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