Transcript Physics 131: Lecture 14 Notes

Physics 151: Lecture 32
Today’s Agenda

Topics
The Pendulum – Ch. 15
Potential energy and SHM
Physics 151: Lecture 32, Pg 1
See text: 15.1 to 15.3
Simple Harmonic Motion
Review


The most general solution is x = Acos(t + )
where A = amplitude
 = frequency
 = phase constant
k
m
k
For a mass on a spring  
m
0
The frequency does not depend on the amplitude !!!
x

The oscillation occurs around the equilibrium point where the force
is zero!

Energy is a constant, it transfers between potential and kinetic.
Physics 151: Lecture 32, Pg 2
Lecture 32, Act 1
Simple Harmonic Motion

You have landed your spaceship on the moon and want to
determine the acceleration due to gravity using a simple
pendulum of length 1.0 m. If he period of the pendulum is
5.0 s what is the value of g on the moon ?
a)
b)
c)
d)
e)
1.3 m/s2.
1.6 m/s2.
0.80 m/s2.
0.63 m/s2.
2.4 m/s2.
Physics 151: Lecture 32, Pg 3
See text: 15.5
General Physical Pendulum


Suppose we have some arbitrarily shaped
solid of mass M hung on a fixed axis, that we
know where the CM is located and what the
moment of inertia I about the axis is.
The torque about the rotation (z) axis for
small  is (sin    )
d 2
 MgR  I
 = -Mgd  -MgR
dt 2

z-axis
R

xCM

d
d 
2
dt 2
  2 
where

MgR
I
Mg
 = 0 cos(t + )
Physics 151: Lecture 32, Pg 4
Lecture 32, Act 2
Physical Pendulum

A pendulum is made by hanging a thin hoola-hoop of
diameter D on a small nail.
What is the angular frequency of oscillation of the hoop
for small displacements ? (ICM = mR2 for a hoop)
(a)

g
D
(b)

2g
D
(c)

g
2D
pivot (nail)
D
Physics 151: Lecture 32, Pg 5
Lecture 32, Act 2
Solution

The angular frequency of oscillation of the hoop for small
displacements will be given by   mgR
I
Use parallel axis theorem: I = Icm + mR2
= mR2 + mR2 = 2mR2
pivot (nail)

mgR
g
g


2R
D
2 mR 2
So

g
D
cm
x R
m
Physics 151: Lecture 32, Pg 6
See text: 15.5
Torsion Pendulum


Consider an object suspended by a
wire attached at its CM. The wire
defines the rotation axis, and the
moment of inertia I about this axis is
known.
The wire acts like a “rotational
spring”.
When the object is rotated, the
wire is twisted. This produces a
torque that opposes the rotation.
In analogy with a spring, the
torque produced is proportional to
the displacement:  = -k
See figure 13.15
wire


I
Physics 151: Lecture 32, Pg 7
See text: 15.4
Torsion Pendulum...

Since  = -k  = Ibecomes
wire
k  I
d 
2
dt
2
d 2
dt 2


  
2
where

k
I
I
Similar to “mass on spring”, except
I has taken the place of m (no surprise)
See figure 13.15
Physics 151: Lecture 32, Pg 8
Lecture 32, Act 3
Period

All of the following pedulum bobs have the same
mass. Which pendulum rotates the fastest, i.e.
has the smallest period? (The wires are identical)
R
R
R
R
A)
B)
C)
D)
Physics 151: Lecture 32, Pg 9
Lecture 32, Act 5
Period

Check each case.
k
5k
 sphere 

I sphere
2MR 2
k
k
 hoop 

I hoop
MR 2 The biggest is (D),
the smallest
k
2k
 disk 

moment of inertia
I disk
MR 2
A)
B)
C)
 stick 
D)
k
I stick
12k
3k


2
MR 2
M 2 R 
R
Physics 151: Lecture 32, Pg 10
See text: 15.3
Energy in SHM

For both the spring and the pendulum, we can derive the
SHM solution using energy conservation.

The total energy (K + U) of a
system undergoing SMH will
always be constant!
U
K

This is not surprising since there
are only conservative forces
present, hence energy is conserved.
E
-A
0
U
A
s
Animation
Physics 151: Lecture 32, Pg 11
See text: Fig. 15.6
SHM and quadratic potentials



SHM will occur whenever the potential is quadratic.
Generally, this will not be the case:
For example, the potential between
H atoms in an H2 molecule looks
U
something like this:
K
E
U
-A
0
U
A
x
x
Physics 151: Lecture 32, Pg 12
See text: Fig. 15.6
SHM and quadratic potentials...
However, if we do a Taylor expansion of this function about the minimum,
we find that for small
displacements, the potential
IS quadratic:
U
U(x) = U(x0 ) + U(x0 ) (x- x0 )
1
+ U (x0 ) (x- x0 )2+....
2
U
x0
U(x) = 0 (since x0 is minimum
of potential)
x = x - x0 and U(x0 ) = 0
1
Then U(x) = U (x0 ) x 2
2
Define
x
x
Physics 151: Lecture 32, Pg 13
See text: Fig. 15.6
SHM and quadratic potentials...
1
U(x) = U (x0) x 2
2
Let k = U (x0)
U
U
x0
Then:
x
1
U(x) = k x 2
2
x
SHM potential !!
Physics 151: Lecture 32, Pg 14
Recap of today’s lecture

Chapter 15 – Pendula
Simple Pendulum
Physical Pendulum
Torsional Pendulum

Next time:
Damped and Driven Oscillations
Physics 151: Lecture 32, Pg 15