PHY 2311

PHYSICS 231Lecture 17: We have lift-off!

Remco ZegersWalk-in hour:Tue 4-5 pm

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

The centripetal acceleration is caused by a changein the direction of the linear velocity vector, nota change in magnitude

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

4 km/s 6 km/s 8 km/slaunchspeed

PHY 2318

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 2319

stone from outer space

A rock of 1.00 kg is dropped from outer space (initial velocity=0) at a distance of 2.50Rearth from the Earth’s center. What will its kinetic energy be when it reaches the surface of the earth, ignoring friction. Rearth=6.38x106 m, Mearth=5.98x1024 kg and G=6.67x10-11 Nm2/kg2.

PHY 23110

Kepler’s laws

Johannes Kepler(1571-1630)

PHY 23111

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 23112

launch speed is 10 km/s

PHY 23113

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 23114

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 23115

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

demo: opening a cellar doordemo: turning a screw

PHY 23116

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 23117

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

demo: balance with unequal masses

PHY 23118

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 23119

Center of Gravity; more general

ii

iii

CG m

xmx

The center of gravity

ii

iii

CG m

ymy

demo center of gravity

PHY 23120

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!