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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.

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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

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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 )

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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

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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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Extra Reading: Acceleration on a Curved Path

q  At every point along the path, the total acceleration is made of by its centripetal and tangential components.

Conceptual understanding only for this course