PHY 231 1 PHYSICS 231 Lecture 17: We have lift-off! Remco Zegers Walk-in hour: Thursday 11:30-13:30...
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Transcript of PHY 231 1 PHYSICS 231 Lecture 17: We have lift-off! Remco Zegers Walk-in hour: Thursday 11:30-13:30...
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PHY 2311
PHYSICS 231Lecture 17: We have lift-off!
Remco ZegersWalk-in hour: Thursday 11:30-13:30 am
Helproom
Comet Kohoutek
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PHY 2312
Previously…
v
r
ac
Centripetal acceleration:ac=v2/r=2r
Caused by force like:•Gravity•Tension•Friction
F=mac for rotating object
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PHY 2313
The gravitational force, revisited
221
r
mmGF
G=6.673·10-11 Nm2/kg2
Newton:
The gravitational force works between every two massiveparticles in the universe, yet is the least well understoodforce known.
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PHY 2314
Gravitation between two objects
AB
The gravitational force exerted by the sphericalobject A on B can be calculated by assuming that all of A’s mass would be concentrated in its center andlikewise for object B.Conditions: B must be outside of A
A and B must be ‘homogeneous’
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PHY 2315
Gravitational acceleration
21
r
mmGF EARTH F=m
g
g=GmEARTH/r2
On earth surface: g=9.81 m/s2 r=6366 kmOn 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+36000kmg=0.22 m/s2
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PHY 2316
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, whatis your weight?
F=mg=70*6.5=455 N
And what is your weight if the space ship would beorbiting the earth?
Weightless!
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PHY 2317
Gravitational potential energySo 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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PHY 2318
Kepler’s laws
Johannes Kepler(1571-1630)
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PHY 2319
Kepler’s First lawEllipticity e(0-1)
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 beingin one of the focus point of the ellipse; planets around thesun.
p+q=constant
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PHY 23110
Kepler’s second law
A line drawn from the sun to the elliptical orbit of a planetsweeps out equal areas in equal time intervals.
Area(D-C-SUN)=Area(B-A-SUN)
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PHY 23111
Kepler’s third law
Consider a planet in circular motion around the sun:
2219
332
2
2
2
/1097.2
4
2
msK
rKrGM
T
T
r
t
sv
r
vM
r
MMG
s
ssun
planetplanetplanetsun
T2
r3
r3=T2/Ks r3=constant*T2
T: period-time it takes to makeone revolution
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PHY 23112
Chapter 8. Torque
It is much easier to swing thedoor if the force F is appliedas far away as possible (d) fromthe rotation axis (O).
Torque: The capability of a force to rotate an object aboutan axis.
Torque =F·d (Nm)
Torque is positive if the motion is counterclockwiseTorque is negative if the motion is clockwise
Top view
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PHY 23113
Decompositions
What is the torque applied to the door?
F//
FL
Force parallel to the rotating door: F//=Fcos600=150 NForce perpendicular to rotating door: FL=Fsin600=260 NOnly FL is effective for opening the door:
=FL·d=260*2.0=520 Nm
F=
Top view
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PHY 23114
Multiple force causing torque.
0.6 m0.3 m
100 N
50 N
Two persons try to gothrough a rotating doorat the same time, one onthe l.h.s. of the rotator andone the r.h.s. of the rotator.If the forces are applied asshown in the drawing, whatwill happen?
Top view
1=F1·d1=-100*0.3=-30 Nm
2=F2·d2=50*0.6 =30 NmNothing will happen! The 2torques are balanced.
+0 Nm
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PHY 23115
Center of gravity.Fpulldpull Vertical direction
(I.e. side view)
Fgravity
dgravity?
=Fpulldpull+Fgravitydgravity
We can assume thatfor the calculationof torque due to gravity,all mass is concentratedin one point:The center of gravity:the average position ofthe massdcg=(m1d1+m2d2+…+mndn) (m1+m2+…+mn)
1 2 3………………………n
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PHY 23116
Center of Gravity; more general
ii
iii
CG m
xmx
The center of gravity
ii
iii
CG m
ymy
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PHY 23117
Object in equilibrium
CG
Fp
-Fp
d-d
Top viewNewton’s 2nd law: F=ma
Fp+(-Fp)=ma=0No acceleration, no movement…
But the block starts to rotate!
=Fpd+(-Fp)(-d)=2FpdThere is movement!
Translational equilibrium: F=ma=0 The center of gravitydoes not move!
Rotational equilibrium: =0 The object does notrotate
Mechanical equilibrium: F=ma=0 & =0 No movement!