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welcome to es 310 lesson 18 today we're
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going to start the second part of the
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class in which we are looking at rigid
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bodies instead of particles and we will
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start back at the beginning with
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kinematics so we're looking at only the
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Motions of those rigid bodies at this
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point uh regardless of the forces acting
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on them we're going to focus today on
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looking at rotational speed because
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that's what the rigid bodies add to our
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systems is the ability to rotate so
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we'll look at rotational speeds and
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accelerations and look at how those are
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related to linear speeds and rot and
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accelerations more information on these
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topics can be found in hibler's Dynamics
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textbook chapter 16 sections 1-
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4 so as an overview we have now
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completed our work on particles so we
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looked at the motion of the particles
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which was the kinematics and then we
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looked at Kinetics in three different
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ways using Newton's law using work
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energy methods and using impulse
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momentum methods we're now going to do
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the same four things with rigid bodies
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So today we're going to start in on our
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motion of rigid bodies we're going to
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continue that for a few lessons and then
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we will look at equations of motions for
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rigid bodies which are luten second laws
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work energy methods for rigid bodies and
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impulse momentum methods for rigid
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bodies then we'll look at a few other
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topics to close out the
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semester so the kinematics of rigid
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bodies is a uh first of all we need to
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Define Rigid bodies so a rigid body is a
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collection of particles that form a
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single body so we can no longer assume
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things are
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particles because the shape and their
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orientation now matter so that they're
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ro they're rotating in some way so
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there's two different types of motion
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for rigid bodies there's translation
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which is what we've been looking at for
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particles right so in Translation any
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line in the B rigid body stays in the
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same orientation but it will move right
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so this vertical line on this body if
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it's re rectilinear translation will
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move along a straight line if it's
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curval linear translation this vertical
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line stays vertical but it moves along
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along a curve
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rotation we can be defined about some
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axis so we Define an axis and then that
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point a point on the body will move in a
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circle around that axis and that is
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rotation when you combine rotation with
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translation you get General planer
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motion which is what we'll finish off
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today with and and focus on in the next
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couple of lessons so let's take a look
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at translation this is what we have been
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doing so it should look familiar this
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equation the position equation is a
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relative position equation right so the
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the position of B is equal to the
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position of a plus the relative position
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from A to B we were we've been used to
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seeing this equation written as RBA is
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equal to RB minus r a but this is the
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same idea here when we take the
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derivatives of this in order to find the
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velocity this term the
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relative position from A to B isn't
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going to change because the body is not
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going to deform so it is zero the
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derivative of it is zero and so you have
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that the velocity of B is equal to the
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velocity of a and you have the
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acceleration of B is equal to the
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acceleration of a so because the body
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isn't deforming it's just translating
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the velocities and accelerations of
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every point in the body are the
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same since this is true that VB is equal
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to V a and v a is equal to a a all of
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the kinematic equations that we saw
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previously still apply and so nothing
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has changed there we can still apply all
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of those various kinematic equations for
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translation now let's take a look at
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rotation this is the new aspects that
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come about when we deal with rigid
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bodies so we start by defining a Theta
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an angular position to define a Theta we
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need some reference line that we then
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Define a Theta from so in this case our
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reference line is from o to P in the
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diagram here and Theta then is
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defined so in this case this is our
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reference line here relative to some
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fixed object out here and Theta is
