physics 201, lecture 6 trivial math review: circle€¦ · 1 physics 201, lecture 6 today’s...
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Physics 201, Lecture 6 Today’s Topics
q Uniform Circular Motion (Section 4.4, 4.5) n Circular Motion n Centripetal Acceleration n Tangential and Centripetal Acceleration
q Relative Motion and Reference Frame (Sec. 4.6)
v Hope you have previewed!.
Trivial Math Review: Circle q A circle can be described by a center and a radius r.
q The circumference (i.e. linear path length along a full circle) of a circle of radius r is 2πr
q A full circular angle is 360o or 2π
q A tangential line is perpendicular to the radial line from center to the tangential point. q Arc distance (arc length ) s = r Δθ
r
tangential line
r
Δθ
s
Review: Kinematical Quantities in Vector Form
q Displacement:
q Velocity (average and instantaneous):
q Acceleration (average and instantaneous):
if rrr −=Δ
dtvd
tva
tva
t
=Δ
Δ=
Δ
Δ=
=Δ 0lim, avg
dtrd
trv
trv
t
=Δ
Δ=
Δ
Δ=
=Δ 0lim, avg
Special Notes q The mathematical treatment for circular motion kinematics in
the next three slides represents some extra readings beyond the textbook contents. It is meant to help you to have a better understanding of kinematical formulas for the circular motion.
q In my judgment, the book treatment is over simplified and possibly less convincing to those who want a deeper understanding.
q In any case, derivation for those formulas is not required for
this course. Please pay more attention to the final results that I will summarize in one slide later.
2
q Delta of two vectors in the same direction:
q Delta of two vectors with the same length:
Math Preparation: Difference of Two Vectors
Δθ ri rf
= - = Δr r̂
Δr = rf −
ri
dθ ∼ 0
ri rf
dr = rf −ri = rdθ •θ̂
90ο -Δθ/2 ∼90ο
Δθ 0
q For vector r = r r, change can be in length and in direction. § keep direction but change in length: § maintain length but change in direction:
Math Preparation: Differential of a Vector
dθ
dr
ri rf
θθ ˆˆ r r drdrd +=Together:
^
rdr ˆdr
θθ ˆ dr
r
r̂
θ̂
radial unit vector
tangential unit vector
r ˆdr
product rule
Velocity in Circular Motion q Recall:
q For circular motion dr =0 v In circular motion, velocity is always in tangential direction, i.e. always perpendicular to radial vector.
q Definition: Angular velocity ω = dθ/dt
Ø and |v| = rω
r
v
θθ ˆˆ r r drdrd +=
θθ ˆ r drd = θθ ˆ r vdtdr
dtd
==
θωθθ ˆˆ r v rdtdr
dtd
===
Uniform Circular Motion q Uniform circular motion is circular motion with constant
angular velocity (ω) .
q Trivial quiz: for a uniform circular motion with ω, how long does it take to complete a full circle?
( 2π/ω)
q For uniform circular motion, period (T) is defined as the time the moving object takes in one full circle.
T = 2π/ω = 2πr/ v q Note: A related quantity: frequency f is defined as f = 1/T
3
Quick Quizzes: Uniform Circular Motion q As shown a particle in uniform circular motion has a period T
and a radius R. (assume it runs in counter-clockwise.) Ø What is the magnitude of its instantaneous velocity when it passes point A? 2πR/T, 2R/T, zero, other Ø What is the magnitude of its average velocity in a time interval when it completes a full circle? 2πR/T, 2R/T, zero, other Ø What is the magnitude of its average speed in a time interval when it completes a full circle? 2πR/T, 2R/T, zero, other
Ø After class exercises: Answer the same questions for time interval from point A to point B.
Acceleration in Uniform Circular Motion q recall:
q For uniform circular motion, r and ω are both constants.
here we used: q In uniform circular motion, a is always pointing towards the center Centripetal Acceleration (ac)
q Properties of centripetal acceleration § Always points to the center
§ ac = r ω2 = v2/r
θω ˆ v r=
)ˆ(ˆ
2 r v a −=== ωθ
ω rdtdr
dtd
)ˆ(ˆ
r−=ωθdtd
(why: see board)
r
v
ac
Summary of Kinematics for Uniform Circular Motion
q Instantaneous velocity is always in tangential direction
(The above is true even for non-uniform circular motion)
q Angular velocity ω is a constant: ω = 2π/T = 2πf
q Instantaneous acceleration is always centripetal
q For circular motions, v and a are never constant ! q Note: vave ≠ rω, and aave ≠ rω2 !
v = r ω θ̂, i.e. v=rω
a = rω 2 (−r̂) , i.e. ac = rω2 =
v2
r
r
v
ac
Exercise: Spin of the Earth q The radius of earth is 6.37x106 m. To a good approximation, the spin of
the earth is uniform with a period T. Quick Quiz: How long is T? Answer: T= 24 hr = 24x3600 = 86400 s !
