Transcript Chapter 7

Chapter 7
Rotational Motion and The Law of Gravity
The radian
• The radian is a unit of angular measure
• The angle in radians can be defined as the ratio of
the arc length s along a circle divided by the radius r
s
 
r
360
1 rad 
 57.3
2

 [rad] 
 [deg rees]
180
Rotation of a rigid body
• We consider rotational motion of a rigid body about
a fixed axis
• Rigid body rotates with all its parts locked together
and without any change in its shape
• Fixed axis: it does not move during the rotation
• This axis is called axis of rotation
• Reference line is introduced
Angular position
• Reference line is fixed in the body, is perpendicular
to the rotation axis, intersects the rotation axis, and
rotates with the body
• Angular position – the angle (in radians or degrees)
of the reference line relative to a fixed direction (zero
angular position)
Angular displacement
• Angular displacement – the change in angular
position.
• Angular displacement is considered positive in the
CCW direction and holds for the rigid body as a
whole and every part within that body
   f   i
Angular velocity
• Average angular velocity
avg
 f  i



t f  ti
t
• Instantaneous angular velocity – the rate of change
in angular position

  lim
t  0 t
Angular acceleration
• Average angular acceleration
 avg
 f  i



t f  ti
t
• Instantaneous angular acceleration – the rate of
change in angular velocity

  lim
t 0 t
Uniform circular motion
• A special case of 2D motion
• An object moves around a circle at a constant speed
• Period – time to make one full revolution
2r
T
v
• An object traveling in a circle, even though it moves
with a constant speed, will have an acceleration
Centripetal acceleration
• Centripetal acceleration is due to the change in the
direction of the velocity


v
a  lim
t  0 t
• Centripetal acceleration is directed toward the
center of the circle of motion
Centripetal acceleration
• The magnitude of the centripetal acceleration is
given by
2
v
ac 
r
Centripetal acceleration
During a uniform circular motion:
• the speed is constant
• the velocity is changing due to centripetal (“center
seeking”) acceleration
• centripetal acceleration is constant in magnitude
(v2/r), is normal to the velocity vector, and points
radially inward
Rotation with constant angular
acceleration
• Similarly to the case of 1D motion with a constant
acceleration we can derive a set of formulas:
Chapter 7
Problem 5
A dentist’s drill starts from rest. After 3.20 s of constant angular acceleration, it
turns at a rate of 2.51 × 104 rev/min. (a) Find the drill’s angular acceleration. (b)
Determine the angle (in radians) through which the drill rotates during this
period.
Relating the linear and angular
variables: position
• For a point on a reference line at a distance r from
the rotation axis:
s  r
• θ is measured in radians
Relating the linear and angular
variables: speed
s
(r )

v  lim
 lim
 r lim
 r
t 0 t
t 0 t
t  0  t
• ω is measured in rad/s
• Period
2r 2
T

v

Chapter 7
Problem 2
A wheel has a radius of 4.1 m. How far (path length) does a point on the
circumference travel if the wheel is rotated through angles of 30°, 30 rad, and
30 rev, respectively?
Relating the linear and angular
variables: acceleration
v
 (r )

at  lim
 lim
 r lim
 r
t  0 t
t 0
t 0 t
t
• α is measured in rad/s2
• Centripetal acceleration
v
(r )
2
 r
ac 

r
r
2
2
Total acceleration
• Tangential acceleration is due to changing speed
• Centripetal acceleration is due to changing direction
• Total acceleration:
a  at  ac
2
2
Centripetal force
• For an object in a uniform circular motion, the
centripetal acceleration is
2
v
ac 
R
• According to the Newton’s Second Law, a force
must cause this acceleration – centripetal force
mv
Fc  mac 
R
2
• A centripetal force accelerates a body by changing
the direction of the body’s velocity without changing
the speed
Centripetal force
• Centripetal forces may have different origins
• Gravitation can be a centripetal force
• Tension can be a centripetal force
• Etc.
Centripetal force
• Centripetal forces may have different origins
• Gravitation can be a centripetal force
• Tension can be a centripetal force
• Etc.
Free-body diagram
Chapter 7
Problem 28
A roller-coaster vehicle has a mass of 500 kg when fully loaded with
passengers. (a) If the vehicle has a speed of 20.0 m/s at point A, what is the
force exerted by the track on the car at this point? (b) What is the in maximum
speed the vehicle can have at point B and still remain on the track?
Newton’s law of gravitation
• Any two (or more) massive bodies attract each other
• Gravitational force (Newton's law of gravitation)

m1m2
F  G 2 rˆ
r
• Gravitational constant G = 6.67*10 –11 N*m2/kg2 =
6.67*10 –11 m3/(kg*s2) – universal constant
Gravitation and the superposition
principle
• For a group of interacting particles, the net
gravitational force on one of the particles is
n 