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defined relative to that that
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reference D Theta is some change in
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Theta and it's in the direct the
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direction of D Theta is the axis of
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rotation so you use your right hand rule
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again your fingers curve in the
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direction of D Theta your thumb points
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upwards it's a positive rotation
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downwards it's a negative but it will
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always be along the axis of
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rotation the angular velocity is defined
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then as the time derivative of theta
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just like the linear velocity was
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defined as the time derivative of
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position
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so Omega which is your angular velocity
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is equal to D Theta DT the angular Exel
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acceleration is called alpha alpha is
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just the time derivative of Omega so
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those are equivalent equations to what
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we've seen before and we have the the
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combined equivalent equation where Alpha
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is equal Alpha * D Theta is equal Omega
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D Omega this is equivalent to a d s is
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equal to V DV right so in all of these
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equations all we've done is replace S
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with
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Theta uh V with
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Omega and a with Alpha and so all of
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these equations are the same you just
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have to remember that the position is
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being replaced by an angular position
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the velocity is being replaced by an
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angular velocity and the acceleration is
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being replaced by an angular
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acceleration if you have a constant
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angular acceleration we have the
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constant acceleration equations again
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with these substitutions
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made so if we know the rotation of a
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body we can find information then about
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the motion of any point in that body so
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the relationship between
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position and angular
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displacement comes about through R this
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is just a geometric definition of an AR
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length relative to the enclosed angle
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that's where this is coming from so DS
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is equal to r d Theta similarly then V
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is equal to R Omega and some portion of
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a is going to equal to R Alpha so to
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keep the directions correct we have to
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remember this cross product right
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because we're crossing R and Omega in
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order to find V and it's Omega crossed
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with
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r um is the order and that order is
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important we would only probably use
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these cross products if we were working
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in 3D most of our motion we're just
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going to use Omega time R because we are
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we're working in one plane so R Omega is
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out of the plane V and R are within the
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plane the acceleration because we're
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rotating things are moving along a curve
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we're going to think in terms of normal
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and tangential components right so the
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tangential component of acceleration if
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you think back was defined as V Dot well
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V Dot is time derivative of V let's take
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our definition of v in terms of angular
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velocity plug it in all right R is not
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going to change R is a constant so R can
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be factored out and you have D Omega
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DT which is defined as Alpha so the
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Accel the tangential acceleration is
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Alpha time r the normal acceleration if
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we remember back to the definition was
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v^2 over row well Row in this case is R
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the dist the radius of our rotation V is
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Omega R so we plug that in we get Omega
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r^ squared over R divide that out we get
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Omega 2 R so our total acceleration is
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the sum of those two parts and in in
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Vector notation with cross products we
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get Alpha crossed with r for the first
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term Omega squar which is not a vector
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times R which is a vector and there's a
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minus sign here to to incorporate the
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fact that the normal is pointing towards
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the center of the
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circle so let's take an example of this
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this is in old cars right that don't
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have power windows this is how you would
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uh roll up and down the window so you
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have a crank that the hand is on here
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and you turn that crank and as you turn
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that crank you shift the angle of this S
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gear this partial gear which causes
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this be uh this beam to rotate upwards
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or downwards depending on which way
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you're
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turning and that is this one's fixed to
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a a fixed point so this is going to move
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the same amount this just keeps it
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aligned right the adding this one if
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this weren't there this whole thing
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would r rotate right so then A and E are
