1. PHYSICS 231 Lecture 19: We have lift-off!
Comet Kohoutek
Remco Zegers
Walk-in hour: Monday 9:15-10:15 am
Helproom
PHY 231

2. Previously… v Centripetal acceleration: ac=v2/r=2r
Caused by force like:
Gravity
Tension
Friction
F=mac for rotating object ac r
The centripetal acceleration is caused by a change
in the direction of the linear velocity vector, not
a change in magnitude
PHY 231

3. quiz (extra credit) ac= 2r. A= B= c rA<rB<rc
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
child A
child B
child C
the same for all three
ac= 2r.
A= B= c
rA<rB<rc
acA<acB<acC A B C
demo
PHY 231

4. The gravitational force, revisited
Newton:
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.
PHY 231

5. Gravitation between two objects
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’
PHY 231

6. Gravitational acceleration
F=mg
g=GmEARTH/r2
On earth surface: g=9.81 m/s2 r=6366 km
On top of mount Everest: r= km g=9.78 m/s2
Low-orbit satellite: r= km g=6.27 m/s2
Geo-stationary satellite: r= km
g=0.22 m/s2
PHY 231

7. 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!
PHY 231

8. launch
speed
4 km/s
6 km/s
8 km/s
PHY 231

9. 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
PHY 231

10. Kepler’s laws
Johannes Kepler
( )
PHY 231

11. 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.
PHY 231

12. launch speed is 10 km/s
PHY 231

13. 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.
PHY 231

14. Kepler’s third law Consider a planet in circular motion r3
around the sun: T2
T: period-time it takes to make
one revolution
r3=T2/Ks r3=constant*T2
PHY 231

15. 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
PHY 231

16. =FL·d=260*2.0=520 Nm Decompositions F= FL F// Top view
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
Top view
What is the torque applied to the door?
PHY 231

17. Multiple force causing torque.
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?
Top view
50 N
0.3 m
0.6 m
1=F1·d1=-100*0.3=-30 Nm
2=F2·d2=50*0.6 =30 Nm
Nothing will happen! The 2
torques are balanced. +
0 Nm 
demo: balance with unequal masses
PHY 231

18. Center of gravity. Fpull dpull 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)
1 2 3………………………n
Fgravity
dgravity?
=Fpulldpull+Fgravitydgravity
PHY 231

19. Center of Gravity; more general
The center of gravity
demo center of gravity
PHY 231

20. Object in equilibrium Fp Top view Newton’s 2nd law: F=ma
Fp+(-Fp)=ma=0
No acceleration, no movement… -d d CG
But the block starts to rotate!
=Fpd+(-Fp)(-d)=2Fpd
There is movement!
-Fp
Translational equilibrium: F=ma=0 The center of gravity
does not move!
Rotational equilibrium: = The object does not
rotate
Mechanical equilibrium: F=ma=0 & =0 No movement!
PHY 231