You see here the topics that will be on your plate on Monday.
It's clearly not possible for me in one exam to cover all these topics, so I will have to make a choice Monday.
Today I also have to make a choice.
I cannot do justice to all these topics in any depth, so I will butterfly over them and some of them I won't even touch at all.
However, what is not covered today may well be on the exam.
You will get loads of equations.
Almost every equation that I could think of will be on your exam.
There's also special tutoring this weekend you can check that on the Web--
Saturday and Sunday.
Let's start with a completely inelastic collision.
Completely inelastic collision.
We have mass m1, we have mass m2.
It's a one-dimensional problem.
This one has velocity v1, and this has velocity v2.
They collide, and after the collision they are together--
because that's what it means when the collision is completely inelastic--
and they have a velocity, v prime.
If there is no net external force on the system, momentum must be conserved.
So I now have that m1 v1 plus m2 v2 must be m1 plus m2 times v prime.
One equation with one unknown--
v prime follows immediately.
You may say, "Gee, you should really have put arrows over here." Well, in the case that it is a one-dimensional collision, you can leave the arrows off because the signs automatically take care of that.
Kinetic energy is not conserved.
Before the collision, your kinetic energy equals one-half m1 v1 squared plus one-half m2 v2 squared.
After the collision, kinetic energy equals one-half m1 plus m2 times v prime squared.
And you can easily prove that this is always less than that in case of a completely inelastic collision.
There's always kinetic energy destroyed, which then comes out in the form of heat.
Let's do now an elastic collision.
And I add the word "completely" elastic, but "elastic" itself is enough because that means that kinetic energy is conserved.
I start with the same initial condition: m1 v1, m2 v2, but now, after the collision, m1 could either go this way or this way, I don't know.
So this could be v1 prime, this could be v1 prime.
m2, however, will always go into this direction.
That's clear, because if you get hit from behind by object one, after the collision, you obviously go in this direction.
Again, we don't have to put the arrows over it, because the signs take care of it in a one-dimensional case.
This will be plus, then.
That could be... you could adopt that as your convention, and if it goes in this direction, then it will be negative.
So if you find, for v1 prime, minus something, it means it's bounced back.
So now we can apply the conservation of momentum if there is no external force on the system.
Internal force is fine.
All right, so now we have m1 v1 plus m2 v2 equals m1 v1 prime plus m2 v2 prime--
conservation of momentum.
Now we get the conservation of kinetic energy, because we know it's an elastic collision.
One-half m1 v1 squared, one-half m2 v2 squared--
that's before the collision.
After the collision, one-half m1 v1 prime squared plus one-half m2 v2 prime squared.
Two equations with two unknowns and in principle, you can solve for v1 prime and for v2 prime, except that it could be time-consuming.
And so on exams, what is normally done when you get a problem like that...
Normally you get a problem whereby either m1 is made m2 or m1 is much, much larger than m2 or m1 is much, much smaller than m2, like banging a basketball onto a Ping-Pong ball or a Ping-Pong ball onto a basketball.
I will do a very simple example whereby I will take now for you m1 equals m2, and I will call that m and I will even simplify the problem by making v2 zero, so the second object is standing still.
One-dimensional collision, one hits that object.
Very special case.
So the conservation of momentum now becomes much simpler.
m times v1--
there is no v2--
equals m v1 prime plus m v2 prime.
Conservation of kinetic energy is one-half m v1 squared equals one-half m v1 prime squared plus one-half m v2 prime squared.
Notice that now I lose my "m"s, which is very convenient.
Here I lose my one-half "m"s, even and this is very easy to solve.
If you square this equation you get something that will look very similar to this.
If you square it, you get v1 squared equals v1 prime squared plus v2 prime squared plus two v1 prime v2 prime.
And compare this equation with this equation--
they're almost identical, except for this term, so this term must be zero.
But we know that v2 prime is not zero.
It got kicked and so it'll go forward.
So what that means, then, is that v1 prime equals zero and if v1 prime equals zero, you see that v2 prime equals v1.
And this is that classic case whereby a ball hits another ball.
This one has no speed.
It hits it with a certain velocity, v1.
They have the same mass.
After the collision, this one stands still and this one takes over the speed.
Famous Newton's Cradle.
You see it often with pendulums.
It is the logo on the 8.01 home page, and I showed you a demonstration here when we discussed that in lectures.
