Uniform Circular Motion – Centripetal Force
Uniform circular motion –acc7
Uniform circular motion
Orbital motion 2
Synchronous orbits 2
Kepler’s Law of Planetary Motion
Prob 5.32
For M = mass of the Sun,
the relation between r3 and
T2 is called Kepler’s Law of
Planetary Motion.
Note: Planetary Motion
Prob 5.32
The planets actually sweep out
elliptical orbits such that the
speed increases when r is small
(“perihelion” for smallest r) and
decreases when r is large
(“aphelion” for largest r).
Note: Planetary Motion
Prob 5.32
Kepler’s 2nd Law: The planets
sweep out equal areas in equal
times.
NOTE: We will generally make
the approximation that the
planetary orbits are circular
(ellipses of small “eccentricity”).
Prob 5.32 a
Prob 5.33
m = 5850 kg
R = 4.15x106 m
r = 4.1x105 m + 4.15x106 m = 4.56x106 m
T = 7200s
Prob 5.33
In orbit:
On the planet’s surface:
We are looking for mgp but we don’t know the planet’s mass, Mp
(in order to make use of the 2nd equation on planet’s surface) .
m = 5850 kg
R = 4.15x106 m
r = 4.1x105 m + 4.15x106 m = 4.56x106 m
T = 7200s
Prob 5.33
In orbit:
On the planet’s surface:
We can use the 1st equation (orbit) to find Mp .
PROBLEM: We don’t know the satellite’s orbital speed, v .
m = 5850 kg
R = 4.15x106 m
r = 4.1x105 m + 4.15x106 m = 4.56x106 m
T = 7200s
Prob 5.33
In orbit:
We DO know the satellite’s orbital speed, v !
Prob 5.33
In orbit:
Combine the two boxes on the right:
(We’ve re-derived Kepler’s Law!)
Prob 5.33
In orbit:
On the planet’s surface:
Combine the two boxes on the right:
Prob 5.33
So, the weight of the satellite is given by:
m = 5850 kg
R = 4.15x106 m
r = 4.1x105 m + 4.15x106 m = 4.56x106 m
T = 7200s
→ mgp = 2.45 x 104 N (gp = 4.2 m/s2 )
Question b
Uniform circular motion
Apparent weightlessness
Apparent weightlessness
Artificial gravity
NOTE: it’s the normal force
exerted by the inner (“floor”)
surface of the spaceship which
acts on the person to provide a
centripetal acceleration.
Artificial gravity 2
Artificial gravity
What about friction?
What about friction?
Consider the rotation
“starting up”.
Artificial gravity
From (a) to (b), the
person’s speed in the x
(“tangential”) direction is
changing. This is an
acceleration and
requires a tangential
force – static friction!
At (c), the person’s
speed is constant, with
uniform motion in the
tangential direction. No
tangential force is
required.
Prob 5.28 b
Uniform circular motion
Vertical circular motion
Vertical circular motion
Vertical circular motion
Prob 5.40 a
Chapter 6
Work and Energy
Work and Energy
We define a new kinematic quantity: Kinetic Energy.
A moving object of mass m has a kinetic energy:
KE = ½mv2
The SI unit of kinetic energy is the Joule (J)
1J = 1kg.m2/s2
Work and Energy
Work and Energy 0
We can change the state of motion of an object
(change v) by applying a net force:
We say that the force “does work” on the object to
change its kinetic energy from
an initial value: KE0 = ½mv02
to a final value: KE = ½mv2
Work and Energy
Work and Energy 1
Using a kinematic equation:
v2 - v02 = 2ax  mv2 - mv02 = 2max

½mv2 - ½mv02 = Fx

KE – KE0 = Fx
Work and Energy
Work and Energy 2
½mv2 - ½mv02 = Fx

KE – KE0 = Fx = W
Fx = W is the “work done by the net force, F in
displacing the object by x”.
W = KE – KE0 = ∆KE
“Work - Energy Theorem”
(work done by Net Force)
(units: Joule J = N.m = 1kg.m2/s2 )
Work and Energy
½mv2 - ½mv02 = KE – KE0 = ∆KE = Fx = W
Work and Energy 2.1
Work and Energy
Prob 6.10
x
If multiple forces are acting on an object, each force
can perform work. The change in kinetic energy is
determined by the work done by the net force.
Work and Energy
Only the component of a force in the direction of the
displacement counts.
Work and Energy 3
For displacement in the x direction, the component of
applied force along the displacement is
Fx = Fcos
Work and Energy
Work and Energy 3.1
If  is the angle between the applied force and the
displacement:
W = Fxcos = the work done by the applied force.
Work and Energy
Prob 6.10
x
If multiple forces are acting on an object, each force performs
work: Wi = Fi xcos .
The change in kinetic energy is determined by the work done
by the net force:
∆KE = Fnet xcos