --> 00:44:38
So now the linear kinetic energy
must be lower

762
00:44:36 --> 00:44:42
because some gravitational
potential energy

763
00:44:39 --> 00:44:45
goes into rotational
kinetic energy.

764
00:44:42 --> 00:44:48
And so it must go slower

765
00:44:43 --> 00:44:49
and so the acceleration
must be less.

766
00:44:45 --> 00:44:51
It's completely intuitive that
it is less than g sine beta.

767
00:44:50 --> 00:44:56
 

768
00:44:52 --> 00:44:58
Let's look
at the frictional force.

769
00:44:54 --> 00:45:00
This is a situation
of pure roll.

770
00:44:57 --> 00:45:03
We know
that the frictional force

771
00:44:59 --> 00:45:05
can never exceed
the maximum value possible.

772
00:45:03 --> 00:45:09
That is a no-no.

773
00:45:04 --> 00:45:10
If the frictional force
becomes the maximum one,

774
00:45:07 --> 00:45:13
it would start to slip.

775
00:45:08 --> 00:45:14
It would no longer be pure roll.

776
00:45:11 --> 00:45:17
In other words, there is
a necessary condition

777
00:45:15 --> 00:45:21
that the frictional force, which
is two-sevenths Mg sine beta,

778
00:45:22 --> 00:45:28
must be less than
the maximum frictional force.

779
00:45:27 --> 00:45:33
But the maximum
frictional force is mu s--

780
00:45:31 --> 00:45:37
static friction coefficient
times this value N--

781
00:45:36 --> 00:45:42
and this value N is
mg cosine theta... beta.

782
00:45:42 --> 00:45:48
And so this is mu s
times Mg cosine beta.

783
00:45:49 --> 00:45:55
I lose my Mg-- no surprise--
and so a necessary condition

784
00:45:54 --> 00:46:00
is that the static friction
coefficient must be larger

785
00:45:58 --> 00:46:04
than two-sevenths times
the tangent of alpha

786
00:46:02 --> 00:46:08
in order to have pure roll.

787
00:46:10 --> 00:46:16
Not alpha-- I don't know
what is wrong with me-- beta.

788
00:46:13 --> 00:46:19
And this, of course,
is pleasing in a way,

789
00:46:15 --> 00:46:21
because what this
is telling you--

790
00:46:17 --> 00:46:23
and we all know that
by our instinct,

791
00:46:19 --> 00:46:25
you feel it in your stomach--

792
00:46:20 --> 00:46:26
that if you make the angle
too large, if you tilt that,

793
00:46:24 --> 00:46:30
there's no way that
you're going to get pure roll.

794
00:46:27 --> 00:46:33
It will start rolling,
of course,

795
00:46:29 --> 00:46:35
but it also starts slipping.

796
00:46:30 --> 00:46:36
For instance, if you freeze
the friction coefficient

797
00:46:34 --> 00:46:40
and you say the friction
coefficient is 0.2,

798
00:46:37 --> 00:46:43
then it means that beta--
the angle beta--

799
00:46:39 --> 00:46:45
must be less than 35 degrees.

800
00:46:41 --> 00:46:47
That's just an example.

801
00:46:43 --> 00:46:49
Well, that's
intuitively pleasing.

802
00:46:46 --> 00:46:52
Remember that during
one of the lectures

803
00:46:50 --> 00:46:56
we calculated
the period of oscillation

804
00:46:54 --> 00:47:00
of an object that was sliding on
an air track without friction.

805
00:47:00 --> 00:47:06
And it was a curve of a circle
with a huge radius.

806
00:47:07 --> 00:47:13
It was hundreds of meters.

807
00:47:08 --> 00:47:14
And we had
a sliding object here

808
00:47:11 --> 00:47:17
and it was sliding
without friction.

809
00:47:13 --> 00:47:19
And we derived that the period

810
00:47:16 --> 00:47:22
was two pi times the square root
of R divided by g,

811
00:47:20 --> 00:47:26
as if this were a pendulum
with length R,

812
00:47:23 --> 00:47:29
and that's in fact what it is.

813
00:47:26 --> 00:47:32
We did that with the air track

814
00:47:29 --> 00:47:35
and the results were
fantastically in agreement

815
00:47:32 --> 00:47:38
with our prediction,
unbelievable accuracy.

816
00:47:35 --> 00:47:41
And then we did it
with a smaller radius--

817
00:47:39 --> 00:47:45
with a ball,
a rolling ball, this one.

818
00:47:43 --> 00:47:49
The radius of that curvature
was 85 centimeters,

819
00:47:48 --> 00:47:54
and we predicted a period of
1.85 seconds-- this equation.

820
00:47:56 --> 00:48:02
And we measured it,
and what did we find?

821
00:47:59 --> 00:48:05
Way higher.

822
00:48:01 --> 00:48:07
We observed something like,
I think, 2.3 seconds.

823
00:48:05 --> 00:48:11
And I asked you at the end
of the lecture, why?

824
00:48:09 --> 00:48:15
And some of you said,
"Well, the radius is smaller."

825
00:48:12 --> 00:48:18
That was no reason,
because the angle--

826
00:48:14 --> 00:48:20
the displacement angle--
was still very small.

827
00:48:17 --> 00:48:23
This maximum angle,
theta maximum, was so small

828
00:48:21 --> 00:48:27
that that could never explain

829
00:48:23 --> 00:48:29
why the period
was so much larger.

830
00:48:25 --> 00:48:31
But now you know, because
when this object rolls down,

831
00:48:29 --> 00:48:35
the kinetic energy is partially
in the linear term

832
00:48:34 --> 00:48:40
and partially in rotation--

833
00:48:36 --> 00:48:42
partially in the linear term
and partially in rotation.

834
00:48:39 --> 00:48:45
If it is sliding, it is all
in the linear term,

835
00:48:43 --> 00:48:49
so it clearly goes faster.

836
00:48:44 --> 00:48:50
Now it has to share
with the rotation.

837
00:48:47 --> 00:48:53
And so that's the reason
why we found the higher period.

838
00:48:50 --> 00:48:56
And that's precisely the
situation that you have here.

839
00:48:54 --> 00:49:00
 

840
00:48:57 --> 00:49:03
I want to repeat

841
00:48:58 --> 00:49:04
that this is what you have on
your plate on Monday.

842
00:49:01 --> 00:49:07
There is no way that I can cover
all of that during one exam.

843
00:49:05 --> 00:49:11
Please hold it--
we have one minute left.

844
00:49:07 --> 00:49:13
There's no way I can cover
all of that during one exam.

845
00:49:11 --> 00:49:17
I'll have to make a choice.

846
00:49:13 --> 00:49:19
I will pick only a few.

847
00:49:14 --> 00:49:20
Neither could I cover today,
at any depth, all these topics.

848
00:49:18 --> 00:49:24
The ones that I did not cover
doesn't mean at all

849
00:49:22 --> 00:49:28
that you will not see them
on the exam.

850
00:49:24 --> 00:49:30
There is a special tutoring
schedule this weekend

851
00:49:27 --> 00:49:33
that you can make use of

852
00:49:28 --> 00:49:34
and I believe that this lecture

853
00:49:30 --> 00:49:36
will be on PIVoT
tomorrow morning.

854
00:49:32 --> 00:49:38
I wish you luck
and I'll see you Monday.

855
00:49:35 --> 00:49:41
 

856
00:49:43 --> 00:49:49