Consider a person standing on the Equator: § What is angular speed of the person? ( ω = 2π /T = 7.27x10-5 rad/s ) § What is the linear speed of that person? ( v =rω = 463.1 m/s )
§ How much is his acceleration ? ( ac = rω2 = 0.034 m/s2 )
4
Non-Uniform Circular Motion q In a generic (non-uniform) circular motion, acceleration usually
has both centripetal and tangential components
è Total acceleration:
a = ac + at
Conceptual understanding only for this course
After Class Quiz q We have just learnt that for a particle in uniform circular
motion, the direction of its acceleration is always centripetal. However, for a generic circular motion, the acceleration can have a centripetal and a tangential component.
Ø what can we say about the velocity in circular motion?
A: For uniform circular motion, the velocity is always perpendicular to radial vector r. (i.e. tangential). But for a generic circular motion, the velocity can have both tangential and centripetal components. B: For any circular motion, the velocity is always tangential.
Relative Motion q All motions are measured in a reference frame. Same motion can be
measured to be differently in different reference frame. § e.g. A passenger sits in a moving bus.
• w.r.t bus, the passenger is stationary (v=0) • w.r.t Earth, the passenger is moving at vbus
q Conversion between reference frames
vobj _wrt _FrameB =vobj _wrt _FrameA +
vFrameA_wrt _FrameB
Relative Motion in 1-D q On a straight road, a bus is moving forward at a speed of 10 m/s
(i.e. vbus_Earth = +10 m/s). in the meanwhile, a man is walking inside the bus.
Quiz 1: If the man is walking forward at 1 m/s w.r.t the bus (i.e. vman_bus = +1.0 m/s), what is the man’s velocity w.r.t. the Earth? Answer: vman_Earth = 11 m/s = 10 + 1 = vman_bus + vbus_Earth
Quiz 2: If the man is walking backward at 1 m/s instead (i.e. vman_bus = -1.0 m/s), what is the man’s velocity w.r.t. the Earth? Answer: vman_Earth = 9 m/s = 10 + (-1 )= vman_bus + vbus_Earth
vobj _wrt _FrameB = vobj _wrt _FrameA + vFrameA_wrt _FrameB
vman_wrt _Earth = vman_wrt _Bus + vBus_wrt _Earth
5
Relative Motion: Galilean Transformation q Conversion between reference frames (Galilean Transformation)
visualization example : A=bus, B=earth, o=rain drops
vobj _wrt _FrameB =vobj _wrt _FrameA +
vFrameA_wrt _FrameB
vo_A
vA_B
vo_B
vo_A
vA_B
vo_B
Same principle but a different configuration
One example
Relative Velocity Example: Rain Trace as Seen Inside a Bus
vrE : velocity rain w.r.t Earth vbE: velocity bus w.r.t Earth vrb: velocity rain w.r.t. bus
bErbrE vvv +=
Rain seen on Earth
vrE
vbE
vrb vrE
vbE
vrb =vrE −vbE
i.e.
Relative Velocity Example: Cross a River
vrE : velocity river w.r.t Earth vbE: velocity boat w.r.t Earth vbr: velocity boat w.r.t. river
rEbrbE vvv +=
Water flow
Exercise: Airplane in Wind q A jet airliner moving at 590 mph due east moves into a region where
the wind is blowing at 140 mph in a direction 60° north of east. What is the speed and direction of the aircraft (w.r.t. Earth)?
q Solution: ( i = east, j = north, J=jet, E=Earth, W= wind) use (vector) relationship vJE = vJW +vWE vJE = (590+70)i + 121.24j= 660i + 121.24j |vJE | = 671mph =sqrt(6602 +121.212), at 10.41o NofE =atan(121.21/660)
vJW = 590 i,
vWE = 140xcos(60o)i + 140xsin(60o)j = 70i + 121.24j vJE = vJW +vWE
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