Physics 121,Lecture 10,Pg 1
Physics 121,Sections 9,10,11,and 12
Lecture 10
Today’s Topics:
? Homework 4,Due Friday Sept,30 @ 6:00PM
? Ch.4,# 52,57,58,59,60,65,80,84,87,and 90.
? Chapter 5,Rotation
? Angular variables
? Centripetal acceleration
? Kepler’s laws
Physics 121,Lecture 10,Pg 2
Chapter 5
Uniform Circular Motion
? What does it mean?
? How do we describe it?
? What can we learn about it?
Physics 121,Lecture 10,Pg 3
Review
( displacement,velocity,acceleration )
Velocity,
Acceleration,
a = dv / dt
? 3-D Kinematics, vector equations:
r = r(t) v = dr / dt a = d2r / dt2
v = dr / dt
vav = ?r / ?t
aav = ?v / ?t
v2
?v-v1
v2
y
x
path
v1
Physics 121,Lecture 10,Pg 4
General 3-D motion with non-zero
acceleration:
? Uniform Circular Motion is one specific case of this,
a
v
path
t
a
a
a = 0
because either or both:
-> change in magnitude of v
-> change in direction of v
a = 0
a = 0
aaa = +
Animation
Physics 121,Lecture 10,Pg 5
What is Uniform Circular Motion (UCM)?
? Motion in a circle with:
?Constant Radius R
?Constant Speed v = |v|
?acceleration?
R
v
x
y
(x,y)
= 0a
a
= const.a
Physics 121,Lecture 10,Pg 6
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,?) [position]
? (vR,?) [velocity]
? In UCM:
?R is constant (hence vR = 0).
?? (angular velocity) is constant.
?Polar coordinates are a natural way to describe UCM!
R
v
x
y
(x,y)
?
Physics 121,Lecture 10,Pg 7
Aside,Polar Unit Vectors
? We are familiar with the Cartesian unit vectors,i j k
x
y
i
j
R
?
r^?^
^ ^
^
^
? Now introduce
“polar unit-vectors” r and ?,
?r points in radial direction
?? points in tangential (ccw) direction
Physics 121,Lecture 10,Pg 8
Polar Coordinates:
? The arc length s (distance along the circumference) is
related to the angle in a simple way:
s = R?,where ? is the angular displacement.
?units of ? are called radians.
? For one complete revolution:
2?R = R?c
??c = 2?
??has period 2?,R
v
x
y
(x,y)
s
?
1 revolution = 2??radians
Physics 121,Lecture 10,Pg 9
Polar Coordinates...
? x = R cos ?
? y = R sin ? R
x
y
(x,y)
?
??? ? ???? ??
-1
1
0
sincos
?
Physics 121,Lecture 10,Pg 10
Polar Coordinates...
? In Cartesian co-ordinates we say velocity ?x/ ?t = v.
?x = vt
? In polar coordinates,angular velocity ??/ ?t = ?.
?? = ?t
?? has units of radians/second.
? Displacement s = vt.
but s = R? = R?t,so,R
v
x
y
s
???t
v = ?R
Physics 121,Lecture 10,Pg 11
Period and Frequency
? Recall that 1 revolution = 2? radians
?frequency (f ) = revolutions / second (a)
?angular velocity (?) = radians / second (b)
? By combining (a) and (b)
?? = 2? f
? Realize that:
?period (T) = seconds / revolution
?So T = 1 / f = 2?/?
R
v
s
?
?= 2? / T = 2?f
v
R
s
Physics 121,Lecture 10,Pg 12
Recap:
R
v
s
???t
(x,y)
x = R cos(?)?= R cos(?t)?
y = R sin(?)?= R sin(?t)
? = tan-1 (y/x)
? = ?t
s = v t
s = R? = R?t
v = ?R
Physics 121,Lecture 10,Pg 13
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
v2
R v1
?v
v1v
2
aav = ?v / ?t
Physics 121,Lecture 10,Pg 14
Acceleration in UCM:
seems like ?v (hence ?v/?t )
points toward the origin !
R
? Even though the speed is constant,velocity is not constant since
the direction is changing.
