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General Physics (PHY 2130)
Lecture 12
•  Rotational kinematics
  Angular speed and acceleration
  Uniform and non-uniform circular motion
  Orbits and Kepler’s laws
http://www.physics.wayne.edu/~apetrov/PHY2130/
Lightning Review
Last lecture:
1.  Rotational kinematics
  angular displacement, angular velocity and angular acceleration
  relations between angular and linear quantities
Review Problem: A high speed dental drill is rotating at 3.14×104 rads/sec. Through
how many degrees does the drill rotate in 1.00 sec? How many revolutions does
that constitute?
Example: A high speed dental drill is rotating at 3.14×104 rads/sec. Through how
many degrees does the drill rotate in 1.00 sec? How many revolutions does that
constitute?
Given:
Solution:
1. Angular acceleration is zero, so
ω  = 3.14×104 rads/sec
Δt = 1 sec
α  = 0
1
θ = θ 0 + ω0 Δt + αΔt 2
2
θ = θ 0 + ω0 Δt
(
)
Δθ = ω0 Δt = 3.14 × 10 4 rads/sec (1.0 sec)
Find:
= 3.14 × 10 4 rads = 1.80 ×10 6 degrees
Δθ = ?
2. One revolution is 360 degrees, so
N = (1.8 ×10 6 )o / 360 o = 5000 revolutions
3
Analogies Between Linear and
Rotational Motion
Rotational Motion About a
Fixed Axis with Constant
Acceleration
Linear Motion with
Constant Acceleration
ω = ωi + αt
v = vi + at
1 2
Δθ = ω i t + αt
2
1 2
Δx = vi t + at
2
2
2
i
ω = ω + 2αΔθ
2
2
i
v = v + 2aΔx
Relationship Between Angular
and Linear Quantities
►  Displacements
►  Speeds
►  Accelerations
Δs
Δθ =
r
Δθ 1 Δs
=
Δt r Δt
or
1
ω= v
r
a = αr
Centripetal Acceleration
►  An
object traveling in a circle,
even though it moves with a
constant speed, will have an
acceleration (since velocity
changes direction)
►  This acceleration is called
centripetal (“center-seeking”).
►  The acceleration is directed
toward the center of the circle
of motion
Centripetal Acceleration
and Angular Velocity
►  The
angular velocity and the
linear velocity are related
(v = ωr)
►  The centripetal acceleration
can also be related to the
angular velocity
Δv Δs
v
=
⇒ Δv = Δs, but
v
r
r
a=
Δv
v Δs
⇒ a=
Δt
r Δt
Thus:
Similar
triangles!
v2
aC =
r
or aC = ω 2 r
Total Acceleration
►  What
happens if linear velocity
also changes?
►  Two-component acceleration:
  the centripetal component of the
acceleration is due to changing
direction
  the tangential component of the
acceleration is due to changing
speed
►  Total
acceleration can be found
from these components:
slowing-down car
2
t
2
C
a = a +a
Vector Nature of Angular
Quantities
►  As
in the linear case,
displacement, velocity
and acceleration are
vectors:
►  Assign a positive or
negative direction
►  A more complete way is
by using the right hand
rule
  Grasp the axis of rotation
with your right hand
  Wrap your fingers in the
direction of rotation
  Your thumb points in the
direction of ω
Forces Causing Centripetal
Acceleration
►  Newton’s
Second Law says that the centripetal
acceleration is accompanied by a force
v2
∑ F = maC = m r
  F stands for any force that keeps an object following a
circular path
► Force
of friction (level and banked curves)
► Tension in a string
► Gravity
Example: level curves
Consider a car driving at 20 m/s (~45
mph) on a level circular turn of
radius 40.0 m. Assume the car’s
mass is 1000 kg.
1. 
2. 
What is the magnitude of
frictional force experienced by
car’s tires?
What is the minimum coefficient
of friction in order for the car to
safely negotiate the turn?
Example:
Given:
masses: m=1000 kg
velocity: v=20 m/s
radius: r = 40.0m
1. Draw a free body diagram, introduce
coordinate frame and consider vertical
and horizontal projections
∑F
y
= 0 = N − mg
N = mg
Find:
1. 
2. 
f=?
µ=?
∑F
x
= ma = − f
2
(20 m s ) = −1.0 ×10 4 N
v2
f = −ma = −m = −1000 kg
r
40 m
 
