Chapter 5: Circular Motion

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Transcript Chapter 5: Circular Motion

Chapter 5
Circular Motion
Circular Motion
• Uniform Circular Motion
• Radial Acceleration
• Banked and Unbanked Curves
• Circular Orbits
• Nonuniform Circular Motion
• Tangential and Angular Acceleration
• Artificial Gravity
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Angular Displacement
y
 is the angular position.
f

i
Angular displacement:
x
   f   i
Note: angles measured CW are negative and angles measured
CCW are positive.  is measured in radians.
2 radians = 360 = 1 revolution
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Arc Length
y
arc length = s = r
f
r

i
x
s
 
r
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 is a ratio of two lengths; it is
a dimensionless ratio!
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Angular Speed
The average and instantaneous angular speeds are:


av 
and   lim
t 0 t
t
 is measured in rads/sec.
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Angular Velocity
The average and instantaneous angular speeds are:


av 
and   lim
t 0 t
t
 is actually a vector quantity.
The direction of  is along the axis of rotation. Since
we are concerned primarily with motion in a plane we
will ignore the vector nature until we get to angular
momentum when we will have to deal with it.
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Linear Speed
y
An object moves along a
circular path of radius r; what
is its average linear speed?
f
r
i

x
Note: Unfortunately the linear speed is
also called the tangential speed. This
can cause confusion.
total distance r
  
vav 

 r
  r av
total time
t
 t 
Also,
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v  r
(instantaneous values).
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Period and Frequency
The time it takes to go one time around a closed path is
called the period (T).
total distance 2r
vav 

total time
T
2
Comparing to v = r:  
 2f
T
f is called the frequency, the number of revolutions (or
cycles) per second.
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Linear and Angular Speed
The linear speed depends
on the radius r at which the
object is located.
y
v  r
v1
Angular speed is independent
of position.
r1
v1 = ωr1
v2
v2 = ωr2
v1 v2
= =ω
r1
r2
r2
x
v1
Everyone sees the same ω, because
they all experience the same rpms.
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v1
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v1
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Centripetal Acceleration
Consider an object moving
in a circular path of radius r
at constant speed.
y
v
Here, v  0. The
direction of v is changing.
v
x
If v  0, then a  0.
Then there is a net force
acting on the object.
v
v
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Centripetal Acceleration
Conclusion: with no net force acting on the object it would
travel in a straight line at constant speed
It is still true that F = ma.
But what acceleration do we use?
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Centripetal Acceleration
The velocity of a particle is tangent to its path.
For an object moving in uniform circular motion, the
acceleration is radially inward.
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Centripetal Acceleration
The magnitude of the radial acceleration is:
2
v
2
ar 
 r  v
r
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Rotor Ride Example
The rotor is an amusement park ride where people stand
against the inside of a cylinder. Once the cylinder is spinning
fast enough, the floor drops out.
(a) What force keeps the people from
falling out the bottom of the cylinder?
y
fs
N
Draw an FBD for a person
with their back to the wall:
x
w
It is the force of static friction.
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Rotor Ride Example
(b) If s = 0.40 and the cylinder has r = 2.5 m, what is the
minimum angular speed of the cylinder so that the people
don’t fall out?
Apply Newton’s 2nd Law:
From (2):
fs  w
1  Fx  N  mar  m 2 r
2  Fy  f s  w  0
 s N   s m 2 r   mg
From (1)
9.8 m/s 2


 3.13 rad/s
0.402.5 m
s r
g
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Unbanked Curve
A coin is placed on a record that is rotating at 33.3 rpm. If s
= 0.1, how far from the center of the record can the coin be
placed without having it slip off?
y
We’re looking for r.
N
Draw an FBD for the coin:
Apply Newton’s 2nd Law:
fs
x
1  Fx  f s  mar  m 2 r
2  Fy  N  w  0
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w
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Unbanked Curve
From (2)
From 1 : f s  m 2 r
f s   s N   s mg   m 2 r
s g
Solving for r: r  2

What is ?
rev  2 rad  1 min 
  33.3


  3.5 rad/s
min  1 rev  60 sec 
 s g 0.19.8 m/s 2 
r 2 
 0.08 m
2

3.50 rad/s 
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Banked Curves
A highway curve has a radius of 825 m. At what angle
should the road be banked so that a car traveling at 26.8
m/s has no tendency to skid sideways on the road? (Hint:
No tendency to skid means the frictional force is zero.)
R
Take the car’s
motion to be into
the page.

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Banked Curves
The normal force is the cause of the centripetal acceleration.
Nsinθ.
We need the radial position of the car itself, not the radius of
the track. In either case the radius is not a uniquely
determined quantity.
y
This is the only path by
which to traverse the
track as there is no
friction in the problem.
 N
x
w
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Banked Curves
y
FBD for the car:

