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Unformatted text preview: {,— UNIFORM CIRCULAR MOTION— KINEMATICS AND DYNAMICS A particle is moving on a circle of radius R at a constant speed S. First, we begin by
describing the motion precisely kinematics. Let us put the circular orbit in the xy—plane with the
center of the  ' Y
(9 at x=0, y=0. Thevery
ﬁrst quantity we deﬁne is the
Period: Time taken to go
around once, T. The speed can then be immediately written as: km
T As you can see when the particle moves around the circle, the radius rotates as a function of
time. That is Why it is customary to describe the motion in terms of . I 1
revolutions per see (us) so T = —sec
.. "S
: (rps) '
. » . 60
or revolutions per minute (nm ) , T = —sec
r ' n m (rpm)
For instance, 15rpm means T=4sec. Speed is an interesting concept but as before it is rather limiting. We need to look deeper.
Position Vector: We notice that the particle moves at ﬁxed distance away from the center but the
radius rotates. Hence, its position vector will be written as: g: = R? ' 1+ (1)
where 1a is a unit vector along the radius which rotates so as to go around once in time T.
Velocity Vector: Velocity is deﬁned as rate of change of position vector so we need to ﬁnd the
displacement vector.
Consider a time interval
At during which P
rotates by angle A3. Wu”. _ displacement during At is RAG so magnitude of instantaneous velocity is kw (At—>0)
At Notice, direction of displacement is perpendicular to F so direction of velocity is along the
tangent to the circle, We deﬁne 7” unit vector along tangent and write
RA®
V =
_’ At
We will soon introducc a formal deﬁnition for rate of change of angle with time, for now let us
introduce as new symbol (greek letter omega) r” A®
w = —
At
and note V : wa ' —> 2 —> and f rotates with time For uniform case rate of rotation is constant. So Eq(2) tells us that magnitude of Vis constant.
DIRECTION CHANGES! _ Acceleration Vector: Since the velocity vector is rotating the object has an accelerati'On. Again .
we need to calculate change in Z and divide by At. Change in magnitude of V: AV = RwA® So magnitude of acceleration is: a = RWA—® = sz
Al and 9 must be perpendicular to 2° . If you look at the K it is continuously turning TOWARD the center SO 9; is along ~41?
SO g = —Rw2f & :9 rotates
So 91 is constant in magnitude but also rotates. This is a special‘case so this acceleration has a special name: CENTRIPETAL
ACCELERATION. ' ‘ Finally we go back and look at A63
w=~—
At This is the rate at which the radius vector sweeps out an angle as it rotates so it is not
surprising that we call it ANGULAR VELOCITY Question? What is the direction of 113. Well / is a positive angle and is a negative angle and rotation is about an axis perpendicular to 0 so it makes sense to say that 1v is perpendicular
to plane of circle. In our case circle is in xy—plane so  :tz". + 2 for counter—clockwise (positive 3's) — 2 for eleckwise (negative 8's ). This is summarized by rightHand Rule: Curl ﬁngers of right hand along direction of motion on the circle, extend
your thumb, it? points it direction of 143 A®A
w=——zl *Al‘ Tabb9 ANGULA‘R VELOCITY L°T“‘ rad/sec vector So to summarize Kinematics:
Position: 1: =' Rf rotates by w rad/sec * (1)
W i ~ * i e * ' *Ve'elocityz' 1) : Rwr“ rotates by W'TEId/SCC" "" (2) ” Centripetal Acceleration 2
a =—Rw219=—Z—f (3) rotates byArad/ sec angglar velocity: Dm amics A particle moving on a circle of radius R at a constant angular velocity Ly has a centripetal acceleration . ‘~
. V 2 _
ac = ~Rw219 = —— F
H‘ R
Newton’s law M g = 2 E requires that for this motion to occur we'must provide a 
, 2 ‘
CENTRIPETAL FORCE Fe = ~MRw2f = ~ M]: a a (5) It is to be noted that Fc must come frOm one or more of the available forces: Weight, normal force, tension, spring force, friction
NOTE: F0 CANNOT BE DRAWN ON A DIAGRAM ...
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 Spring '10
 Shawhan
 Physics

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