2. Use definition of friction force:
v2
f = µmg = m = 10 4 N , thus
r
1.0 × 10 4 N
µ=
≈ 1.02
2
1000 kg 9.8 m s
 
Lesson: µ for rubber on dry concrete is 1.00!
rubber on wet concrete is 0.2!
driving too fast…
ConcepQuestion
Is it static or kinetic friction that is responsible for the fact
that the car does not slide or skid?
1. Static
2. Kinetic
Example: banked curves
Consider a car driving at 20 m/s (~45
mph) on a 30° banked circular
curve of radius 40.0 m. Assume
the car’s mass is 1000 kg.
1. 
2. 
What is the magnitude of
frictional force experienced by
car’s tires?
What is the minimum coefficient
of friction in order for the car to
safely negotiate the turn?
A component of the normal force adds to the
frictional force to allow higher speeds
v2
tan θ =
rg
Example:
Given:
masses: m=1000 kg
velocity: v=20 m/s
radius: r = 40.0m
angle: α = 30°
Find:
1. 
2. 
f=?
µ=?
1. Draw a free body diagram,
introduce coordinate frame and
consider vertical and
horizontal projections
v2
∑ Fx = −m r cos 30 = − f − mg sin 30
v2
f = m cos 30  − mg sin 30  = 3760 N
r
2
v
∑ Fy = m r sin 30 = N − mg cos 30
v2
N = m sin 30 + mg cos 30  = 1.3 ×10 4 N
r
2. Use definition of friction force:
f = µ s N , thus minimal µs is
µs =
fs
3760 N
=
≈ 0.28
4
N 1.3 ×10 N
 
 
Lesson: by increasing angle of banking,
one decreases minimal µ or friction with
which one can take curve!
Tension as a source of
centripetal acceleration
►  The
horizontal component
of the tension causes the
centripetal acceleration
aC = g tan θ
Forces in Accelerating Reference
Frames
►  Distinguish
real forces from fictitious forces
►  Centrifugal force is a fictitious force
►  Real forces always represent interactions
between objects
Circular Orbits
►  Consider
r
Earth
an object of mass m in a
circular orbit about the Earth.
►  The
only force on the satellite is the
force of gravity:
Gms M e
v2
∑ F = Fg = r 2 = ms ar = ms r
Solve for the speed of the satellite:
Gms M e
v2
= ms
2
r
r
GM e
v=
r
18
Escape Speed
►  The
escape speed is the speed needed for an
object to soar off into space and not return
vesc
►  For
2GM E
=
RE
the earth, vesc is about 11.2 km/s
►  Note, v is independent of the mass of the object
Example: How high above the surface of the Earth does a satellite need to be so
that it has an orbit period of 24 hours?
GM e
v
=
From previous slide:
r
Also need,
Combine these expressions and solve for r:
(
−11
2
2
)(
2πr
v=
T
⎛ GM e 2 ⎞
r = ⎜
T ⎟
2
⎝ 4π
⎠
24
)
1
3
⎛ 6.67 ×10 Nm /kg 5.98 ×10 kg
2 ⎞
⎟⎟
(
)
r = ⎜⎜
86400
s
2
4π
⎝
⎠
= 4.225 ×10 7 m
1
3
r = Re + h ⇒ h = r − Re = 35,000 km
20
Kepler’s Laws
► All
planets move in elliptical orbits with
the Sun at one of the focal points.
► A line drawn from the Sun to any planet
sweeps out equal areas in equal time
intervals.
► The square of the orbital period of any
planet is proportional to cube of the
average distance from the Sun to the
planet.
Kepler’s Laws, cont.
► Based
on observations made by Brahe
► Newton later demonstrated that these laws
were consequences of the gravitational
force between any two objects together
with Newton’s laws of motion
Kepler’s First Law
►  All
planets move in
elliptical orbits with
the Sun at one
focus.
  Any object bound to
another by an
inverse square law
will move in an
elliptical path
  Second focus is
empty
Kepler’s Second 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
Kepler’s Third Law
► The
square of the orbital period of any
planet is proportional to cube of the average
distance from the Sun to the planet.
2
T = Kr
3
or
⎛ GM e 2 ⎞
r = ⎜
T ⎟
2
⎝ 4π
⎠
1
3
  For orbit around the Sun, KS = 2.97x10-19 s2/m3
  K is independent of the mass of the planet