Transcript No Slide Title

PHYSICS 231
Lecture 19: We have lift-off!
Comet Kohoutek
Remco Zegers
Walk-in hour: Monday 9:15-10:15 am
Helproom
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Previously…
v
ac
r
Centripetal acceleration:
ac=v2/r=2r
Caused by force like:
•Gravity
•Tension
•Friction
•F=mac for rotating object
The centripetal acceleration is caused by a change
in the direction of the linear velocity vector, not
a change in magnitude
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quiz (extra credit)
3 children are sitting on a rotating disc in a playground.
The disc starts to spin faster and faster.
Which of the three is most likely to start sliding first?
top view
a) child A
A
b) child B
c) child C
d) the same for all three
B
C
ac= 2r.
A= B= c
rA<rB<rc
acA<acB<acC
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demo
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The gravitational force, revisited
Newton:
m1 m2
F G 2
r
G=6.673·10-11 Nm2/kg2
The gravitational force works between every two massive
particles in the universe, yet is the least well understood
force known.
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Gravitation between two objects
A
B
The gravitational force exerted by the spherical
object A on B can be calculated by assuming that all
of A’s mass would be concentrated in its center and
likewise for object B.
Conditions: B must be outside of A
A and B must be ‘homogeneous’
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Gravitational acceleration
m1 m EARTH
F G
2
r
F=mg
g=GmEARTH/r2
On earth surface: g=9.81 m/s2 r=6366 km
On top of mount Everest: r=6366+8.850km g=9.78 m/s2
Low-orbit satellite: r=6366+1600km g=6.27 m/s2
Geo-stationary satellite: r=6366+36000km
g=0.22 m/s2
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Losing weight easily?
You are standing on a scale in a stationary space ship
in low-orbit (g=6.5 m/s2). If your mass is 70 kg, what
is your weight?
F=mg=70*6.5=455 N
And what is your weight if the space ship would be
orbiting the earth?
Weightless!
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launch
speed
4 km/s
6 km/s
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8 km/s
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Gravitational potential energy
So far, we used: PEgravity=mgh Only valid for h near
earth’s surface.
More general: PEgravity=-GMEarthm/r
PE=0 at infinity distance from the center of the earth
See example 7.12 for consistency between these two.
Example: escape speed: what should the minimum initial
velocity of a rocket be if we want to make sure it will not
fall back to earth?
KEi+PEi=0.5mv2-GMEarthm/REarth
KEf+PEf=0
v=(2GMearth/REarth)=11.2 km/s
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Kepler’s laws
Johannes Kepler
(1571-1630)
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Kepler’s First law
Ellipticity e(0-1)
p+q=constant
An object A bound to another object B by a force that goes
with 1/r2 moves in an elliptical orbit around B, with B being
in one of the focus point of the ellipse; planets around the
sun.
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launch speed is 10 km/s
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Kepler’s second law
Area(D-C-SUN)=Area(B-A-SUN)
A line drawn from the sun to the elliptical orbit of a planet
sweeps out equal areas in equal time intervals.
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Kepler’s third law
r3
T2
r3=T2/Ks
Consider a planet in circular motion
around the sun:
M sun M planet M planet v 2planet
G

2
r
r
s 2r
v

t
T
 4 2  3
2
r  K s r 3
T  
 GM sun 
K s  2.97  1019 s 2 / m 2
r3=constant*T2
T: period-time it takes to make
one revolution
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Chapter 8. Torque
Top view
It is much easier to swing the
door if the force F is applied
as far away as possible (d) from
the rotation axis (O).
Torque: The capability of a force to rotate an object about
an axis.
Torque
=F·d
(Nm)
Torque is positive if the motion is counterclockwise
Torque is negative if the motion is clockwise
demo: opening a cellar door
demo: turning a screw
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Decompositions
F=
FL
F//
Top view
What is the torque applied to the door?
Force parallel to the rotating door: F//=Fcos600=150 N
Force perpendicular to rotating door: FL=Fsin600=260 N
Only FL is effective for opening the door:
=FL·d=260*2.0=520 Nm
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Multiple force causing torque.
100 N
0.3 m
Top view
Two persons try to go
through a rotating door
at the same time, one on
the l.h.s. of the rotator and
one the r.h.s. of the rotator.
If the forces are applied as
shown in the drawing, what
will happen?
50 N
0.6 m
1=F1·d1=-100*0.3=-30 Nm
2=F2·d2=50*0.6 =30 Nm

0 Nm +
Nothing will happen! The 2
torques are balanced.
demo: balance with unequal masses
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dpull
Center of gravity.
Fpull
1 2 3………………………n
Fgravity
dgravity?
=Fpulldpull+Fgravitydgravity
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Vertical direction
(I.e. side view)
We can assume that
for the calculation
of torque due to gravity,
all mass is concentrated
in one point:
The center of gravity:
the average position of
the mass
dcg=(m1d1+m2d2+…+mndn)
(m1+m2+…+mn)
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Center of Gravity; more general
The center of gravity
m x

m
i
xCG
i
i
i
i
m y

m
i
yCG
i
i
i
i
demo center of gravity
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Top view Fp
-d d
CG
-Fp
Object in equilibrium
Newton’s 2nd law: F=ma
Fp+(-Fp)=ma=0
No acceleration, no movement…
But the block starts to rotate!
=Fpd+(-Fp)(-d)=2Fpd
There is movement!
Translational equilibrium: F=ma=0 The center of gravity
does not move!
Rotational equilibrium:
=0
The object does not
rotate
Mechanical equilibrium: F=ma=0 & =0 No movement!
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