Transcript hp1f2013_class05_NewtonsLawsApplications

Honors Physics 1
Class 05 Fall 2013
Newton’s Laws
Applications
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Activity: Application
The Spinning Terror ride
The spinning terror is a large vertical drum which spins
so fast that everyone stays pinned to the wall when
the floor drops out. For a typical ride the radius of the
drum is 2 m.
What is the minimum angular velocity if the coefficient
of friction between the patron and the wall is 0.3?
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Spinning Terror
v2
The radial acceleration that must act on the rider is
 R 2 .
R
The normal force is therefore N  Ma  MR 2 .
The maximum static friction that can act on the
rider is therefore f max  N .
Since we require M to be in vertical equilibrium,
we need f max  Mg .
So N   MR 2   Mg .
g
Thus  2 
and the smallest value of  that
R
allows the ride to run is:
min 
g
10
rad

4
R
0.3  2
s
which is about 0.6 turns per second.
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Application example:
Mass on a spring
Equilibrium position x  0; Starting point=x0
F  kx
Write the F  ma equation.
Assume a solution of the form: x(t )  Ae t .
See what conditions have to be met by A and 
to solve the relation F  ma and satisfy initial
conditions.
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Mass on a spring
F   kx and F  ma  m
d 2x
dt
2
  kx
d 2x
k
which gives 2  x  0, a common differential equation.
m
dt
We will try a solution of the form: x(t )  Ae t
k
 2 Ae t  Ae t  0 and assuming A  0,
m
k
k
2
 
   i
 i
m
m
So a general solution is: x  Aeit  Be it
(each one works independently)
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Mass on a spring: initial conditions
x  Aeit  Be it
(Initial conditions: x(0)  x0 and v(0)  0)
dx
(0)  0  i A  i B so A  B
dt


x(t )  A eit  e it  2 A cos t 
x(0)  x0  2 A
k
The mass oscillates with amplitude x0 at frequency  
.
m
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Application example:
Falling through a viscous fluid
Assume that the density of the fluid is very
small compared to the density of the falling
object. (e.g – a human body in air)
Assume that the body falls under the action of
constant gravity and drag force only.
Assume that the drag force is linear in speed:
FD  Cv
Is there a terminal velocity?
If there is, find the terminal velocity.
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Falling...
The falling object is acted on by buoyant force, gravity, and drag.
F   Mg  FB  Cv  Ma
If the body is in equilibrium (not accelerating) then Ma=0.
Mg  FB
Cvterminal  Mg  FB so vterminal 
C
Mg
and for a body falling in air: vterminal 
C
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Taylor series
( x  a)2
f ( x)  f (a )  f '(a )( x  a )  f ''(a )
 ...
2!
1 3 1 5
sin( x)  x  x  x  ...
3!
5!
x2 1 4
cos( x)  1 
 x  ...
2 4!
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