Announcement:

LABS start this week !

Homework #2 : due Fri. (Sept. 15) by 5.00 PM on
webassign Problems from Chapter 3 and 4.
Physics 151: Lecture 6, Pg 1
Physics 151: Lecture 6
Today’s Agenda

Today’s topics
Finish circular motion
Discuss relative motion
text sections 4.4-4.6
Physics 151: Lecture 6, Pg 2
See text: 4-4
Review: Uniform Circular Motion (UCM)

Motion in a circle with:
y
Constant Radius R
v
(x,y)
a
 Constant Speed v = |v|
R
x
acceleration ?
a
=0
a
= const.
Physics 151: Lecture 6, Pg 3
Polar Coordinates...



In Cartesian co-ordinates we say velocity dx/dt = v.
 x = vt
In polar coordinates, angular velocity d/dt = .
  = t
 has units of radians/second.
y
Displacement s = vt.
v
but s = R = Rt, so:
R
v = R
t
s
x
frequency (f) = /2 
period (T) = 1 / f = 2/
Physics 151: Lecture 6, Pg 4
Lecture 6, ACT 1
Uniform 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 ?
Physics 151: Lecture 6, Pg 5
See text: 4-4
Acceleration in UCM:

Even though the speed is constant, velocity is not constant
since the direction is changing: must be some acceleration !
 Consider average acceleration in time t
aav = v / t
v
v2
R
v1
v2
v1
Physics 151: Lecture 6, Pg 6
Acceleration in UCM:


This is called Centripetal Acceleration.
Now let’s calculate the magnitude:
v
v2
v2
R
R
v1
Similar triangles:
v1
v R

v
R
But R = vt for small t
v vt

So:
v
R
v v 2

t
R
v2
a
R
Physics 151: Lecture 6, Pg 7
See text: 4-4
Centripetal Acceleration

UCM results in acceleration:
Magnitude:
a = v2 / R = 2 R
Direction:
-^
r (toward center of circle)
a
R

Physics 151: Lecture 6, Pg 8
Lecture 6, ACT 1
Uniform 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) 20 m
(b) 200 m
(c) 2,000 m
(d) 20,000 m

Physics 151: Lecture 6, Pg 9
Example:
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
a = 0.00272 m/s2
Physics 151: Lecture 6, Pg 10
Moon...


So we find that amoon / g = .000278
Newton noticed that RE2 / R2moon = .000273
amoon
g
R

RE
This inspired him to propose that FMm  1 / R2
(more on gravity later)
Physics 151: Lecture 6, Pg 11
Lecture 6, ACT 2
Uniform Circular Motion
A satellite is in a circular orbit 600 km above the Earth’s
surface. The acceleration of gravity is 8.21 m/s2 at this
altitude. The radius of the Earth is 6400 km.
Determine the speed of the satellite, and the time to
complete one orbit around the Earth.
Answer:
• 7,580 m/s
• 5,800 s
Physics 151: Lecture 6, Pg 12
Lecture 6, ACT 3
Uniform Circular Motion
A stunt pilot performs a circular
dive of radius 800 m. At the
bottom of the dive (point B in the
figure) the pilot has a speed of 200
m/s which at that instant is
increasing at a rate of 20 m/s2.
What acceleration does the pilot
have at point B ?
Physics 151: Lecture 6, Pg 13
Lecture 6, ACT 3
The Pendulum
1m
 = 30°
Which statement best describes
the motion of the pendulum bob
at the instant of time drawn ?
•
•
A) vr = 0 ar = 0
v  0 a  0
the bob is at the top of its swing.
which quantities are non-zero ?
B) vr = 0 ar = 0
v = 0 a  0
C) vr = 0 ar  0
v = 0 a  0
Physics 151: Lecture 6, Pg 14
Lecture 6, ACT 3
The Pendulum
Solution
O
1m
 = 30°
a
a
ar
NOT uniform circular motion :
Is circular motion so must be ar  0
Speed is increasing so a not zero
At the top of the swing, the bob
temporarily stops, so v = 0.
C) vr = 0 ar  0
v = 0 a  0
What are components of v and a at  = 0 ?
Animation
Physics 151: Lecture 6, Pg 15
Inertial Reference Frames:

A Reference 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 IRF’s in this course.

Valid IRF’s 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.
Animation
Physics 151: Lecture 6, Pg 16
Text Ch 4, sect 6
Lecture 6 – ACT 4
Relative Motion
Consider an airplane flying on a windy day.
 A pilot wants to fly from New Haven to Bradley airport.
She knows that Bradley is 120 miles due north of
New Haven and there is a wind blowing due east at
30 mph. She takes off from New Haven Airport at
noon. Her plane has a compass and an air-speed
indicator to help her navigate. She uses her compass
at the start to aim her plane north, and her air speed
indicator tells her she is traveling at 120 mph with
respect to the air.

After one hour she is:
(A) at Bradley
(B) east of Bradley
(C) southeast of Bradley
Physics 151: Lecture 6, Pg 17
See text: Ex. 4.9 and 4.10
Lecture 6 – ACT 4
Relative Motion

The plane is moving north in the IRF attached to the air:
 Vp, a is the velocity of the plane w.r.t the air.
Air
Vp,a
See example 4-9,10 (Boat Crossing a River)
Physics 151: Lecture 6, Pg 18
See text: Ex. 4.9 and 4.10
Lecture 6 – ACT 4
Relative Motion

But the air is moving east in the IRF attached to the ground.
 Va,g is the velocity of the air w.r.t the ground (i.e. wind).
Air
Vp,a
Va,g
Physics 151: Lecture 6, Pg 19
See text: Ex. 4.9 and 4.10
Lecture 6 – ACT 4
Relative Motion

What is the velocity of the plane in an IRF attached to the
ground?
 Vp,g is the velocity of the plane w.r.t the ground.
Vp,g
Physics 151: Lecture 6, Pg 20
See text: Ex. 4.9 and 4.10
Relative Motion...

Vp,g = Vp,a + Va,g Is a vector equation relating the airplanes
velocity in different reference frames.
Va,g
Vp,a
Vp,g
The north component of vp,g is just her air speed. So she gets
far enough north. The wind takes her further east.
Answer is (B) – due east.
Physics 151: Lecture 6, Pg 21
Lecture 6, ACT 5
Relative 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) 50 2  25s
b) 50 1  50s
c) 50
3  29s
d) 50
2  35s
2m/s
50m
1m/s
Physics 151: Lecture 6, Pg 22
Lecture 6, ACT 5
Solution
y
Choose x axis along riverbank and y axis across river
x

The time taken to swim straight across is (distance across) / (vy )

Since you swim straight across, you must be tilted in the water so that
your x component of velocity with respect to the water exactly cancels
the velocity of the water in the x direction:
1m/s
y
2m/s
2 2 12
 3 m/s
1m/s
x
Physics 151: Lecture 6, Pg 23
Lecture 6, ACT 5
Solution

So the y component of your velocity with respect to the water is

So the time to get across is
3 m/s
50 m
 29s
3m s
3 m/s
50m
Answer (c)
y
x
Physics 151: Lecture 6, Pg 24
Recap for today:

Circular Motion (Text Ch. 4.4)

Relative Motion (Text Ch 4.6)

Reading assignment for Wed.
Read about Forces: Ch 5.1-3

Homework #2 : due Fri. (Sept. 15) by 5.00 PM on webassign
Problems from Chapter 3 and 4
Physics 151: Lecture 6, Pg 25