F1,net   F1i
i2
Chapter 7
Problem 33
Objects with masses of 200 kg and 500 kg are separated by 0.400 m. (a) Find
the net gravitational force exerted by these objects on a 50.0-kg object placed
midway between them. (b) At what position (other than infinitely remote ones)
can the 50.0-kg object be placed so as to experience a net force of zero?
Gravity force near the surface of Earth
• Earth can be though of as a nest of shells, one
within another and each attracting a particle outside
the Earth’s surface
• Thus Earth behaves like a particle located at the
center of Earth with a mass equal to that of Earth

 GmEarth  ˆ
mEarthm1 ˆ
F1, Earth  G 2
j   2
m1 j  g m1 ˆj
REarth
 REarth 
g = 9.8 m/s2
• This formula is derived for stationary Earth of ideal
spherical shape and uniform density
Gravity force near the surface of Earth
In reality g is not a constant because:
Earth is rotating,
Earth is approximately an ellipsoid
with a non-uniform density
Gravitational potential energy
• Gravitation is a conservative force (work done by it
is path-independent)
• For conservative forces potential energy can be
introduced
• Gravitational potential energy:
Gm1m2
U (r )  
r
Gravitational potential energy
Gm1m2
U (r )  
r
Escape speed
• Accounting for the shape of Earth, projectile motion
(Ch. 3) has to be modified:
2
v
ac 
 g  v  gR
R
Escape speed
• Escape speed: speed required for a particle to
escape from the planet into infinity (and stop there)
Ki  U i  K f  U f
2
m1v Gm1m planet

 00
2
R planet
vescape 
2Gm planet
R planet
Escape speed
• If for some astronomical object
vescape 
2Gmobject
Robject
 3 10 m / s  c
8
• Nothing (even light) can escape from the surface of
this object – a black hole
Escape speed
Chapter 7
Problem 56
Show that the escape speed from the surface of a planet of uniform density is
directly proportional to the radius of the planet.
Kepler’s laws
Tycho Brahe/
Tyge Ottesen
Brahe de Knudstrup
(1546-1601)
Johannes Kepler
(1571-1630)
Three Kepler’s laws
• 1. The law of orbits: All planets move in elliptical
orbits, with the Sun at one focus
• 2. The law of areas: A line that connects the planet
to the Sun sweeps out equal areas in the plane of the
planet’s orbit in equal time intervals
• 3. The law of periods: The square of the period of
any planet is proportional to the cube of the
semimajor axis of its orbit
First Kepler’s law
• All planets move in elliptical orbits with the Sun at
one focus, whereas the second focus is empty
• Any object bound to another by an inverse square
law will move in an elliptical path
Second Kepler’s law
• A line drawn from the Sun to any planet will sweep
out equal areas in equal times
• Area from A to B and C to D are the same
Third Kepler’s law
• For a circular orbit and the Newton’s Second law
GMm
GM
2
2
 (m)( r )   3
2
r
r
F  ma
• From the definition of a period
T
2

T 
2
4

2
2
4 3
T 
r
GM
2
2
Satellites
• For a circular orbit and the Newton’s Second law
2

GMm
v 
 (m) 
F  ma
2
r
 r 
• Kinetic energy of a satellite
2
GMm
U
mv


K
2r
2
2
• Total mechanical energy of a satellite
GMm
GMm
GMm
E  K U 


 K
2r
r
2r
Chapter 7
Problem 45
The Solar Maximum Mission Satellite was placed in a circular orbit about 150
mi above Earth. Determine (a) the orbital speed of the satellite and (b) the time
required for one complete revolution.
Questions?
Answers to the even-numbered problems
Chapter 7
Problem 10
50 rev
Answers to the even-numbered problems
Chapter 7
Problem 30
(a) 4.39 × 1020 N toward the Sun
(b) 1.99 × 1020 N toward the Sun
(c) 3.55 × 1020 N toward the Sun
Answers to the even-numbered problems
Chapter 7
Problem 36
(a) 5.59 × 103 m/s
(b) 3.98 h
(c) 1.47 × 103 N