?Consider average acceleration in time ?t aav = ?v / ?t
?v
Physics 121,Lecture 10,Pg 15
Acceleration in UCM:
? Even though the speed is constant,velocity is not constant
since the direction is changing.
?As we shrink ?t,?v / ?t dv / dt = a
a = dv / dt
We see that a points
in the - R direction.
R
Physics 121,Lecture 10,Pg 16
Acceleration in UCM:
? This is called Centripetal Acceleration.
? Now let’s calculate the magnitude:
v1v
2
?v
v2
v1R
?R
? ?v
v
R
R?Similar triangles:
But ?R = v?t for small ?t
So,? ?vv v tR? ??vt vR?
2
a vR? 2
Physics 121,Lecture 10,Pg 17
Centripetal Acceleration
? UCM results in acceleration:
?Magnitude,a = v2 / R = ?? R since v = ?R
?Direction,- r (toward center of circle)
R
a
?
^
Physics 121,Lecture 10,Pg 18
Lecture 10,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) 2000 m
(d) 20,000 m
Physics 121,Lecture 10,Pg 19
Example,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 a here?
Physics 121,Lecture 10,Pg 20
Example:
? First calculate the angular velocity of the propeller:
? so 3500 rpm means ? = 367 s-1
? Now calculate the acceleration,
?a = ?2R = (367s-1)2 x (0.8m) = 1.1 x 105 m/s2
= 11,000 g
direction of a is toward propeller hub (-r ).
1 1 160 2 0 105 0 105 r pm s -1? ? ?r ot x s x r adr ot r adsm i n m i n,,?
^
Physics 121,Lecture 10,Pg 21
Lecture 10,ACT 2
Centripetal 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
Physics 121,Lecture 10,Pg 22
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
Physics 121,Lecture 10,Pg 23
Moon...
? So ? = 2.66 x 10-6 s-1.
? Now calculate the acceleration,
?a = ?2R = 0.00272 m/s2 =,000278 g
?direction of a is toward center of earth (-r ).
1
27 3
1
86400 2 2 66 10
6
.,
r ot
day x
day
s x
r ad
r ot x? ?
? s -1
^
? Calculate angular frequency:
Physics 121,Lecture 10,Pg 24
Moon...
? So we find that amoon / g =,000278
? Newton noticed that RE2 / R2moon =,000273
R RE
amoon g
? This inspired him to propose that FMm? 1 / R2
?(more on gravity later)
Physics 121,Lecture 10,Pg 25
Problem,Motion in a Circle
? A boy ties a rock of mass m to the end of a string and
twirls it in the vertical plane,The distance from his
hand to the rock is R,The speed of the rock at the top
of its trajectory is v.
?What is the tension T in the string at the top of the
rock’s trajectory?
R
v
T
Physics 121,Lecture 10,Pg 26
Motion in a Circle...
? Draw a Free Body Diagram.
? We will use FTOT = ma (surprise)
? First find FTOT in x direction:
FTOT = mg + T Tmg
x
Physics 121,Lecture 10,Pg 27
Motion in a Circle...
FTOT = mg + T
? Now remember that
a = v2 / R (in x direction)
mg + T = mv2 / R
T = mv2 / R - mg
R
T
v
mg
x
Physics 121,Lecture 10,Pg 28
Motion in a Circle...
? What is the minimum speed of the mass at the
top of the trajectory such that the string does not
go limp?
?i.e,find v such that T = 0.
mv2 / R = mg + T
v2 / R = g
? Notice that this does
not depend on m.
R
mg
v
T= 0
v Rg?
Physics 121,Lecture 10,Pg 29
Lecture 10,ACT 3
Motion in a Circle
? A skier of mass m goes over a mogul having a radius
of curvature R,How fast can he go without leaving
the ground?
R
mg N
(a) (b) (c) Rgv =m R gv = mRgv =
Physics 121,Lecture 10,Pg 30
Problem,Rotating puck & weight.
? A mass m1 slides in a circular path with speed v
on a horizontal frictionless table,It is held at a
radius R by a string threaded through a
frictionless hole at the center of the table,At the
other end of the string hangs a second mass m2.
