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Approximate Grade Cutoffs Ø 75 -‐ 100 A Ø 55 – 74 B Ø 35 – 54 C Ø 25 – 34 D Ø 0 – 24 F
You can pick them up a/er class today or Wednesday T. S%egler 12/1/2014 Texas A&M University
Chapter 14
T. S%egler 12/1/2014 Texas A&M University
Today • Simple pendulums • Physical pendulums • Ch 14 Examples
Start Ch 15 (if we get to it)
Last Time
• Simple harmonic mo%on of a mass-‐spring system
• Differen%al equa%ons and SHM
• Oscilla%ons in terms of amplitude, period, frequency and angular frequency
• Simple harmonic mo%on using energy
To double the total energy of a mass-‐spring system oscilla%ng in simple harmonic mo%on, the amplitude must increase by a factor of
A. 4.
B.
C. 2.
D.
E.
=2 1.414.
=4 2 1.189.
=2 2 2.828.
0
02
0
( ) cos( )( ) sin( )
( ) cos( )
x t A tv t A ta t A t
ω φ
ω ω φ
ω ω φ
= +
= − +
= − +
21
pend
ff T
mgd gI L
ω π
ω
=
=
= =
12kA2 = 1
2mv2 + 1
2kx2 = const.
Clicker QuesEon Review from last Monday
T. S%egler 12/1/2014 Texas A&M University
SHM and Pendulums
• For the physical pendulum the CoM is a distance d from the pivot, then the torque at an angle θ is:
! = !d(mgsin" )
• Using the small angle approxima%on we obtain:
! = !dmg" so I! ! "dmg"
• Comparing to the SHM equa%on again gives:
2
2
1 22
d mgddt I
mgd mgd If TI I mgd
θα θ
ω ππ
⎛ ⎞= = −⎜ ⎟⎝ ⎠
⇒ = ⇔ = ⇔ =
T. S%egler 12/1/2014 Texas A&M University
SHM and Pendulums
• The simple pendulum is a special case of the physical pendulum. • All the mass is located at a point aTached to the end of string of length L.
Using this d = L and I = mL2
2
mgd mgL gI mL L
ω = = =
! =gL! f = 1
2"gL!T = 2" L
g
T. S%egler 12/1/2014 Texas A&M University
Wri%ng down Newton's Second Law in the x direc%on for some system results in the equa%on on the right. What is the oscilla%on frequency of this system?
a2 d2xdt2
= !b2xa) a/b b) a*b c) b/a
Prelecture: Physical Pendula Problem 1
T. S%egler 12/1/2014 Texas A&M University
Fx =max
Fx =md 2xdt2
d 2xdt2
+! 2x = 0
Prelecture: Physical Pendula Problem 2
A uniform s%ck and a mass on a string are used to make two pendula that have the same length. Which one swings with the longer period?
a) the mass on the string b) the s%ck c) can't tell without knowing how the masses compare
T. S%egler 12/1/2014 Texas A&M University
! =mgdI
=mgLmL2
=gL
T = 2!"
=4! 2Lg
! =mgdI
=mgL13mL2
=3gL
T = 2!"
=4! 2L3g
(a)
(b)
Checkpoint: Physical Pendula Problem 1
A simple pendulum is used as the %ming element in a clock as shown. An adjustment screw is used to make the pendulum shorter (longer) by moving the weight up (down) along the sha_ that connects it to the pivot. If the clock is running too fast, the weight needs to be moved a)Up b)Down
T. S%egler 12/1/2014 Texas A&M University
T = 2! Lg
Checkpoint: Physical Pendula Problem 2
A torsion pendulum is used as the %ming element in a clock as shown. The speed of the clock is adjusted by changing the distance of two small disks from the rota%on axis of the pendulum. If we adjust the disks so that they are closer to the rota%on axis, the clock runs a) faster b) slower
! ="I
f = 12#
"I
T = 2# I"
Angular SHM: κ = torsion constant
T. S%egler 12/1/2014 Texas A&M University
Checkpoint: Physical Pendula Problem 3 Consider the two pendula shown above. In Case 1 a s%ck of mass M is pivoted at one end and used as a pendulum. In Case 2 a point par%cle of mass M is aTached to the center of the same s%ck. In which case is the period of the pendulum the longest? a) Case 1 b) Case 2 c) Same
T. S%egler 12/1/2014 Texas A&M University
! =mgdI
T = 2!"
