Mon. Tues. Wed. Fri.
5.5 -.7 Curving Motion
6.1-.4 (.21) Introducing Energy & Work Quiz 5
3pm – Visit from Columbia Rep
RE 5.b EP 5, HW5: Ch5 Pr’s 16, 19, 45, 48, 64(c&d)
dt
all
system
Ch. 5 – Rate of change (if any) of momentum
Special Cases Equilibium
0 dt
all
system
Uniform Circular Motion
0dt
pd
The process for force problems
Define and stick with your system
Identify what objects are interacting with it and determine the corresponding forces acting upon it
Add the forces to get the net force acting on the system
Equate with rate of change of momentum via momentum principle
Solve for unknowns
𝐹 𝑠𝑦𝑠𝑡𝑒𝑚 ←=𝑑𝑝 𝑠𝑦𝑠𝑑𝑡
𝑎𝑙𝑙
Earth
𝐹 𝑝𝑒𝑟𝑠𝑜
𝑛←
𝐸𝑎𝑟𝑡ℎ
𝑝 𝑝𝑒𝑟𝑠𝑜𝑛
Changing Momentum: Magnitude and Direction
dt
pdF objnet
dt
pdpp
dt
pd
dt
ppd ˆˆ
ˆ
Speeding/slowing changing direction
Special Case: Uniform Circular Motion (only direction changing)
dt
pdpp
dt
pd
dt
ppd
dt
pd ˆˆ
ˆ
similarly
comparing
d
r
dt
rd
dt
rdv
dt
rdrr
dt
rd
dt
rdv
ˆˆ
dt
dr
dt
rd
dt
d
dt
rd
ˆ
Rate of change of position vector’s direction
vr1
Should point… Parallel to momentum vector
Perpendicular to momentum vector
rdrd
Speeding/slowing changing direction
Should point… Parallel to momentum vector
Perpendicular to momentum vector
Changing Momentum: Magnitude and Direction
dt
pdF objnet
dt
pdpp
dt
pd
dt
ppd ˆˆ
ˆ
Special Case: Uniform Circular Motion
dt
pdpp
dt
pd
dt
ppd
dt
pd ˆˆ
ˆ
d
dt
d
dt
rd
ˆ
Rate of change of position vector’s direction
Rate of change of momentum vector’s direction
Equals
dt
pd
dt
rd ˆˆ
vr1
vr1
pdˆ
direction? Opposite of r. r̂
rr
vp
dt
pdˆ
rp
See Vpython example
Example: Geosynchronous Orbit There’s only one orbital radius for satellites that ‘stay put’ in the sky – orbit with the same period as the Earth spins: T = 1 day. What’s the orbital radius?
-G𝑀
𝐸𝑀
𝑆
|𝑟𝑠←𝐸|2𝑟 𝑠 ← 𝐸 =-|𝑝 |
|𝑣|
|𝑟𝑠←𝐸|𝑟 𝑠 ← 𝐸
Application: Circular Gravitational Orbits
𝐹 𝑛𝑒𝑡=𝑑𝑝
𝑑𝑡
System: Satellite
Gravitational Force
Circular Motion
G𝑀
𝐸𝑀
𝑆
|𝑟𝑠←𝐸|=|𝑝 ||𝑣|
𝑣 𝑝
𝑀𝐸
𝑀𝑆
≈ 𝑀𝑆𝑣 G𝑀
𝐸𝑀
𝑆
|𝑟𝑠←𝐸|= 𝑀𝑆|𝑣||𝑣|
G𝑀
𝐸
|𝑟𝑠←𝐸|=𝑣2
v = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒 =
𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑎𝑛𝑐𝑒
𝑃𝑒𝑟𝑖𝑜𝑑 =
2𝜋𝑟𝑠←𝐸
𝑇 G
𝑀𝐸
|𝑟𝑠←𝐸|=
2𝜋𝑟𝑠←𝐸
𝑇 2
𝑟𝑠←𝐸 = G𝑀𝐸
𝑇
2𝜋 2
13
= 6.7 × 10−11𝑁𝑚
2
𝑘𝑔2 6 × 1024𝑘𝑔86,400𝑠
2𝜋
2
13
= 4.2 × 107𝑚
Kepler’s 3rd Law of Planetary Motion
𝑣 𝑝
𝑀𝐸
𝑀𝑆
≈ 𝑀𝑆𝑣
v = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒 =
𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑎𝑛𝑐𝑒
𝑃𝑒𝑟𝑖𝑜𝑑 =
2𝜋𝑟𝑠←𝐸
𝑇
𝑟𝑠←𝐸3 = G𝑀𝐸
𝑇
2𝜋 2
𝑟𝑠←𝐸 = G𝑀𝐸
𝑇
2𝜋 2
13
G𝑀
𝐸
|𝑟𝑠←𝐸|=
2𝜋𝑟𝑠←𝐸
𝑇 2
𝑟𝑠←𝐸3 =
G𝑀𝐸
2𝜋 2 𝑇2
G𝑀
𝐸
|𝑟𝑠←𝐸|=𝑣2
Orbital radius’s relation to its period Note: depends on mass of object orbited not orbiting
Application: Circular Gravitational Orbits The Moon travels in a nearly circular orbit around the Earth, at nearly constant speed.