?What is the tension in the string T?
?What is the speed of the sliding mass v?
m1
m2
v
R
Physics 121,Lecture 10,Pg 31
Problem,Rotating puck & weight...
? Draw FBD of hanging mass:
?Since R is constant,a = 0.
so T = m2g
m2
m2g
T
m1
m2
v
R
T
Physics 121,Lecture 10,Pg 32
Problem,Rotating puck & weight...
? Draw FBD of sliding mass:
m1
T = m2g
v gR mm? 2
1
m1g
N
m1
m2
v
R
T
Use F = T = m1a
where a = v2 / R
m2g = m1v2 / R
Physics 121,Lecture 10,Pg 33
Forces Causing Centripetal Acceleration
r
va t
c
2
?
r
vmaF t
c
2
??
v
F
r
mg
m
f
n
Physics 121,Lecture 10,Pg 34
Forces Causing Centripetal Acceleration
? Turn on a sloped road …
?
mg
n
?s in
2
nrvmaF tc ???
mg
n sin ?
n cos ?
Physics 121,Lecture 10,Pg 35
Accelerated Reference Frames:
The Accelerometer
? Your first job is with General Motors,You are
working on a project to design an accelerometer,
The inner workings of this gadget consist of a
weight of mass m that is hung inside a box that is
attached to the ceiling of a car,You design the
device with a very light string so that you can
mathematically ignore it,The idea is that the
angle the string makes with the vertical is
determined by the car’s acceleration,Your
preliminary task is to think about calibration of the
accelerometer,First you calibrate the
measurement for when the car travels on a flat
road,
Physics 121,Lecture 10,Pg 36
Accelerated Reference Frames:
The Accelerometer
a
?
i
1
We need to solve for the angle the plum bob makes with
respect to vertical,
First we will solve by using Newton’s Second Law and checking
x and y components,Then we will consider other possible
solution methods,
Physics 121,Lecture 10,Pg 37
Accelerometer...
2,Draw a free-body diagram for the mass:
We wish to solve for ? in terms of the acceleration a.
We will use SF = ma in two cartesian dimensions.
m
T
mg (gravitational force)
?
a
x
y
?
Physics 121,Lecture 10,Pg 38
Accelerometer...
3,Solving,
i,FX = TX = T sin ? = ma
j,FY = TY ? mg
= T cos ??? mg = 0
? Eliminate T
T
mg
m
ma
j
i
?
?
T sin ???= ma
T cos ???= mg
t a n ? ? ag
Physics 121,Lecture 10,Pg 39
Accelerometer...
4,No Numbers involved
5,This answer has the right units (none)
It does give 1 in terms of the acceleration,
t a n ? ? ag
Physics 121,Lecture 10,Pg 40
Accelerometer – Other Thoughts
? Think of this problem from the point of view of the person
inside the car.
? This person sees the plumb bob making the same angle
with respect to the ground,but detects no acceleration.
t a n ? ? ag
a
?
Physics 121,Lecture 10,Pg 41
Accelerometer...
? There must be some other force to balance the x component
i,FX = TX + F? = T sin ? + F? = 0
j,FY = TY ? mg
= T cos ??? mg = 0
? And we must put F? = -ma to get the same answer as before.
? F? is known as a fictitious force.
t a n ? ? ag
?
a T
mg
Physics 121,Lecture 10,Pg 42
Lecture 10,ACT 4
Accelerated Reference Frames
You are a passenger in a car and not wearing your seatbelt,
Without increasing or decreasing speed,the car makes a sharp
left turn,and you find yourself colliding with the right-hand door,
Which is a correct description of the situation?
A) Before and after the collision there is a rightward force pushing
you into the door.
B) Starting at the time of the collision,the door exerts a leftward
force on you.
C) Both of the above.
D) Neither of the above.
Physics 121,Lecture 10,Pg 43
Newton’s Law of Universal Gravitation
? Every particle in the Universe attracts every other
particle with a force that is directly proportional the
product of their masses and inversely proportional to
the square of the distance between them.
? Constant of Universal gravitation:
2
21
r
mmGF ?
