WEBVTT
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This is Giancoli Answers with Mr. Dychko.
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At the bottom of this pilot's
looped trajectory,
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we have the lift force going upwards
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and gravity going down
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then lift force is much greater
than gravity because
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it has to lift the plane back up
into this circle
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as it goes back up along this
curved path here.
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Well, let's take care of some bookkeeping
first as we often do.
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So, we have the speed that
this plane is going is
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840 kilometers an hour but we'll convert
that into meters per second by
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multiplying by a 1000 and
dividing by 3600
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and we see that the meters or
kilometers cancel
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and the hours cancel and we are left with
meters per second.
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That's the same as dividing by 3.6
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so we get 233.33 meters per second
is the speed of the jet.
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The maximum centripetal
acceleration it can have
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at the bottom of this loop is 6*g*'s
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and we'll convert that into meters
per second squared.
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So, we have 9.8 meters per
second squared per *g*;
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the *g*'s cancel giving us 58.8 meters
per second squared.
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We know that centripetal acceleration is
*v* squared over *r*
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and we'll solve this for *r*
to find out what
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radius of curvature this path should have
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so that the acceleration is 6*g*'s.
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So, we'll times by *r* over *a c*
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and the *a c* cancels on the left
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and the *r* cancels on the right
leaving us with
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*r* equals *v* squared over
centripetal acceleration.
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So that's 233.33 meters per second
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squared divided by 58.8 meters
per second squared
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gives us about 930 meters
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is the radius of this loop.
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And then for part *B*,
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we use that free-body diagram
to talk about
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the lift force upwards minus
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gravity downwards has to equal
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*ma*, that's gonna be the net force
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and *a* is
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centripetal acceleration
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which we have already calculated is
58.8 meters per second squared.
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So, the lift force is the
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apparent weight of the pilot;
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it's the normal force that they experience
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being pushed up by the seat
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and we'll add *F g* both sides here
to solve for the lift force.
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And lift force is
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*m* times acceleration plus gravity
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so that's *ma c* plus *mg*
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and you can factor out the *m*'s
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and you get 78 kilograms times 58.8 meters
per second squared plus
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9.8 meters per second squared
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and
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let's figure out what that makes.
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78 times 58.8
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plus 9.8
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makes about 5400
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with two significant figures.
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That's the apparent weight
at the bottom of the loop.
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At the top of the loop, the pilot will
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will have gravity downwards
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and then it's kind of a
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misnomer, it's kind of confusing to
call this a lift force because
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it's actually gonna be a force that's
pushing down towards the ground
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to keep the plane going
downwards in this loop.
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So, it goes down along this path
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but I'll just keep calling it a lift force;
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it's the force you know that
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the wings are, you know, the
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air is pushing on the wings
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and then the wings push the plane
down with this force.
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OK
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So they are both pointing
in the same direction
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in the direction of acceleration
and so we will call
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the radial direction towards
the center positive.
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So that makes gravity positive
and the lift force positive
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equals mass times acceleration;
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acceleration is *v* squared over *r*
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because we can see the acceleration
is in a circleâ€¦
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it's radial so it's *v* squared over *r*
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and substituting for gravity is *mg*.
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And then we solve for this lift force
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and we'll subtract *mg* from
both sides here
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and we get lift force is *mv* squared
over *r* minus *mg*;
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factor out the *m*
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and you get *m* times
*v* squared over *r*.
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78 kilograms times
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233.33 squared over 925.9
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and then minus 9.8
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and we get about 3800 newtons
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is the effective weight at
the top of the loop.