N
x
w
Apply Newton’s Second Law:
v2
1  Fx  N sin   mar  m
r
2  Fy  N cos   w  0
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Banked Curves
Rewrite (1) and (2):
v2
1 N sin   m
r
2 N cos   mg
Divide (1) by (2):
v
 26.8 m/s 
tanθ =
=
= 0.089
2
gr  9.8 m/s   825 m 
2
2
θ = 5.1°
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Circular Orbits
Consider an object of mass m in a circular
orbit about the Earth.
v
r
Earth
The only force on the satellite is the force
of gravity. That is the cause of the
centripetal force.
Gms M e
v2
 F  Fg  r 2  ms ar  ms r
Solve for the speed of the satellite:
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Gms M e
v2
 ms
2
r
r
GM e
v
r
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Circular Orbits
Consider an object of mass m in a
circular orbit about the Earth.
v
r
Earth
v=
GM e
r
GMe is fixed - v is proportional to
1
r
v and r are tied together. They are not independent
for orbital motions.
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Circular Orbits
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
r
From previous slide:
Combine these expressions and solve for r:


v
Also need,

 GM e 2 
r 
T 
2
 4

 6.67 10 Nm /kg 5.98 10 kg
2



r  
86400
s
2
4


 4.225 107 m
11
2
2
24
2r
T
1
1
3
3
r  Re  h  h  r  Re  35,000 km
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Circular Orbits
Kepler’s Third Law
 GM e 2 
r 
T 
2
 4

 GM 2 
r

It can be generalized to:
 2T 
 4

1
1
3
3
Where M is the mass of the central body.
For example, it would be Msun if speaking of the planets in the
solar system. For non-circular orbits (elliptical) the mean
radius is used and so the mean velocity is obtained.
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Nonuniform Circular Motion
Nonuniform means the speed (magnitude of velocity)
is changing.
a
at
There is now an acceleration
tangent to the path of the particle.
ar
v
The net acceleration of the body is a  ar  at
2
2
This is true but useless!
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Nonuniform Circular Motion
a
at
at changes the magnitude of v.
ar
Changes energy - does work
ar changes the direction of v.
Doesn’t change energy does NO WORK
Can write:
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 F  ma
 F  ma
r
r
t
t
The accelerations are only
useful when separated into
perpendicular and parallel
components.
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Loop Ride
Example: What is the minimum speed for the car so that it
maintains contact with the loop when it is in the pictured
position?
FBD for the car at
the top of the loop:
r
y
Apply Newton’s 2nd Law:
x
N
w
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v2
 Fy   N  w  mar  m r
v2
N wm
r
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Loop Ride
The apparent weight at the top of loop is:
You lose contact when N = 0
N = 0 when
v2
N wm
r
 v2

N  m  g 
 r

 v2

N  m  g   0
 r

v  gr
This is the minimum speed needed to make it around the
loop.
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Loop Ride
Consider the car at the bottom of the loop; how does the
apparent weight compare to the true weight?
FBD for the car at the
bottom of the loop:
y
N
x
w
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Apply Newton’s 2nd Law:
v2
 Fy  N  w  mac  m r
v2
N wm
r
 v2


N  m  g 
 r

Here, N  mg
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Linear and Angular Acceleration
The average and instantaneous angular acceleration are:


 av 
and   lim
t 0 t
t
 is measured in rads/sec2.
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Linear and Angular Acceleration
Recalling that the tangential velocity is vt = r means the
tangential acceleration is
vt

r
 r
at t 
t
t
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Linear and Angular Kinematics
Angular
Linear (Tangential)
v  v0  at
   0   t
1
2
x  x0  v0 t  at
2
v 2  v02  2ax
1
2
   0  0 t  t
2
 2  02  2
With vt  r and at  r
“a” and “at” are the same thing
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Dental Drill 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?
Given:  = 3.14104 rads/sec; t = 1 sec;  = 0
Want .
1
   0  0 t  t 2
2
   0  0 t


  0 t  3.14  10 4 rads/sec 1.0 sec 
 3.14  10 4 rads  1.80  106 degrees
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Dental Drill Example
A high speed dental drill is rotating at 3.14104 rads/sec.
What is that in rpm’s?
rad
rev
rev
4 rev
1
π × 10
×
= 2 10
= 5000
s
2πrad
s
s
rev 60sec
5000
×
= 300,000rpm
s
min
4
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Car Example
Your car’s wheels are 65 cm in diameter and are rotating at 
= 101 rads/sec. How fast in km/hour is the car traveling,
assuming no slipping?
v
X
total distance 2r N 2r
v


 r
T N T
total time
 101 rads/sec 32.5 cm 
 3.28 103 cm/sec  118 km/hr
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Artificial Gravity
A large rotating cylinder in
deep space (g0).
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Artificial Gravity
y
y
N
x
x
N
Top position
Bottom position
Apply Newton’s 2nd Law to each:
2
F

N

ma

m

r
 y
r
2
F


N


ma


m

r
 y
r
N is the only force acting. It causes
the centripetal acceleration.
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Space Station Example
A space station is shaped like a ring and rotates to simulate
gravity. If the radius of the space station is 120m, at what
frequency must it rotate so that it simulates Earth’s gravity?
Using the result from the previous slide:
2
F

N

ma

m

r
 y
r
N
mg



mr
mr
g
 0.28 rad/sec
r
The frequency is f = (/2) = 0.045 Hz (or 2.7 rpm).
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Summary
• A net force MUST act on an object that has circular
motion.
• Radial Acceleration ar=v2/r
• Definition of Angular Quantities (, , and )
• The Angular Kinematic Equations
• The Relationships Between Linear and Angular
Quantities
v  r and a  r
t
•
t
Uniform and Nonuniform Circular Motion
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