All right, let's move on, and let's do something now on torques, angular momentum, rotation and let's discuss the Atwood machine.
The Atwood machine is a clever device that allows you to measure, to a reasonable degree of accuracy, the gravitational acceleration.
Here's a pulley, and the pulley has mass m, has radius R.
It's solid, so it's a solid disk--
rotates, frictionless, about point P, radius R.
And there is a rope here, near massless--
we ignore the mass.
Mass m2 is here and mass m1 is here, and let's assume that m2 is larger than m1.
So this will be accelerated in this direction, this will be accelerated in this direction, and this will start to rotate with angular velocity omega, which will be a function of time.
And now the first thing we want to do
is to make up "free-body" diagrams.
"Free-body" diagrams, for this one, is easy.
m1 g down... and T1 up.
For this one, we have m2 g down and we have T2 up.
For the pulley, it's a little bit more complicated.
This is that point P.
If here is a tension T1 that's pulling down on the pulley--
so this is T1--
and this T2 is pulling down on the pulley--
so there's T2--
it has a mass, so it has weight mg.
The sum of all forces on the pulley must be zero; otherwise it would accelerate itself down, which it doesn't.
And so there has to be a force up--
I'll call it n.
And that force n has to cancel out all these three forces.
So that must be T1 plus T2 plus mg.
We will not need it any further in our calculations, but there has to be a force to hold that in place, so to speak.
So now we're going to calculate the acceleration under the condition that the rope does not slip.
That means there is friction with the pulley--
not friction here, but here.
Otherwise, the rope would slip, the rope would slip.
What it means, if there is no slip, that if the rope moves one centimeter that the wheel also turns at the circumference one centimeter.
That's what it means when there is no slip.
That means the velocity of the rope--
v of r, v of the rope--
must be omega times R of the pulley.
That's what "no slip" means.
So the acceleration--
which is the derivative of that velocity of the rope--
A, is omega dot times R, which is alpha times R.
Omega is the angular velocity and alpha is the angular acceleration.
This is a condition... it is an important condition for no slip.
So let's now start at object number one and write down Newton's second law.
I call this the positive direction for object one and I call this the positive direction for object two--
just easier for me.
So we get T1 minus m1 g must be m1 a--
I don't know what T1 is, I don't know what a is.
Second equation, for this one.
I call this the positive direction.
m2 g minus T2 must be m2 a--
One unknown has been added, so I need more.
Of course I need more.
I also have to think about the pulley.
The pulley... the net force on the pulley is zero.
That's why this stays in place, but it is going to rotate because this force T2 is larger than this T1.
There is a torque relative to that point P, and torque is defined as r cross F.
The torque relative to point P, the magnitude is this position vector times this force.
That is a torque in the blackboard.
What is in the blackboard, I will call positive.
The torque, due to this force, is out of the blackboard, and I will call that negative.
Since this angle is 90 degrees, I simply get that the torque relative to point P equals the radius of the cylinder--
the radius of that pulley--
That's the positive part, and the negative part is the radius times T1.
Notice that this force and this force go through P, do not contribute to the torque.
And that now equals the moment of inertia about that point P, times alpha.
But since we have no slip, alpha is a divided by R.
So it's the moment of inertia about point P times a divided by R.
But since it is a rotating disk which is rotating about its center of mass...
I know the moment of inertia, I looked that up--
if you need it during your exam, you will find that in the exam--
that is one-half MR squared...
one-half MR squared times a divided by R and I lose one R, and so I find, then, that T2 minus T1 equals one-half Ma.
Notice I also lose my second R.
And so now I have a third equation and I can solve for T2, I can solve for T1 and I can solve for a.
And you can do that as well as I can.
If you find the result, you should always do a little bit of testing to make sure that your result makes sense.
And what you should do is you should say...
You should make sure that m2 g is larger than T2.
That's a must--
otherwise, it's not being accelerated down.
You should also check that T1 comes out larger than m1 g.
That's also a must.
And you should also check that T2 be larger than T1.
Otherwise, the pulley wouldn't rotate in clockwise direction.
It would also be useful, which is a trivial check, to stick in your results m1 equals m2.
That should give you that the acceleration should be zero and it should give you that T1 equals T2.
Those are obvious things and that can be done very simply.
It takes you no more than ten seconds.
And if it's... any one of these is not met, then somewhere you slipped up and it will give you an opportunity to go over the problem again.