221110673.6 kgmNG ??? ?
m1
m2
r12
F21 F12
Physics 121,Lecture 10,Pg 44
Newton’s Law of Universal Gravitation
? The gravitational force is a field force that always
exists between two particles regardless of the
medium that separates them.
? The force varies as one over the square of the
distance between the particles and therefore
decreases rapidly with increasing separation.
? The gravitational force is actually very weak as
shown be the size of G.
It is easily observed for a falling ball of mass m
because the mass of Earth ME is so large in the
product mME
221110673.6 kgmNG ??? ?
Physics 121,Lecture 10,Pg 45
Newton’s Law of Universal Gravitation
? The gravitational force exerted by a spherical
object on a particle outside the sphere is the
same as if the entire mass of the sphere were
concentrated at its center:
2
E
E
R
mMGF ?
Physics 121,Lecture 10,Pg 46
Kepler’s Laws
? Much of Sir Isaac’s motivation to deduce the laws of
gravity was to explain Kepler’s laws of the motions of the
planets about our sun,
? Ptolemy,a Greek in Roman times,famously described a
model that said all planets and stars orbit about the
earth,This was believed for a long time.
? Copernicus (1543) said no,the planets orbit in circles
about the sun,
? Brahe (~1600) measured the motions of all of the known
planets and the position of 777 stars (ouch !)
? Kepler,his student,tried to organize all of this,He came
up with his famous three laws of planetary motion.
Physics 121,Lecture 10,Pg 47
Copernicus Brahe Kepler
1473-1543 1546-1601 1571-1630
Physics 121,Lecture 10,Pg 48
Keplers’ Laws
? After 20 years of work on Tycho Brahe’s data,Kepler
formulated his three laws:
1,All planets move in elliptical orbits
with the Sun at one of the focal
points
2,A line drawn from the Sun to any
planet sweeps out equal areas in
equal time intervals
3,The square of the orbital period of
any planet is proportional to the
cube of the average distance
from the planet to the Sun
? It was later shown that all three of these
laws are a result of Newton’s laws of
gravity and motion.
Physics 121,Lecture 10,Pg 49
Kepler’s First Law
? In the Solar System,the Sun is at one focus of
the elliptical orbit of each planet and the other
focus is empty
Sun
Planet
Physics 121,Lecture 10,Pg 50
Kepler’s First Law
p q
? The sum p + q is the same for every point on the
ellipse
Physics 121,Lecture 10,Pg 51
Kepler’s Second Law
? A line drawn from the Sun to any planet sweeps out
equal areas in equal time intervals
? Due to angular momentum conservation
A
B
C
D
Physics 121,Lecture 10,Pg 52
Kepler’s Third Law
? A planet of mass Mp is moving about the Sun Ms in a
circular orbit,Newton’s second law gives:
r
vM
r
MMG pps 2
2 ?
The speed v = 2?r/T,with T is the period of the planet.
r
Tr
r
MG s 2
2
)/2( ??
3
21923322 1097.244
m
s
GMKrKrGMT ssss
???????
?
?
???
?? ??
Physics 121,Lecture 10,Pg 53
Kepler’s Third Law
Ms
Mp
2
322 44
GT
r
GKM ss
?? ??
Kepler’s third law gives us a method for measuring
the mass of the Sun.
33
2
2 4 rKr
GMT ss ????
?
???
?? ?
Physics 121,Lecture 10,Pg 54
Geosynchronous Orbit
h
r
R
v
F
The gravitational force produces
the centripetal acceleration:
r
mv
r
mMG E 2
2 ?
T
r
T
dv ?2??
3/1
2
2
4 ?
?
?
?
???
??
EGM
Tr
?
T = 24h and ME = 5.98 ? 1024 kg
The height above Earth is app,22 000 miles
Physics 121,Lecture 10,Pg 55
Recap of today’s lecture
? Homework 4,Due Friday Sept,30 @ 6:00PM
?Ch.4,# 52,57,58,59,60,65,80,84,87,and 90.
