- Good morning.
Billy, please read the problem
and Bobby, please translate.
♫ Flipping Physics ♫
- A turntable is turning at
45 revolutions per minute.
- Angular velocity is 45
revolutions per minute,
which we need to multiply by
one minute over 60 seconds,
and two pi radians over one revolution
to get 1.5 pi radians per second.
- Mints are located 0.030
meters, 0.080 meters,
and 0.130 meters from
the center of the record.
- Radius one equals 0.030 meters.
Radius two equals 0.080 meters.
And radius three equals 0.130 meters.
- Determine what you can
about the coefficient of static friction
between the turntable and the mints.
- The coefficient of static
friction equals question mark.
- Bo, please begin solving the problem.
- Well, we need to start
with free body diagrams,
so let's pause the mints
and add a side view.
And, actually, the free body
diagram will be the same
for all three mints so
we only need to draw one.
The force normal is up.
The force of gravity is down.
And there has to be a
force pulling the mints in
toward the center of the record player
so that is the centripetal force.
- Class, what are the three
things I asked you to remember
about the centripetal force?
- It's not a new force.
- It's never in a free body diagram.
- Yeah, so the centripetal force is never
in a free body diagram so that's
not the inward force there.
- And the in direction is positive
and the out direction is negative.
- Correct.
So, Bo, what force is
acting inward on the mints
to keep the mints moving in a circle?
- Okay.
That is the force the
surface of the record causes
on the mints, which is the
force of static friction.
- That is correct.
Bo, please keep going.
- Now we can sum the forces.
Let's start with the net
force in the y direction
equals force normal
minus force of gravity,
which equals mass times
acceleration in the y direction.
The acceleration of the mints
in the y direction equals zero
so the force normal equals
the force of gravity
or mass times acceleration due to gravity.
- We can put that in our equation holster
and then sum of the
forces in the x direction.
- Actually, we need to sum
the forces in the in direction
because the mints are moving in circles.
So the net force in the in direction
equals the force of static friction.
- Which is positive in
this particular example
because the force of static
friction is always inward.
- Which then equals mass times
centripetal acceleration,
but I don't know what to do now.
- Force of static friction equals
the coefficient of static
friction times force normal.
And centripetal acceleration
equals radius times
angular velocity squared.
- And we can plug in
from our equation holster
force normal equals mass times
acceleration due to gravity,
and then,
- [All] everybody brought
mass to the party.
- Yep.
♫ Everybody brought mass
- Billy, please finish the problem.
- Oh come on, I was almost done.
- [Mr. P.] Okay, Bo, go ahead.
- We now know the static
coefficient friction
equals radius times the
angular velocity squared,
all divided by acceleration
due to gravity.
We can plug in the numbers
for all three radii.
The first one is 0.03
times 1.5 pi squared,
divided by 9.81, which is 0.067910 or 0.068
with two significant digits.
The second one is 0.08
times 1.5 pi squared,
divided by 9.81, which is 0.18109 or 0.18.
The third one is 0.13
times 1.5 pi squared,
divided by 9.81, which is 0.29428 or 0.29.
- That does not really make sense to me.
How could the coefficient of
static friction be different
for each mint?
I mean, the surface of the record player
does not change, so...
- Bobby, that's a good point.
So, the coefficient of static friction
between each mint and the
turntable should be the same.
And remember, the problem did not say
determine the coefficient
of static friction.
It said determine what you can
about the coefficient of static friction.
- Does this have to do with the fact
that we used the maximum
force of static friction
in our equations?
- Right, the force of
static friction is actually
less than or equal to the
coefficient of static friction
times force normal.
- Okay, so then, the
force of static friction
acting on each mint is larger
when the radius of the circle is larger.
- So, what we found is
the minimum coefficient of
static friction necessary
to keep each mint on the turntable
and as the radius increases,
it takes a larger coefficient
of static friction to keep
each mint on the turntable.
- So then, the answer is that
the coefficient of static friction
must be equal to or greater than
the largest coefficient
of static friction we got
because our mints always
stayed on the turntable.
- So our answer is that the
coefficient of static friction
is greater than or equal to 0.29.
- Very good reasoning.
Thank you very much for
learning with me today.
I enjoyed learning with you.
