Transcript Simple Pendulum

Torque and Simple Harmonic
Motion
8.01
Week 13D2
Today’s Reading Assignment Young
and Freedman: 14.1-14.6
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Announcements
Problem Set 11 Due Thursday Nov 1 9 pm
Sunday Tutoring in 26-152 from 1-5 pm
W013D3 Reading Assignment Young and Freedman:
14.1-14.6
2
Simple Pendulum
Table Problem: Simple Pendulum
by the Torque Method
(a) Find the equation of motion for θ(t)
using the torque method.
(b) Find the equation of motion if θ is
always <<1.
Table Problem: Simple Pendulum
by the Energy Method
1. Find an expression for the mechanical
energy when the pendulum is in
motion in terms of θ(t) and its
derivatives, m, l, and g as needed.
2. Find an equation of motion for θ(t)
using the energy method.
Simple Pendulum: Small Angle
Approximation
Equation of motion
2
d

lmg sin   ml 2 2
dt
Angle of oscillation is small
sin  
Simple harmonic oscillator
d 2
g
 
2
l
dt
d 2x
k
 x
2
m
dt
Analogy to spring equation
Angular frequency of oscillation
Period
T0 
2
0
0  g / l
 2 l / g
Simple Pendulum:
Approximation to Exact Period
Equation of motion:
2
d

lmg sin   ml 2 2
dt
Approximation to exact period:
T ; T0 (sin0 / 0 )3/8  T0  T
Taylor Series approximation:
T ; T0
1 2
0
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Concept Question: SHO and the
Pendulum
Suppose the point-like object of a simple pendulum is pulled
out at by an angle 0 << 1 rad. Is the angular speed of the
point-like object equal to the angular frequency of the
pendulum?
1.Yes.
2.No.
3.Only at bottom of the swing.
4.Not sure.
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Demonstration
Pendulum: Amplitude Effect on
Period
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Table Problem: Torsional Oscillator
A disk with moment of inertia I cm about
the center of mass rotates in a horizontal
plane. It is suspended by a thin, massless
rod. If the disk is rotated away from its
equilibrium position by an angle  , the rod
exerts a restoring torque given by
 cm  
At t = 0 the disk is released from rest at
an angular displacement of  0 . Find the
subsequent time dependence of the
angular displacement  (t)
.
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Worked Example: Physical
Pendulum
A general physical pendulum
consists of a body of mass m
pivoted about a point S. The center
of mass is a distance dcm from the
pivot point. What is the period of
the pendulum.
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Concept Question: Physical
Pendulum
A physical pendulum consists
of a uniform rod of length l and
mass m pivoted at one end. A
disk of mass m1 and radius a
is fixed to the other end.
Suppose the disk is now
mounted to the rod by a
frictionless bearing so that is
perfectly free to spin. Does the
period of the pendulum
1. increase?
2. stay the same?
3. decrease?
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Physical Pendulum
Rotational dynamical equation
r
r
 S  IS
Small angle approximation
sin  
Equation of motion
Angular frequency
Period
lcm mg
d 2



2
IS
dt
0 
2
lcm mg
IS
IS
T
 2
0
lcm mg
Demo: Identical Pendulums,
Different Periods
Single pivot: body rotates about center of mass.
Double pivot: no rotation about center of mass.
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Small Oscillations
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Small Oscillations
Potential energy function U (x) for object of mass m
Motion is limited to the region
x1  x  x2
Potential energy has a minimum at x  x0
Small displacement from minimum, approximate
potential energy by
2
1
d
U
U (x) ; U (x0 )  (x  x0 )2 2 (x0 )
2!
dx
1
U (x) ; U (x0 )  keff (x  x0 )2
2
Angular frequency of small oscillation
 0  keff / m 
d 2U
(x0 ) / m
2
dx
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Concept Question: Energy Diagram 1
A particle with total mechanical energy
E has position x > 0 at t = 0
1) escapes to infinity in the – x-direction
2) approximates simple harmonic motion
3) oscillates around a
4) oscillates around b
5) periodically revisits a and b
6) not enough information
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Concept Question: Energy Diagram 2
A particle with total mechanical energy
E has position x > 0 at t = 0
1) escapes to infinity
2) approximates simple harmonic motion
3) oscillates around a
4) oscillates around b
5) periodically revisits a and b
6) not enough information
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Concept Question: Energy Diagram 3
A particle with total mechanical energy
E has position x > 0 at t = 0
1) escapes to infinity
2) approximates simple harmonic motion
3) oscillates around a
4) oscillates around b
5) periodically revisits a and b
6) not enough information
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Concept Question: Energy Diagram 4
A particle with total mechanical energy
E has position x > 0 at t = 0
1) escapes to infinity
2) approximates simple harmonic motion
3) oscillates around a
4) oscillates around b
5) periodically revisits a and b
6) not enough information
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Concept Question: Energy Diagram 5
A particle with total mechanical energy
E has position x > 0 at t = 0
1) escapes to infinity
2) approximates simple harmonic motion
3) oscillates around a
4) oscillates around b
5) periodically revisits a and b
6) not enough information
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Table Problem: Small Oscillations
A particle of effective mass m is acted on by a potential energy given by
  x 2  x 4
U (x)  U 0  2      
  x0   x0  
where U 0 and x0 are positive constants
a)Sketch
U (x) / U 0 as a function of
x / x. 0
b)Find the points where the force on the particle is zero. Classify them as
U (x) / Uat0 these equilibrium
stable or unstable. Calculate the value of
points.
c)If the particle is given a small displacement from an equilibrium point, find
the angular frequency of small oscillation.
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Appendix
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Simple Pendulum: Mechanical Energy
Velocity
vtan
d
l
dt
2
Kinetic energy
1 2 1  d 
K f  mvtan m  l 
2
2  dt 
Initial energy
E0  K 0  U 0  mgl(1  cos0 )
2
Final energy
1  d 
E f  K f  U f  m  l   mgl(1 cos )
2  dt 
Conservation of energy
2
1  d 
m  l   mgl(1 cos )  mgl(1 cos0 )
2  dt 
Simple Pendulum: Angular Velocity
Equation of Motion
Angular velocity
d
2g

(cos  cos 0 )
dt
l
Integral form

d
(cos  cos 0 )


2g
dt
l
Can we integrate this to get the period?
Simple Pendulum: Integral Form
 
Change of variables
“Elliptic Integral”

Power series approximation
1 b sin a
2
Solution
2
1 2
 
b  sin  2 
bsin a  sin  2
1 cos
 sin 2  2
2
0
da
1  b sin a 
2
2
12


1 2 2
3 4 4
 1 b sin a  b sin a  
2
8


1
2   sin 2  0 2     
2
g
T
l
g
dt
l
Simple Pendulum: First Order
Correction
Period
l
T  2
g

 2 

1 2
1 2
 1  4 sin  0 2        1  4 sin  0 2  
Approximation
First order correction






sin 2  0 2  02 4
l
T  2
g


1 2
1 2
 1  16  0   T0  1  16  0 
1 2
 T0  T1 where T1   0 T0
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