All right, let's now do another problem--
simple harmonic oscillation of a physical pendulum.
I have here a rod--
mass M, length l--
rotating here about an axis perpendicular to the blackboard without friction.
Here is the center of the rod, and let this angle be theta.
There is a torque relative to point P.
There is also a force that goes through point P.
I'm not even interested in that force.
I know that it's here, mg.
Since I'm going to take the torque relative to point P, I don't worry about the force, but there has to be a force through point P.
Otherwise this ruler would be accelerated down with acceleration g, if this is the only force there were.
But we know that's not the case, it's going to swing.
So there has to be a force to P.
I don't want to know what it is, but there has to be one.
So the torque relative to point P--
is the position vector, r of P, from here to here crossed with this force.
And so that makes it one-half l times mg times the sine of the angle, and that's... the angle is theta so that is times the sine of theta.
The cross product has the sine of the angle in it.
So this is the torque relative to point P which also must be the moment of inertia about point P times alpha.
No different from what we just had with the pulley.
So alpha equals omega dot.
It's also theta double dot.
This omega is the angular velocity--
it's d theta/dt and the derivative gives me the angular acceleration.
There has to be a minus sign here, that's important because the torque is restoring--
the same situation we had when we had a spring.
We were oscillating a spring, the spring force is minus kx.
The minus sign here plays exactly the same role.
So I can write down here minus I of p theta double dot.
Now, if we take a small angle approximation--
"small angle approximation"--
then the sine of theta is approximately theta, if theta is in radians.
And so I can replace now the sine of theta here by theta and so now I find, if I bring this to the other side, I get theta double dot plus one-half lMg divided by the moment of inertia about point P times theta equals zero.
Needless to say that I am happy like a clam at high tide because I see here an equation which clearly tells me that we have a simple harmonic oscillation--
theta double dot plus a constant times theta is zero.
And so we must get as a solution that theta must be some maximum angle times the cosine of omega t plus or minus phi.
This omega has nothing to do with that omega.
This is the angular frequency.
This is related to the period T of the oscillation, which is two pi divided by omega.
This is a constant.
This omega is not a constant.
It is unfortunate, in physics, that we use the same symbol.
The angular velocity is zero when the object stands still; is a maximum when the object is here.
That omega is always the same.
It's related to how long it takes to make one oscillation.
Both are called omega, both are radians per second.
Couldn't be more confusing.
All right, if we find I of P then we can solve for the frequency, angular frequency and we can solve for the period.
So let's do the... calculate the moment of inertia about point P.
In order to do that, we have to apply the Parallel-Axis Theorem because you will probably be given the moment of inertia for rotation about the axis through the center of mass parallel to this axis.
And then you will have to add the mass times the distance between these two axes squared to apply the Parallel Axis Theorem.
So this is the moment of inertia about the center of mass plus m times the distance squared.
And the distance between these two axes is one-half l, so this is one-quarter l squared.
I look up in a table what the moment of inertia is for rotation of a rod about its center--
a perpendicular rod is perpendicular to the axis--
and that is 1/12 Ml squared plus one-quarter Ml squared give me one-third Ml squared.
So I know now what I of P is, and so now I can solve for the value for omega--
the angular frequency.
I'll do that here so that we keep everything on one blackboard.
So this term here, which is omega squared...
Omega squared equals one-half lMg divided by one-third Ml squared.
I lose one l, I lose one M--
very common, always lose your "M"s in gravity--
and so this is three-halves g divided by l.
And so the period of one oscillation is two pi divided by omega--
omega is the square root of this--
so that becomes the square root of two-thirds l over g.
And that is the period of the oscillation of this ruler.
We worked on something similar in lectures and we even measured the period and we found very, very good agreement with the theoretical prediction.
You can ask, now, what is the kinetic energy of rotation of this rod, which changes with time? Kinetic energy of rotation is one-half I omega squared.
Remember, the linear kinetic energy is one-half mv squared.
The m becomes I when you go to rotation, and v becomes omega.
So the kinetic energy of rotation equals one-half I omega squared--
you'll find this equation on your exam.
This is I about that point P.
You know what theta is, as a function of time.
So omega equals d theta/dt, so you can find what omega is.
You know what I of P is, we just calculated it.
And so you know what the kinetic energy of rotation is at any moment in time.
It will change.
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