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Physics 111: Lecture 3, Pg 1

Physics 111: Lecture 3Physics 111: Lecture 3

Today’s AgendaToday’s Agenda

� Reference frames and relative motion

� Uniform Circular Motion


Inertial Reference Frames:Inertial Reference Frames:

� A Reference FrameReference Frame is the place you measure from.� It’s where you nail down your (x,y,z) axes!

� An Inertial Reference Frame (IRF) is one that is not accelerating.� We will consider only IRFs in this course.

� Valid IRFs can have fixed velocities with respect to each other. � More about this later when we discuss forces.� For now, just remember that we can make measurements

from different vantage points.

Cart ontrack on

track


Relative MotionRelative Motion

� Consider a problem with twotwo distinct IRFs:� An airplane flying on a windy dayAn airplane flying on a windy day..

� A pilot wants to fly from Champaign to Chicago. Having asked a friendly physics student, she knows that Chicago is 120 miles due north of Urbana. She takes off from Willard Airport at noon. Her plane has a compass and an air-speed indicator to help her navigate.

� The compass allows her to keep the nose of the plane pointing north.

� The air-speed indicator tells her that she is traveling at 120 miles per hour with respect to the airwith respect to the air.


Relative Motion...Relative Motion...

� The plane is moving north in the IRF attached to the air:

� VVp, a is the velocity of the plane w.r.t. the air.

AirAir

VVp,a



� But suppose the air is moving east in the IRF attached to the ground.

� VVa,g is the velocity of the air w.r.t. the ground (i.e. wind).

VVa,g

AirAir

VVp,a



� What is the velocity of the plane in an IRF attached to the ground?

� VVp,g is the velocity of the plane w.r.t. the ground.

VVp,g



� VVp,g = VVp,a + VVa,g Is a vector equation relating the airplane’s velocity in different reference frames.

VVp,g

VVa,g

VVp,a

Tractor


Lecture 3, Lecture 3, Act 1Act 1Relative MotionRelative Motion

� You are swimming across a 50m wide river in which the current moves at 1 m/s with respect to the shore. Your swimming speed is 2 m/s with respect to the water. You swim across in such a way that your path is a straight perpendicular line across the river.� How many seconds does it take you to get across ?

(a)

(b)

(c)

50 3 29=

2 m/s

1 m/s50 m

35250 =

50150 =


Lecture 3, Lecture 3, Act 1Act 1solution solution

� The time taken to swim straight across is (distance across) / (vy )

Choose x axis along riverbank and y axis across rivery

x

� Since you swim straight across, you must be tilted in the water so thatyour x component of velocity with respect to the water exactly cancels the velocity of the water in the x direction:

2 m/s 1m/sy

x

1 m/s

2 1

3

2 2−= m/s


Lecture 3, Lecture 3, Act 1Act 1solutionsolution

� So the y component of your velocity with respect to the water is

� So the time to get across is

y

x

3m/s

503

29m

m ss=

50 m

3m/s


Uniform Circular MotionUniform Circular Motion

� What does it mean?

� How do we describe it?

� What can we learn about it?


What is UCM?What is UCM?

� Motion in a circle with:

� Constant Radius R

� Constant Speed v = |vv|

R

vv

x

y

(x,y)

Puck on ice


How can we describe UCM?How can we describe UCM?

� In general, one coordinate system is as good as any other:� Cartesian:

» (x,y) [position]

» (vx ,vy) [velocity]

� Polar:» (R,θ) [position]

» (vR ,ω) [velocity]

� In UCM:

� R is constant (hence vR = 0).

� ω (angular velocity) is constant.� Polar coordinates are a natural way to describe UCM!Polar coordinates are a natural way to describe UCM!

RR

vv

x

y

(x,y)

θ


Polar Coordinates:Polar Coordinates:

� The arc length s (distance along the circumference) is related to the angle in a simple way:

s = Rθ, where θ is the angular displacement.� units of θ are called radians.

� For one complete revolution:

2πR = Rθc

� θc = 2πθ has period 2π.

1 revolution = 21 revolution = 2π π radiansradians

RR

vv

x

y

(x,y)s

θ


Polar Coordinates...Polar Coordinates...

x = R cos θ y = R sin θ

π/2 π 3π/2 2π

-1

1

0

sincos

RR

x

y

(x,y)

θ

θ


Polar Coordinates...Polar Coordinates...

� In Cartesian coordinates, we say velocity dx/dt = v.� x = vt

� In polar coordinates, angular velocity dθ/dt = ω.� θ = ωt

� ω has units of radians/second.