? Chapter 5,Rotation
?Angular variables
?Centripetal acceleration
?Kepler’s laws
Physics 121,Sections 9,10,11,and 12
Lecture 10
Today’s Topics:
? Homework 4,Due Friday Sept,30 @ 6:00PM
? Ch.4,# 52,57,58,59,60,65,80,84,87,and 90.
? Chapter 5,Rotation
? Angular variables
? Centripetal acceleration
? Kepler’s laws
Physics 121,Lecture 10,Pg 2
Chapter 5
Uniform Circular Motion
? What does it mean?
? How do we describe it?
? What can we learn about it?
Physics 121,Lecture 10,Pg 3
Review
( displacement,velocity,acceleration )
Velocity,
Acceleration,
a = dv / dt
? 3-D Kinematics, vector equations:
r = r(t) v = dr / dt a = d2r / dt2
v = dr / dt
vav = ?r / ?t
aav = ?v / ?t
v2
?v-v1
v2
y
x
path
v1
Physics 121,Lecture 10,Pg 4
General 3-D motion with non-zero
acceleration:
? Uniform Circular Motion is one specific case of this,
a
v
path
t
a
a
a = 0
because either or both:
-> change in magnitude of v
-> change in direction of v
a = 0
a = 0
aaa = +
Animation
Physics 121,Lecture 10,Pg 5
What is Uniform Circular Motion (UCM)?
? Motion in a circle with:
?Constant Radius R
?Constant Speed v = |v|
?acceleration?
R
v
x
y
(x,y)
= 0a
a
= const.a
Physics 121,Lecture 10,Pg 6
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,?) [position]
? (vR,?) [velocity]
? In UCM:
?R is constant (hence vR = 0).
?? (angular velocity) is constant.
?Polar coordinates are a natural way to describe UCM!
R
v
x
y
(x,y)
?
Physics 121,Lecture 10,Pg 7
Aside,Polar Unit Vectors
? We are familiar with the Cartesian unit vectors,i j k
x
y
i
j
R
?
r^?^
^ ^
^
^
? Now introduce
“polar unit-vectors” r and ?,
?r points in radial direction
?? points in tangential (ccw) direction
Physics 121,Lecture 10,Pg 8
Polar Coordinates:
? The arc length s (distance along the circumference) is
related to the angle in a simple way:
s = R?,where ? is the angular displacement.
?units of ? are called radians.
? For one complete revolution:
2?R = R?c
??c = 2?
??has period 2?,R
v
x
y
(x,y)
s
?
1 revolution = 2??radians
Physics 121,Lecture 10,Pg 9
Polar Coordinates...
? x = R cos ?
? y = R sin ? R
x
y
(x,y)
?
??? ? ???? ??
-1
1
0
sincos
?
Physics 121,Lecture 10,Pg 10
Polar Coordinates...
? In Cartesian co-ordinates we say velocity ?x/ ?t = v.
?x = vt
? In polar coordinates,angular velocity ??/ ?t = ?.
?? = ?t
?? has units of radians/second.
? Displacement s = vt.
but s = R? = R?t,so,R
v
x
y
s
???t
v = ?R
Physics 121,Lecture 10,Pg 11
Period and Frequency
? Recall that 1 revolution = 2? radians
?frequency (f ) = revolutions / second (a)
?angular velocity (?) = radians / second (b)
? By combining (a) and (b)
?? = 2? f
? Realize that:
?period (T) = seconds / revolution
?So T = 1 / f = 2?/?
R
v
s
?
?= 2? / T = 2?f
v
R
s
Physics 121,Lecture 10,Pg 12
Recap:
R
v
s
???t
(x,y)
x = R cos(?)?= R cos(?t)?
y = R sin(?)?= R sin(?t)
? = tan-1 (y/x)
? = ?t
s = v t
s = R? = R?t
v = ?R
Physics 121,Lecture 10,Pg 13
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
v2
R v1
?v
v1v
2
aav = ?v / ?t
Physics 121,Lecture 10,Pg 14
Acceleration in UCM:
seems like ?v (hence ?v/?t )
points toward the origin !
R
? Even though the speed is constant,velocity is not constant since
the direction is changing.
