I'd now like to consider
a very unusual problem,
and we saw a demo of this.
Suppose you have a yoyo,
and the yoyo an inner radius
b and an outer radius
R. And the yoyo
is rolling without
slipping along the ground.
And what we have here is
we're pulling the yoyo
with a string that's
wrapped around the spool,
and I want to find F max such
that it rolls without slipping.
If I don't pull it
hard enough, the wheel
will roll without slipping.
And if I pull it
harder, then this F max,
the wheel will start to slip.
So now let's analyze
this problem,
and let's use our
dynamics approach.
Let's choose an i
hat, a j hat, k hat.
As before, we'll
define an angle theta.
And now we'll apply
both Newton's second law
for linear motion and our
torque about the center of mass.
Now, we have to
consider our forces.
So I'm going to put the
forces on this diagram.
We have the normal force
from the ground pointing up.
We have the gravitational
force pointing down.
And remember, it's
rolling without slipping.
So the wheel is--
the contact point is
instantaneously at rest.
But because the center
of mass of the wheel
is accelerating, once again,
in order to keep a equal to R
alpha, then we need
some type of torque
that will produce
a non-zero alpha.
And that's going to come from
a non-zero static friction.
So once again, we have a case,
unlike a wheel which is just
rolling along a
horizontal plane,
because we're pulling
this yoyo's string,
the static friction is not 0.
Static friction depends
on everything else
that's happening in the system.
And now we can apply Newton's
second law, F equals ma.
And so in our x direction,
we have F minus F equals ma.
And if we look at the torque
about the center of mass
is Icm alpha.
Then this is our x equation.
Now here, both the pulling
force and the static friction
both exert torques about
the center of mass.
The static friction will exert
a torque in the direction k hat,
and the pulling force will exert
a torque in the minus k hat
direction.
So our torques--
the normal force
does not produce any torque
about the center of mass,
nor does gravity.
So what we have is fs times
the radius of the wheel--
that is the torque due
to static friction--
minus b times the
pulling force F,
and that's equal to Icm alpha.
So we have our two
dynamic equations.
And again, let's
recopy our rolling
without slipping
condition, now expressed
in terms of acceleration.
So now what I'd
like to do is solve
for this force F.
And the way I'll do
it is I'll write down,
from this condition, alpha
in terms of a/R. And from
this equation, I have that--
so alpha is a/R. And
from the top equation,
I have that a is F minus
f static over mR, times R.
So I can replace the
alpha in this equation,
and I get fsR minus bF is
Icm F minus fs over mR.
And now I just need
to collect terms.
I'll bring the static friction
term to this side and my f term
to that side.
And so I get fs
times R. And when
I bring the static friction
term over to that other side,
I get Icm over mR.
Let's just check dimensions.
Icm is mR squared.
So these terms have the
dimensions of length,
so looks like I'm OK.
And now I bring the
F to the other side,
and I get F times b plus Icm
over mR. I can now solve for F,
and I get f static
R plus Icm m/R--
it's a complicated answer--
divided by b plus
Icm divided by mR.
Now, if I want to ask
the question, what
is the maximum force I can
pull in which it just slips,
what's physically happening
is the harder I pull,
the bigger the
static friction is.
I pull harder, static
friction gets bigger.
I pull harder, static
friction gets bigger.
But static friction can
only have a maximum value.
That hasn't changed.
The maximum value
of static friction
is the coefficient of
static friction times
a normal force,
which, in this case,
is just normal balance
in gravity, mg.
So this is the maximum value
that static friction can have.
And so now I get F max--
this is, again, a little
complicated-- mu s mg times R
plus Icm divided by mR divided
by b plus Icm divided by mR.
And that is the
maximum force that I
can pull this yoyo with that it
still rolls without slipping.
