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
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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  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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Concept Question: SHO and the
Pendulum
Solution 2:
The angular frequency is a constant of the motion and by
definition is ω0 = 2π/T . For small angle the pendulum
approximates a simple harmonic oscillator with ω0 = (g/l)1/2.
The angular speed ω by definition is the magnitude of the
component of the angular velocity, ωz = dθ/dt. Note that
sometimes the symbol ω may be used for both quantities. This
is a result of the fact that for uniform circular motion, angular
frequency and angular speed are equal because the period T
= 2πR/v and the speed and angular speed are related by v =
Rω . Therefore T = 2π/ω. So for this special case ω = ω0.
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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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Concept Question: Physical
Pendulum
Answer 3. When the disk is fixed to the rod, an internal
torque will cause the disk to rotate about its center of mass.
When the pendulum reaches the bottom of its swing, the
decrease in potential energy will be result in an increase in
the rotational kinetic energy of both the rod and the disk and
the center of mass translation kinetic energy of the rod-disk
system. When the disk is mounted on the frictionless bearing
there is no internal torque that will make the disk start to
rotate about its center of mass when the pendulum is
released. Therefore when the pendulum reaches the bottom
of its swing, the same decrease in potential energy will be
transferred into a larger smaller in rotational kinetic energy of
just the rod since the disc is not rotating and a greater
increase in the center of mass translation kinetic energy of
the rod-disk system.
So when the disk bearings are frictionless, the center of mass
of the rod-disk system is traveling faster at the bottom of its
arc hence will take less time to complete one cycle and so the
period is shorter compared to the fixed disk.
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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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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 1
Solution 1. Because the
energy is greater than the
potential energy for values of
x that approach negative
infinity (on the left in the
above figure), the particle
can escape to infinity in the
negative x-direction with a
positive kinetic energy.
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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 2
Solution 5. Now the range
of motion for the particle is
limited to the regions in
which the kinetic energy is
either zero or positive, so the
particle is confined to the
regions where the potential
energy is less than the
energy. Hence the particle
periodically revisits a and b.
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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 3
Solution 6. Now the range of
motion for the particle is
limited to the regions in which
the kinetic energy is either
zero or positive, so the particle
is confined to either the region
around a and or the region
around b but since we do not
know where the particle has
started, we do not have
enough information to state
where the particle will be.
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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 4
Solution 3. Now the range of
motion for the particle is limited to
the regions in which the kinetic
energy is either zero or positive, so
the particle is confined to move
around the region surrounding a .
The motion will be periodic but not
simple harmonic motion because
the potential energy function is not
a quadratic function and only for
quadratic potential energy
functions will the motion be simple
harmonic. Hence the particle
oscillates around a.
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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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Concept Question: Energy Diagram 5
Solution 2. Now the particle
oscillates around the region
surrounding a. Since the energy
is so close to the minimum of
the potential energy, we can
approximate the potential
energy as a quadratic function
and hence the particle motion
approximates simple harmonic
motion.
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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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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
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

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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