?Consider average acceleration in time ?t aav = ?v / ?t
?v
Physics 121,Lecture 10,Pg 15
Acceleration in UCM:
? Even though the speed is constant,velocity is not constant
since the direction is changing.
?As we shrink ?t,?v / ?t dv / dt = a
a = dv / dt
We see that a points
in the - R direction.
R
Physics 121,Lecture 10,Pg 16
Acceleration in UCM:
? This is called Centripetal Acceleration.
? Now let’s calculate the magnitude:
v1v
2
?v
v2
v1R
?R
? ?v
v
R
R?Similar triangles:
But ?R = v?t for small ?t
So,? ?vv v tR? ??vt vR?
2
a vR? 2
Physics 121,Lecture 10,Pg 17
Centripetal Acceleration
? UCM results in acceleration:
?Magnitude,a = v2 / R = ?? R since v = ?R
?Direction,- r (toward center of circle)
R
a
?
^
Physics 121,Lecture 10,Pg 18
Lecture 10,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) 2000 m
(d) 20,000 m
Physics 121,Lecture 10,Pg 19
Example,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 a here?
Physics 121,Lecture 10,Pg 20
Example:
? First calculate the angular velocity of the propeller:
? so 3500 rpm means ? = 367 s-1
? Now calculate the acceleration,
?a = ?2R = (367s-1)2 x (0.8m) = 1.1 x 105 m/s2
= 11,000 g
direction of a is toward propeller hub (-r ).
1 1 160 2 0 105 0 105 r pm s -1? ? ?r ot x s x r adr ot r adsm i n m i n,,?
^
Physics 121,Lecture 10,Pg 21
Lecture 10,ACT 2
Centripetal 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
Physics 121,Lecture 10,Pg 22
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
Physics 121,Lecture 10,Pg 23
Moon...
? So ? = 2.66 x 10-6 s-1.
? Now calculate the acceleration,
?a = ?2R = 0.00272 m/s2 =,000278 g
?direction of a is toward center of earth (-r ).
1
27 3
1
86400 2 2 66 10
6
.,
r ot
day x
day
s x
r ad
r ot x? ?
? s -1
^
? Calculate angular frequency:
Physics 121,Lecture 10,Pg 24
Moon...
? So we find that amoon / g =,000278
? Newton noticed that RE2 / R2moon =,000273
R RE
amoon g
? This inspired him to propose that FMm? 1 / R2
?(more on gravity later)
Physics 121,Lecture 10,Pg 25
Problem,Motion in a Circle
? A boy ties a rock of mass m to the end of a string and
twirls it in the vertical plane,The distance from his
hand to the rock is R,The speed of the rock at the top
of its trajectory is v.
?What is the tension T in the string at the top of the
rock’s trajectory?
R
v
T
Physics 121,Lecture 10,Pg 26
Motion in a Circle...
? Draw a Free Body Diagram.
? We will use FTOT = ma (surprise)
? First find FTOT in x direction:
FTOT = mg + T Tmg
x
Physics 121,Lecture 10,Pg 27
Motion in a Circle...
FTOT = mg + T
? Now remember that
a = v2 / R (in x direction)
mg + T = mv2 / R
T = mv2 / R - mg
R
T
v
mg
x
Physics 121,Lecture 10,Pg 28
Motion in a Circle...
? What is the minimum speed of the mass at the
top of the trajectory such that the string does not
go limp?
?i.e,find v such that T = 0.
mv2 / R = mg + T
v2 / R = g
? Notice that this does
not depend on m.
R
mg
v
T= 0
v Rg?
Physics 121,Lecture 10,Pg 29
Lecture 10,ACT 3
Motion in a Circle
? A skier of mass m goes over a mogul having a radius
of curvature R,How fast can he go without leaving
the ground?
R
mg N
(a) (b) (c) Rgv =m R gv = mRgv =
Physics 121,Lecture 10,Pg 30
Problem,Rotating puck & weight.
? A mass m1 slides in a circular path with speed v
on a horizontal frictionless table,It is held at a
radius R by a string threaded through a
frictionless hole at the center of the table,At the
other end of the string hangs a second mass m2.
