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...
PHY 2311
PHYSICS 231Lecture 17: We have lift-off!
Remco ZegersWalk-in hour: Thursday 11:30-13:30 am
Helproom
Comet Kohoutek
PHY 2312
Previously…
v
r
ac
Centripetal acceleration:ac=v2/r=2r
Caused by force like:•Gravity•Tension•Friction
F=mac for rotating object
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.
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’
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
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!
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
PHY 2318
Kepler’s laws
Johannes Kepler(1571-1630)
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
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)
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
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
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
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
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
PHY 23116
Center of Gravity; more general
ii
iii
CG m
xmx
The center of gravity
ii
iii
CG m
ymy
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!