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going to move up and that's going to
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cause the window to move up all right so
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let's see when A and E though A and E
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are actually moving tangential to this
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circle right because they're both pinned
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it here so the velocity of a and e is
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sort of at an angle here and this
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angle if we look at the geometry here
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the angle in there is the same as the
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angle down here all right so this angle
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in here is
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Theta okay so now let's take a look here
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so we've we're putting in motion at the
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top in onto this gear and we're trying
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to figure out how fast the window goes
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up and down based on
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that so let's take a look at cog c first
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which is this little gear
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here cog
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c has an angular velocity Omega of
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0.5 and it has a radius of 20 mm or
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02 so the velocity of a point on the
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outer surface of cog c all right is
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going to equal Omega *
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R which is equal to.5 * 02 which is
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equal to 0
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01 all right so that is the velocity of
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the point on the outer surface of C
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well the outer surface of C is meshed
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with the outer surface of s this uh spur
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gear so they're going to move together
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because the gears are locked together so
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the
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velocity of the spur gear s right so the
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outer a point on the outer surface of
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the spur gear is going to have the same
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velocity as the Cog because they're mesh
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together so that's going to equal
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01 and then we have the radius of the
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spur gear is equal to
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0.5 so the angular speed Omega of the
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spur gear is going to equal V over R
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which is 0.01 /
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.5 which is
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02 so this spur gear is moving with an
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angular velocity of 02 well this Rod is
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physic is rigidly attached to that spur
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gear so this Rod is also moving at an
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angle velocity of
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02 so this point out here point
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a so we have
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Rod what ba has a mo angular velocity of
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02 and a radius or length in this case
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of
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2 so the speed of point
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a is going to equal Omega R which is 02
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*
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.2 which equals
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04 and the units would be me/ second all
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right so that's the speed the direction
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of that speed is at this angle right
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because it's tangential or it's normal
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this is a right angle it's normal to the
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radius or the length of the rod in this
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case so then the velocity then of the of
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point a let's write it as a vector would
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equal uh
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0.4 and the X part would be the negative
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s of theta I hat and the Y part would be
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positive cosine of theta J hat and we're
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told we're looking at the specific Theta
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of
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30° the wi
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window moves in the y direction solely
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right because it's con it's constrained
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on the side so it doesn't move side to
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side so the velocity of the window is
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going to equal the Y part of
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va which is equal to
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04 cine of 30 then equal to
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034 m/ second so as what you see then is
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that we used this V equals Omega R over
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and over and over again in various forms
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as we work our way through the various
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gears and shaft systems and then we
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recognize the window is in the y
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direction VA is in the nor is normal to
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the radius Direction so it's in the
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tangential
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Direction um and so we find the Y part
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of that tangential velocity and find our
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velocity of the window so let's take
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another look at example Le here here we
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have a series of Gears that are meshed
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but what this includes is the the
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kinematic equations that we have been
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seeing so we start at Point a we're
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given Alpha of
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a is equal to 90 so knowing Alpha a we
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can find the acceleration of a point on
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the outside edge of a where it's meshed
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with B so the acceleration of a is going
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to
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equal Alpha a * r
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and that the acceleration of that point
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is also going to be equal to the
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acceleration of the outer point on
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B which is equal to Alpha B * RB this is
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r
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a all right so that will allow us to
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find Alpha B so from here we get Alpha B
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is equal to we move the RS over Alpha a