= 2! Imgd
T1 = 2!
13mL2
mg L2
= 2! 2L3g
T2 = 2!
13mL2 +m L
2!
"#
$
%&
2!
"
##
$
%
&&
2mg L2
= 2! 7L12g
Damped OscillaEons
• Real-world systems have some dissipative forces that decrease the amplitude.
• The decrease in amplitude is called damping and the motion is called damped oscillation.
• The figure at the right illustrates an oscillator with a small amount of damping.
• The mechanical energy of a damped oscillator decreases continuously.
• When , the system is critically damped and if b is larger than this it is overdamped
• A critically or overdamped oscillator returns to equilibrium without oscillating.
b = 2 km
T. S%egler 12/1/2014 Texas A&M University
Example Pendulum on Mars (14.48)
A certain simple pendulum has a period on the earth of 1.60 s. What is its period on the surface of Mars where gravity is gmars = 3.71 m/s2?
0
02
0
( ) cos( )( ) sin( )
( ) cos( )
x t A tv t A ta t A t
ω φ
ω ω φ
ω ω φ
= +
= − +
= − +
21
pend
ff T
mgd gI L
ω π
ω
=
=
= =
T. S%egler 12/1/2014 Texas A&M University
A simple pendulum consists of a point mass suspended by a massless, unstretchable string.
If the mass is doubled while the length of the string remains the same, the period of the pendulum
A. becomes 4 %mes greater.
B. becomes twice as great.
C. becomes greater by a factor of .
D. remains unchanged.
E. decreases.
2
Clicker QuesEon
0
02
0
( ) cos( )( ) sin( )
( ) cos( )
x t A tv t A ta t A t
ω φ
ω ω φ
ω ω φ
= +
= − +
= − +
21
pend
ff T
mgd gI L
ω π
ω
=
=
= =
12kA2 = 1
2mv2 + 1
2kx2 = const.
T. S%egler 12/1/2014 Texas A&M University
Example Physical pendulum (small angle approxima%on)(14.54)
A 1.80 kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for small angle oscilla%ons is 0.940 s. a) What is the moment of iner%a of the wrench about an axis through the pivot?
b) If the wrench is ini%ally displaced 0.400 rad from its equilibrium posi%on, what is the angular speed of the wrench as it passes through the equilibrium posi%on?
0
02
0
( ) cos( )( ) sin( )
( ) cos( )
x t A tv t A ta t A t
ω φ
ω ω φ
ω ω φ
= +
= − +
= − +
21
pend
ff T
mgd gI L
ω π
ω
=
=
= =
T. S%egler 12/1/2014 Texas A&M University
Example Physical pendulum (14.57)
The two pendulums shown each consist of a uniform solid ball of mass M supported by a rigid massless rod, but the ball for pendulum A is very small compared to that of pendulum B. Find the period of each pendulum for small displacements. Which ball takes longer to complete a swing?
0
02
0
( ) cos( )( ) sin( )
( ) cos( )
x t A tv t A ta t A t
ω φ
ω ω φ
ω ω φ
= +
= − +
= − +
21
pend
ff T
mgd gI L
ω π
ω
=
=
= =
T. S%egler 12/1/2014 Texas A&M University
Example Pendulums and collisions(14.95) In the figure the upper ball is released from rest, collides with the sta%onary lower ball, and s%cks to it. The strings are both 50.0 cm long. The upper ball has mass 2.00 kg, and ini%ally 10 cm higher than the lower ball (mass 3.00 kg). Find the frequency and maximum angular displacement of the mo%on a_er the collision.
0
02
0
( ) cos( )( ) sin( )
( ) cos( )
x t A tv t A ta t A t
ω φ
ω ω φ
ω ω φ
= +
= − +
= − +
21
pend
ff T
mgd gI L
ω π
ω
=
=
= =
T. S%egler 12/1/2014 Texas A&M University
Example Energy and momentum in SHM
A block of mass M is aTached to a horizontal spring with spring constant k and is moving in SHM. As it passes through its equilibrium point a lump of puTy of mass m is dropped from a small height and s%ck to it. Find the new amplitude and period.
T. S%egler 12/1/2014 Texas A&M University
21
pend
ff T
mgd gI L
ω π
ω
=
=
= =
12kA2 = 1
2mv2 + 1
2kx2 = const.