?What is the tension in the string T?
?What is the speed of the sliding mass v?
m1
m2
v
R
Physics 121,Lecture 10,Pg 31
Problem,Rotating puck & weight...
? Draw FBD of hanging mass:
?Since R is constant,a = 0.
so T = m2g
m2
m2g
T
m1
m2
v
R
T
Physics 121,Lecture 10,Pg 32
Problem,Rotating puck & weight...
? Draw FBD of sliding mass:
m1
T = m2g
v gR mm? 2
1
m1g
N
m1
m2
v
R
T
Use F = T = m1a
where a = v2 / R
m2g = m1v2 / R
Physics 121,Lecture 10,Pg 33
Forces Causing Centripetal Acceleration
r
va t
c
2
?
r
vmaF t
c
2
??
v
F
r
mg
m
f
n
Physics 121,Lecture 10,Pg 34
Forces Causing Centripetal Acceleration
? Turn on a sloped road …
?
mg
n
?s in
2
nrvmaF tc ???
mg
n sin ?
n cos ?
Physics 121,Lecture 10,Pg 35
Accelerated Reference Frames:
The Accelerometer
? Your first job is with General Motors,You are
working on a project to design an accelerometer,
The inner workings of this gadget consist of a
weight of mass m that is hung inside a box that is
attached to the ceiling of a car,You design the
device with a very light string so that you can
mathematically ignore it,The idea is that the
angle the string makes with the vertical is
determined by the car’s acceleration,Your
preliminary task is to think about calibration of the
accelerometer,First you calibrate the
measurement for when the car travels on a flat
road,
Physics 121,Lecture 10,Pg 36
Accelerated Reference Frames:
The Accelerometer
a
?
i
1
We need to solve for the angle the plum bob makes with
respect to vertical,
First we will solve by using Newton’s Second Law and checking
x and y components,Then we will consider other possible
solution methods,
Physics 121,Lecture 10,Pg 37
Accelerometer...
2,Draw a free-body diagram for the mass:
We wish to solve for ? in terms of the acceleration a.
We will use SF = ma in two cartesian dimensions.
m
T
mg (gravitational force)
?
a
x
y
?
Physics 121,Lecture 10,Pg 38
Accelerometer...
3,Solving,
i,FX = TX = T sin ? = ma
j,FY = TY ? mg
= T cos ??? mg = 0
? Eliminate T
T
mg
m
ma
j
i
?
?
T sin ???= ma
T cos ???= mg
t a n ? ? ag
Physics 121,Lecture 10,Pg 39
Accelerometer...
4,No Numbers involved
5,This answer has the right units (none)
It does give 1 in terms of the acceleration,
t a n ? ? ag
Physics 121,Lecture 10,Pg 40
Accelerometer – Other Thoughts
? Think of this problem from the point of view of the person
inside the car.
? This person sees the plumb bob making the same angle
with respect to the ground,but detects no acceleration.
t a n ? ? ag
a
?
Physics 121,Lecture 10,Pg 41
Accelerometer...
? There must be some other force to balance the x component
i,FX = TX + F? = T sin ? + F? = 0
j,FY = TY ? mg
= T cos ??? mg = 0
? And we must put F? = -ma to get the same answer as before.
? F? is known as a fictitious force.
t a n ? ? ag
?
a T
mg
Physics 121,Lecture 10,Pg 42
Lecture 10,ACT 4
Accelerated Reference Frames
You are a passenger in a car and not wearing your seatbelt,
Without increasing or decreasing speed,the car makes a sharp
left turn,and you find yourself colliding with the right-hand door,
Which is a correct description of the situation?
A) Before and after the collision there is a rightward force pushing
you into the door.
B) Starting at the time of the collision,the door exerts a leftward
force on you.
C) Both of the above.
D) Neither of the above.
Physics 121,Lecture 10,Pg 43
Newton’s Law of Universal Gravitation
? Every particle in the Universe attracts every other
particle with a force that is directly proportional the
product of their masses and inversely proportional to
the square of the distance between them.
? Constant of Universal gravitation:
2
21
r
mmGF ?