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which is known R
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A and so we get Alpha b equals Alpha a *
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R A over
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RB now the shaft of B is attached to the
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shaft of C right so from B to C Alpha B
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is equal to Alpha C right because
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they're on the same shaft so if B is
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accelerating with an angular
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acceleration of alpha c b c will have
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the same angular acceleration then we
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have a point on the gears where C is
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meshed with
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d so knowing Alpha C Alpha C can give us
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[Music]
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AC so AC is equal to Alpha C *
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RC and that the acceleration of the
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point on the outer side of C is equal to
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the acceleration of the point on the out
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outside of D because they're meshed so
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that's a d which is equal to Alpha d r d
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and we're looking for
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the angular
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velocity uh information about D and so
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if we can find the angular acceleration
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information about D we can figure out
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through kinematics what the velocity is
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so solve this for Alpha D is equal to
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Alpha c r c over Rd but Alpha C is equal
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to this right so this is going to equal
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Alpha a r a RC over RB
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Rd and all of those are known so we can
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plug in all those numbers we get
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90 times r a is 15 so
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0.015 time RC is 25
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.025 divided by RB
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.05 time Rd
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.075 plug all those numbers in and we we
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get that Alpha D is equal to
00:18:01
9 so now knowing Alpha D we want to
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figure out information about angular
00:18:07
velocities it looks like so it says we
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start from rest so Omega KN is equal to
00:18:14
zero and we're trying to get to a
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velocity of 600 RPM so Omega final is
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600 revolutions per
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minute but we need that in radians per
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second to keep our units
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uh the same so there are in 1 minute
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there are 60
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seconds and in one revolution there are
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2 pi
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radians multiply that out we get
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62.8 radians per
00:18:46
second so we have an initial veloc Omega
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angular velocity a final
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Omega um we have the alpha the
00:18:56
acceleration equals 9 which is constant
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so we can use our constant acceleration
00:19:01
equations we want to know determine the
00:19:04
time that it takes to do this and
00:19:07
determine the number of revolutions it
00:19:09
takes to do this so if we go back to our
00:19:11
kinematic
00:19:12
equations here they are in the first one
00:19:16
we're looking for time we know Omega
00:19:18
Omega KN and Alpha so we're going to use
00:19:20
the first one to find time once we know
00:19:23
time we know Omega KN we know time we
00:19:26
know Alpha we can find our our angular
00:19:30
position the number of Revolutions in
00:19:32
other words all right so the first
00:19:35
equation
00:19:37
was the
00:19:39
Omega equal Omega KN + Alpha
00:19:43
T so that's equal
00:19:46
to0 this is
00:19:49
62.8 + 9 * T gives us a t = to
00:19:54
6.98 seconds and then we use the second
00:19:58
equation which is the Theta minus Theta
00:20:01
KN is equal to Omega t + 12 Alpha t^ 2
00:20:08
this is zero so we plug everything in
00:20:11
and we
00:20:13
get 219
00:20:16
radians divide that by 2 pi you get
00:20:20
34.9
00:20:21
Revolutions in order for it to go from
00:20:24
zero to 600
00:20:27
RPM
00:20:28
now let's take a quick look at Absolute
00:20:32
rigid body planer Motion in our next
00:20:35
lessons we're going to be looking at
00:20:36
relative planer motion So today we're
00:20:39
going to look at
00:20:40
Absolute think back to when we were
00:20:42
looking at relative motion like with our
00:20:45
pulley systems and things like that this
00:20:47
is going to parallel that except now
00:20:49
we're
00:20:50
incorporating rotation as well so planer
00:20:53
motion involves translation and rotation
00:20:56
together to solve these types of
00:20:59
problems we're going to pick some point
00:21:02
on our body sometimes that Point's given
00:21:04
sometimes we just pick it and give some
00:21:07
information about that point relative to
00:21:10
some fixed origin so that's our position
00:21:13
s then we have an angle also drawn to a
00:21:17
fixed line on our body from a reference
00:21:20
and through the geometry of our body we
00:21:22
can relate s to Theta and then we can
00:21:25
take time derivatives of this
00:21:27
relationship to find velocities
00:21:30
accelerations angular velocities and
00:21:33
angular accelerations so this will make
00:21:35
more sense through an example so we have
00:21:37
a series of bars here so there's some
00:21:40
motor that pulls this bar around the
00:21:42
first bar ab and that causes B to move
00:21:46
which causes C to move and c will slide
00:21:48
along the slot back and forth depending
00:21:51
on what this is H doing so in this case
00:21:54
A and B are equal so we've got 3 for a
00:22:00
so a is equal to 3 and b is equal
00:22:04
to3 if a and b are equal then if this
00:22:07
angle is Theta this angle is also Theta
00:22:10
it's an equilateral triangle all right
00:22:13
so our goal then the position of C is
00:22:16
what we're trying to describe so let's
00:22:17
call this
00:22:19
distance from our fixed point to c s and
00:22:23
our goal is to relate s to Theta and
00:22:26
then we can take derivatives so to
00:22:28
relate s to Theta we look at these
00:22:30
triangles well let's split it right
00:22:33
those two triangles are the same right
00:22:34
because this is 3 that's 3 that's Theta
00:22:37
that's Theta so s is going to equal 2
00:22:42
times whatever this this Edge is of one
00:22:45
of these triangles well that edge
00:22:50
is3 times it's adjacent to our angle so
00:22:54
cosine of theta so s is point 6 cosine
00:22:59
Theta that relates s to Theta now let's
00:23:02
take some time derivatives so we get s
00:23:05
dot which is V is equal to 6 the
00:23:08
constant the time derivative of cosine
00:23:11
of theta derivative of the cosine is the
00:23:14
negative s of theta but Theta itself is
00:23:17
dependent on T so the inner derivative
00:23:19
is Theta dot then we take another time
00:23:23
derivative s double dot which is a we
00:23:26
get 6 is still a constant derivative of
00:23:31
the first so the derivative of negative
00:23:33
s is negative
00:23:38
cosine of theta times the inner
00:23:40
derivative of theta dot times the second
00:23:43
Theta dot plus 6 * negative s the 1 time
00:23:49
the derivative of the second Theta
00:23:51
double
00:23:52
dot so now we know all of these things
00:23:55
right so we have that Theta is is equal
00:23:58
to 30 Omega which is Theta dot is equal
00:24:02
to 10 and Alpha which is Theta double do
00:24:06
is equal to two so we can plug all of
00:24:09
those things into these Expressions we
00:24:11
get S is equal
00:24:14
to.
00:24:16
519 s dot is equal to V which is equal
00:24:20
to
00:24:21
-3 and s double dot is equal to a t
00:24:25
which is equal to5
00:24:28
52.6 so this is in
00:24:30
[Music]
00:24:31
meters this is in me/ second and this is
00:24:35
in me/ second squared