� Displacement s = vt.

but s = Rθ = Rωt, so:RR

vv

x

y

sθ=ωt

v = ωR

Tetherball


Period and FrequencyPeriod and Frequency

� Recall that 1 revolution = 2π radians� frequency (f) = revolutions / second (a)� angular velocity (ω) = radians / second (b)

� By combining (a) and (b)� ω = 2π f

� Realize that:� period (T) = seconds / revolution� So T = 1 / f = 2π/ω

RR

vv

s

ω = 2π / T = 2πf

ω


Recap:Recap:

RR

vv

sθ=ωt

(x,y)

x = R cos(θ) = R cos(ωt) y = R sin(θ) = R sin(ωt)

θ = arctan (y/x)

θ = ωt

s = v t

s = Rθ = Rωt

v = ωR


Aside: Polar Unit VectorsAside: Polar Unit Vectors

� We are familiar with the Cartesian unit vectors: i j ki j k

� Now introduce“polar unit-vectors” rr and θθ :� rr points in radial direction� θθ points in tangential direction

RR

x

y

ii

j j θθ

rr̂^θθ ̂

^

^̂ ^̂

^̂^̂

(counter clockwise)


Acceleration in UCM:Acceleration in UCM:

� Even though the speedspeed is constant, velocityvelocity is notnot constant since the direction is changing: must be some acceleration!

� Consider average acceleration in time ∆t aav = ∆vv / ∆t

vv2

ω∆t

vv1

vv1vv2

∆vv

RR



seems like ∆vv (hence ∆vv/∆t )

points at the origin!

RR

� Even though the speedspeed is constant, velocityvelocity is notnot constant since the direction is changing.

� Consider average acceleration in time ∆t aav = ∆vv / ∆t

∆vv



� Even though the speedspeed is constant, velocityvelocity is notnot constant since the direction is changing.� As we shrink ∆t, ∆vv / ∆t dvv / dt = aa

aa = dvv / dt

We see that a a points

in the - R R direction.

RR



� This is called This is called Centripetal Acceleration.� Now let’s calculate the magnitude:

vv2

vv1

vv1vv2

∆vv

RR∆∆RR

∆ ∆vv

RR

=Similar triangles:

But ∆R = v∆t for small ∆t

So:∆∆vt

vR

=2∆ ∆v

vv tR

=

avR

=2


Centripetal AccelerationCentripetal Acceleration

� UCM results in acceleration:� Magnitude: a = v2 / R� Direction: - r r (toward center of circle)

Raa

ω

^̂


Useful Equivalent:Useful Equivalent:

( )RR

a2

=

We know that and v = ωR

Substituting for v we find that:

avR

=2

ω

a = ω2R


Lecture 3, Lecture 3, Act 2Act 2Uniform Circular MotionUniform Circular Motion

� A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ?

(a) 500 m

(b) 1000 m

(c) 2000 m


Lecture 3, Lecture 3, Act 2Act 2SolutionSolution

avR

g= =2

9

2

2

2

2

s

m8199

s

m90000

g9

vR

.×==

m1000m819

10000R ≈=

.D R m= ≈2 2000

2km


Example: Propeller TipExample: Propeller Tip

� The propeller on a stunt plane spins with frequency f = 3500 rpm. The length of each propeller blade is L = 80cm. What centripetal acceleration does a point at the tip of a propeller blade feel?

f

L

what is aa here?


Example:Example:

� First calculate the angular velocity of the propeller:

� so 3500 rpm means ω = 367 s-1

� Now calculate the acceleration. � a = ω2R = (367s-1)2 x (0.8m) = 1.1 x 105 m/s2

= 11,000 g

� direction of aa points at the propeller hub (-r r ).

1 11

602 0105 0105 rpm s-1= = ≡rot

x s

x rad

rot

rad

smin

min. .π

^̂


Example: Newton & the MoonExample: Newton & the Moon

� What is the acceleration of the Moon due to its motion around the Earth?

� What we know (Newton knew this also):� T = 27.3 days = 2.36 x 106 s (period ~ 1 month)

� R = 3.84 x 108 m (distance to moon)

� RE = 6.35 x 106 m (radius of earth)

R RE


Moon...Moon...

� Calculate angular velocity:

� So ω = 2.66 x 10-6 s-1.

� Now calculate the acceleration. � a = ω2R = 0.00272 m/s2 = 0.000278 g� direction of aa points at the center of the Earth (-rr ).

1

27 3

1

864002 2 66 10 6

..

rot

dayx

day

sx

rad

rotxπ = − s-1

^


Moon...Moon...� So we find that amoon / g = 0.000278

� Newton noticed that RE2 / R2 = 0.000273

� This inspired him to propose that FMm ∝ 1 / R2

� (more on gravity later)

R RE

amoon g


Lecture 3, Lecture 3, Act 3Act 3Centripetal AccelerationCentripetal Acceleration

� The Space Shuttle is in Low Earth Orbit (LEO) about 300 km above the surface. The period of the orbit is about 91 min. What is the acceleration of an astronaut in the Shuttle in the reference frame of the Earth? (The radius of the Earth is 6.4 x 106 m.)

(a) 0 m/s2

(b) 8.9 m/s2

(c) 9.8 m/s2


� First calculate the angular frequency ω:

� Realize that:


RO

300 kmRO = RE + 300 km = 6.4 x 106 m + 0.3 x 106 m = 6.7 x 106 m

RE

1-s 00115.0 rot

rad2 x

s60

1 x

rot

91

1 == πω min

min


� Now calculate the acceleration:


a = ω2R

a = (0.00115 s-1)2 x 6.7 x 106 m

a = 8.9 m/s2


Recap for today:Recap for today:

� Reference frames and relative motion.

� Uniform Circular Motion

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