221110673.6 kgmNG ??? ?
m1
m2
r12
F21 F12
Physics 121,Lecture 10,Pg 44
Newton’s Law of Universal Gravitation
? The gravitational force is a field force that always
exists between two particles regardless of the
medium that separates them.
? The force varies as one over the square of the
distance between the particles and therefore
decreases rapidly with increasing separation.
? The gravitational force is actually very weak as
shown be the size of G.
It is easily observed for a falling ball of mass m
because the mass of Earth ME is so large in the
product mME
221110673.6 kgmNG ??? ?
Physics 121,Lecture 10,Pg 45
Newton’s Law of Universal Gravitation
? The gravitational force exerted by a spherical
object on a particle outside the sphere is the
same as if the entire mass of the sphere were
concentrated at its center:
2
E
E
R
mMGF ?
Physics 121,Lecture 10,Pg 46
Kepler’s Laws
? Much of Sir Isaac’s motivation to deduce the laws of
gravity was to explain Kepler’s laws of the motions of the
planets about our sun,
? Ptolemy,a Greek in Roman times,famously described a
model that said all planets and stars orbit about the
earth,This was believed for a long time.
? Copernicus (1543) said no,the planets orbit in circles
about the sun,
? Brahe (~1600) measured the motions of all of the known
planets and the position of 777 stars (ouch !)
? Kepler,his student,tried to organize all of this,He came
up with his famous three laws of planetary motion.
Physics 121,Lecture 10,Pg 47
Copernicus Brahe Kepler
1473-1543 1546-1601 1571-1630
Physics 121,Lecture 10,Pg 48
Keplers’ Laws
? After 20 years of work on Tycho Brahe’s data,Kepler
formulated his three laws:
1,All planets move in elliptical orbits
with the Sun at one of the focal
points
2,A line drawn from the Sun to any
planet sweeps out equal areas in
equal time intervals
3,The square of the orbital period of
any planet is proportional to the
cube of the average distance
from the planet to the Sun
? It was later shown that all three of these
laws are a result of Newton’s laws of
gravity and motion.
Physics 121,Lecture 10,Pg 49
Kepler’s First Law
? In the Solar System,the Sun is at one focus of
the elliptical orbit of each planet and the other
focus is empty
Sun
Planet
Physics 121,Lecture 10,Pg 50
Kepler’s First Law
p q
? The sum p + q is the same for every point on the
ellipse
Physics 121,Lecture 10,Pg 51
Kepler’s Second Law
? A line drawn from the Sun to any planet sweeps out
equal areas in equal time intervals
? Due to angular momentum conservation
A
B
C
D
Physics 121,Lecture 10,Pg 52
Kepler’s Third Law
? A planet of mass Mp is moving about the Sun Ms in a
circular orbit,Newton’s second law gives:
r
vM
r
MMG pps 2
2 ?
The speed v = 2?r/T,with T is the period of the planet.
r
Tr
r
MG s 2
2
)/2( ??
3
21923322 1097.244
m
s
GMKrKrGMT ssss
???????
?
?
???
?? ??
Physics 121,Lecture 10,Pg 53
Kepler’s Third Law
Ms
Mp
2
322 44
GT
r
GKM ss
?? ??
Kepler’s third law gives us a method for measuring
the mass of the Sun.
33
2
2 4 rKr
GMT ss ????
?
???
?? ?
Physics 121,Lecture 10,Pg 54
Geosynchronous Orbit
h
r
R
v
F
The gravitational force produces
the centripetal acceleration:
r
mv
r
mMG E 2
2 ?
T
r
T
dv ?2??
3/1
2
2
4 ?
?
?
?
???
??
EGM
Tr
?
T = 24h and ME = 5.98 ? 1024 kg
The height above Earth is app,22 000 miles
Physics 121,Lecture 10,Pg 55
Recap of today’s lecture
? Homework 4,Due Friday Sept,30 @ 6:00PM
?Ch.4,# 52,57,58,59,60,65,80,84,87,and 90.
? Chapter 5,Rotation
?Angular variables
?Centripetal acceleration
?Kepler’s laws