what is the direction of ?
?
𝑑𝑝
𝑑𝑡
A geosynchronous satellite has an orbital radius of 4.2×107m.
If the moon’s period were 30 days (it’s really about 27), what would be its
orbital radius?
a) 13×107m
b) 41×107m
c) 126×107m
d) 690×107m
Circular Motion Example: Say you had a 1 km radius space-station, with what period must it spin to provide a normal force equal to that on Earth?
r
Homework: spinning carnival ride – friction holds you up
𝐹 𝑊𝑎𝑙𝑙.𝑁
𝐹 𝑊𝑎𝑙𝑙.𝑓
𝐹 𝐸𝑎𝑟𝑡ℎ.𝑔
≤ 𝜇𝑠𝐹𝑊𝑎𝑙𝑙.𝑁
(Know and care what interactions provide net force) Person rides on a “swing” ride at a carnival. The man rides in a swing at the end of a 12.0 m chain which hangs out at 65 up from vertical. Assuming uniform circular motion, and a combined, chair+rider, mass of 220 kg, (a) what is the tension in the chain? (b) what is tangential speed of the chair?
𝑭𝒄𝒉𝒂𝒊𝒏
𝒗=? M = 220kg
= 65o
𝑭𝑬𝒂𝒓𝒕𝒉
Circular Motion: Banked Curves
Why is the track banked? 𝒙
𝒚 𝐹𝑐←
𝑅.𝑦
𝐹𝑐 ← 𝑅. 𝑥
𝐹 𝑛𝑒𝑡=𝑑𝑝
𝑑𝑡
𝑦 : 𝐹𝑐 ← 𝐸 + 𝐹𝑐 ← 𝑅.y= 0
-𝑚𝑔 + 𝐹𝑐 ← 𝑅.y= 0
𝑚
𝐹 𝑐←
𝐸
𝐹𝑐 ← 𝑅cos(θ) = 𝑚𝑔
𝐹𝑐 ← 𝑅 =𝑚𝑔
cos(θ)
y-co
mp
on
ent
of
forc
es a
nd
ch
ange
in m
om
en
tum
x-
com
po
nen
t o
f fo
rces
an
d
chan
ge in
mo
me
ntu
m
Aerial View
Curve’s “kissing” circle r
𝑥 : 𝐹𝑐← 𝑅.x =|𝑝 |
|𝑣|
|𝑟|
𝑝 ≈ 𝑚𝑣
𝐹𝑐 ← 𝑅.x =𝑚𝑣2
|𝑟|
𝐹𝑐 ← 𝑅 sin()=𝑚𝑣2
|𝑟|
𝑚𝑔
cos(θ )sin()=𝑚
𝑣2
|𝑟| 𝑔tan()=𝑣2
|𝑟|
For a given radius of curvature and
expected speed, there’s a particular
banking angle.
Circular Motion: Banked Curves
𝒙
𝒚 𝐹𝑐←
𝑅.𝑦
𝐹𝑐 ← 𝑅. 𝑥
𝑚
𝐹 𝑐←
𝐸
𝒈𝐭𝐚𝐧()=𝒗𝟐
|𝒓|
For a given radius of curvature and expected speed, there’s a particular banking angle.
If you’re taking the interchange between I-10 West and I-125 South, you may well be going 65 mph around a curve with a radius of curvature of 1/8 mi. If it was designed for this speed, at what angle should it be banked?
If your tires have a coefficient of static friction of 0.9, how fast could you take the curve without sliding sideways?
Circular Motion: Generalized – “Kissing” Circles
rk.1
rk.2
rk.3
kiss
kiss
objnet
objnet
RR
vpp
dt
dt
pdpp
dt
pd
dt
ppdF
ˆˆ
ˆˆ
ˆ
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