It's very general--
I have a rope, and the rope is wrapped around an object,
which could be a disk.
Let's call this point Q, let's call this point P, and this is
the center of the object.
The object could be a sphere or a cylinder, and moment of
inertia about that point C-- about the axis of rotation
perpendicular to the paper--
equals kmR squared.
If this were a solid sphere, then k would be 2/5, if it
were a solid cylinder or a disk, then it would be 1/2,
and if it were a hollow sphere,
then k would be [? 2/3 ?]
All these combinations you can look up in tables.
The question now is what is the acceleration of the center
of mass as it goes down?
What is the tension T [UNINTELLIGIBLE]
and not to be confused with the period?
What is omega as a function of time as the
thing starts to unroll?
You can add to that, what is the speed of the center of
mass as a function of time?
The thing you're going to rotate around with angular
velocity omega--
this is clearly a situation of no slip.
If I make an enlargement here--
that means if it rotates over an angle d theta, and if the
rope never slips here, then the ropes moves
over an angle ds.
It's immediately obvious that d theta times R, which is the
radius, equals ds.
Divide left and right by dt and you'll find omega R equals
v. This is the tangential velocity of the rope here, but
that is also the same as v c because there is no slip.
If you take the derivative of this so that omega dot becomes
alpha, then you get alpha times R equals the
acceleration of the center of mass.
I will repeatedly use today that alpha equals the
acceleration off the center of mass divided by R. That is a
necessary condition for no slip.
Let's put all the forces in here.
Here we have mg, here we have the tension T, and that's all
there is on this object.
Newton's Law must hold: F equals ma.
This holds for all the sources on the system with the
acceleration of the center of mass.
What we see if we take mg positive down is mg minus T,
which is up, must be m times a c.
That is my first equation, and I have two unknowns--
I have T and a as unknowns.
Now we get that to the torque relative to point C equals I
relative to point C times alpha as the moment of inertia
through this [? axel. ?]
The torque relative to point C is T times this perpendicular
distance R, so I can write down that this is R times T.
That is also I C, which we know is kmr squared times
alpha, but alpha is a C divided by R. This is my
second equation--
I have two equations with two unknowns in which you can
solve easily for a, and you can solve easily for T.
I will now leave you with the massaging, but I will give you
the answers, because they are kind of interesting: a equals
g divided by 1 plus k.
The tension equals kmg divided by 1 plus k.
What you see here is that the acceleration, independent of
the mass, is only a function of geometry--
it only depends on k.
I will jump a surprise on you with something that you may
not have known, and I don't blame you if
you don't know this.
If I have a pure roll situation--
that means if this object has a pure roll, and we discussed
earlier what we mean by p pure roll if this is radius R--
it's going around with angular velocity omega, and it has a
velocity of the center of mass v c, then for one thing, v C
equals omega R. There's something else which is
interesting: if I look at the rotation about this point P--
something I will not prove--
then this motion of this object as it goes into pure
roll, is also a perfect rotation of this whole object
about this point with exactly that same omega.
I don't prove it, and maybe you are will to do that, but I
will assume that you will take my word for it.
That is the reason, as perhaps you remember where I mentioned
earlier, that the velocity here equals 0, and that the
velocity here equals 2 v C. If you were to take a position
somewhere here, then this is now the R that you have to
take, this is the pivot point you have to take, and so v at
that location relative to point P, equals omega R. You
take the same omega, and so you get a
vector in that direction.
It's a very complicated motion, but what is
interesting is that it is also a pure rotation about this
point with angular velocity omega, except that now the
distances R depend on where you are on the circumference.
If we take this for granted, there is something that you
can do that is very nice: we've taken the torque
relative to point C--
why not take the torque relative to point Q, which is
also the same as the torque relative to point P?
You will see there's no difference, because it is R
cross F. Since T goes through P and Q, you only deal with
this force, and it is a perpendicular distance.
The torque, relative to point P, which is also the torque
relative to point Q equals mgR, as I just explained.
That, of course, is the moment of inertia relative to that
point Q or through that point P times alpha, and for alpha
you can write down a divided by R, if you want that.
What is now this moment of inertia?
We use the parallel axis theorem.
We get mgr equals the moment of inertia about the center of
mass, which is kmR squared plus the distance through the
[? new ?] axis squared times the mass, so--
plus m R squared.
This is simply an application of the parallel axis theorem.
I have this times alpha now, and so this becomes times a
divided by R.
What you have here--
maybe to your surprise--
one equation with one unknown.
There's no T in here, even, and when you work this out,
you will find exactly the same that you found before.
Maybe you'll find this even easier--
you find a equals g divided by 1 plus k.
If you want to calculate what T is, you can go to the
previous part of the problem, and you substitute a in the
equation for T, and obviously you'll find
exactly the same answer.
Let us take some cases whereby k is 2/5, just to give you
some feeling for it--
a would be roughly 0.7 g, and the tension would be
roughly 0.29 g.
This would be a solid sphere.
If I take k equals 1/2, which would be a solid cylinder, a
would be 2/3 g--
which is very close, by the way--
and T would 1/3 mg.
That is also very close to that value.
It doesn't make all much difference to whether you take
a solid sphere or whether you take a solid disk.
That's interesting all by itself.
Now we start this system--
so we have this object, we drop it from a position y
equals 0, and we drop it over a distance h.
We now want to know what the angular velocity is when this
object reaches this height h.
You can call this w for y equals h, or you can also
calculate the velocity of the center of mass for y equals h.
I'll first do it in the clumsy way, and then I'll do it in a
better way.
The motion in this direction is accelerated--
it's constant acceleration in a in the y direction, so y of
t equals y 0, which is 0, plus 1/2 at squared.
That equals h-- that's the distance that it travels.
You find immediately that t equals the square root of 2h
divided by a.
What is the velocity of the center of mass at
any moment in time?
That is at, because it starts at 0 speed.
That is a times this t, which is the square root of 2h over
a, and so that is the square root of 2 ah.
The the speed of the center of mass after this thing has
unrolled over a distance h would be the square root of 2
g h divided by 1 plus k.
Now we also have the velocity.
What is omega when y equals h?
Well, v equals omega R, and omega equals this number
divided by R, so I get 1 over R times the square root of 2 g
h divided by 1 plus k.
Notice that the velocity that it reaches when it has
unrolled over a distance h is again independent of mass.
It only has the geometry in k, and this
is the angular velocity.
Again, the angular velocity only depends on geometry,
which is maybe not all that intuitive.