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Today:
Tue Oct 27
• Hooke’s Law
• Assign 10 Friday
– Use examples
• Periodic Motion
• Mass-Spring System
Fapplied = kx
ω spring =
k
m
f spring
1
=
2π
Note: T, f, ω are Independent
k
m
Tspring
m
= 2π
k
of Amplitude!
Each Red dot represents a 1kg mass on a massless turntable. The blue
dot is 2kg. The inner circle has a radius half that of the outer circle.
If the masses start in Configuration A and move to Configuration B
while the turntable is spinning, how does the final angular speed of
the turntable compare to the initial? (assume no external torque)
(A)
(B)
1. Greater
2. Smaller
3. Same
LINITIAL= LFINAL
Moment of Inertia stays same
IA = (1kg)R2 +(1kg)R2 + (2kg)(½R)2
IB = (2kg)R2 +(1kg)(½R)2 + (1kg)(½R)2
IAωA = IB ωB
IA = IB ; ωA = ωB
A wheel of radius 0.30m rolls without slipping 15m along the
sidewalk. How many revolutions did the the wheel undergo?
(1) 0.72 rev
(5) 15 rev
(2) 6.2 rev
(6) 50 rev
(3) 8.0 rev
Rolling w/o slipping
x=rθ
15 m =0.3 θ
θ = 50 rad
50 rad (1 rev/2π rad) = 7.96 rev
(4) 9.7 rev
Or
Circumference = 2π r = 1.885m
Distance/Circumference = # rev
15m/ 1.885 m = 7.96 rev
Carousel Example
• Playground carousel
• As people redistribute, moment of inertia changes
• As moment of inertia changes, angular speed changes
A solid cylinder and a hoop roll down an incline. Both are the same
mass and radius. Which reaches the bottom of the incline first?
1. The cylinder
2. The hoop
3. Both at same time
"An object's moment of inertia determines how much it
resists rotational motion.
The cylinder has a smaller moment of inertia, so it’s easier
“to get going”.
Less energy goes into rolling the cylinder, so it has more
translational kinetic energy at the bottom.
Turns out that mass and radius don’t matter – solid cylinders
will always beat hoops.
Rolling Racers: four objects with same mass are placed on a ramp
and left to roll without slipping. Starting from rest, which one is
traveling the ramp length faster?
1. Spherical shell (I=2/3
MR2)
2. Solid sphere (I=2/5 MR2)
3. Hollow cylinder (I=MR2)
4. Solid cylinder (I=1/2 MR2)
The solid sphere has a smaller moment of inertia, so it’s easier “to
get going”.
https://www.youtube.com/watch?v=b44WbCS9xnc
What are the units for the spring constant, k?
1. N
2. m/N
3. N*m
4. N/m
5. kg*m
6. kg*N
F
k=
x
Repetitive Motion
• Examples?
• central force (circular motion)
• restoring force (back and forth)
• Oscillations
• How can we describe motion and properties of object?
• Simple Harmonic Motion/Oscillations (SHM/SHO)
• Why?
Spring
• Rubber band – simple spring
– Stretch it – pulls back
– More you stretch it – the harder it pulls back
– Some rubber bands harder to stretch than others
• When compress a spring – still pushes back
! Sect 16.1
! Sect 16.1
Springs - Hooke’s Law
• Robert Hooke (1635-1703)
Fapplied = kx
• x = position relative to equilibrium
has sign (direction)
• k is spring constant (units = N/m)
Cutnell & Johnson, Physics, 6th ed, Fig. 10.1
Applied Force & Spring Restoring Force
• Force Applied by you is
balanced by a Spring
Restoring Force
Fapplied = − Fspring
Fspring = −kx
Cutnell & Johnson, Physics, 6th ed, Fig. 10.5
Fspring
FApplied
Spring Constant of a Chromosome
What is the spring constant of this newt chromosome?
Fapplied = kx
Force
•
10 µm
Poirier & Marko, J. Muscle Res. Cell Mot., 23:409, 2002
Extension (µm)
F=kx
1 nN = k 1µm
k = 1x10-9 N / 1x10-6 m
k = 1x10-3 N/m
Spring Constants from Graphs: Which of
the curves represents the stiffest spring?
Force
3
2
1) 1
2) 2
3) 3
4) All the same
1
Displacement
Fapplied = kx
Stiffest spring, largest force for fixed displacement
Stiffest spring, largest spring constant
Spring constant is equal to slope of Force vs Displacement
Suppose you attach a mass to three identical springs. Compared to a
mass attached to just one spring, it will: (assume pull 1cm in each case)
1) Require a larger force to stretch three springs than one.
2) Require the same force to stretch three springs as it takes to stretch one.
3) Require less force to stretch three springs than one.
k1
k1
k1
k1
Free Body Diagram for block
FSPRING
FAPPLIED
FSPRINGS
FAPPLIED
Effective spring constant 3k. The 3 springs in parallel act as 1
single spring 3 times as stiff.
Hanging Masses
• Can apply a known force by hanging a mass from a spring.
• FAPPLIED = mg
• Can find spring constant FAPPLIED = kx
x
kx
mg
A mass of 0.20 kg hangs from a spring,
stretching it 0.10 m from it's equilibrium
position. What is the spring constant of the
spring?
1. 0.051 N/m
2. 0.50 N/m
3. 2.0 N/m
4. 19.6 N/m
Fapplied = kx
mg = kx
(0.2kg)(9.8 m/s2) = k (0.10m)
http://loncapa.phy.ohiou.edu/res/ohiou/physlets/shm/shm_mass_spring_novec.html
Each of the spring is shown in its unstretched
position, then stretched by hanging an object
from it. Rank the spring constants. The grid
allows a length reference.
1. kA > kB > kC > kD
2. kA = kB > kC = kD
3. kC = kD > kA = kB
4. kA = kD > kB = kC
5. kA > kB = kD > kC
6. kB = kC > kA = kD
Look at how far it stretches from equilibrium.
A is stiffer than B (kA>kB) since it stretches less than B with the
same mass.
A and D have the same spring constant. Twice as much force
(double mass) stretches D twice as far as A.
Thera-bands ™
Bands 30.5 cm long and 15.2 cm wide but come in different thicknesses
– Yellow: 0.045” = 1.14 mm; k = 71 N/m for Δx < 10 cm
–Red:
0.057” = 1.45 mm; k = 95 N/m for Δx < 15 cm
–Green:
0.069” = 1.75 mm; k = 120 N/m for Δx < 15 cm
– Blue:
0.085” = 2.16 mm; k = 140 N/m for Δx < 20 cm
For small stretching (Δx less than about 10 to 20 cm), the F vs Δx lines
are straight. For larger deflections, the lines are again straight, but the
spring constant for each band is smaller than it was for small stretching.
60
Fapplied (N)
45
Yellow
Red
Green
Blue
30
15
http://www.thera-band.com/
store/products.php?
0.00
0.18
0.35
0.53
0.70
ProductID=44
Dx (m)
M. Hollis & P. Fletcher-Cook, Practical Exercise Therapy 4th ed. p. 111
0
Frequency and Period
A mass is oscillating up and down on a spring.
It completes 10 complete oscillations in 15 seconds. What is the
frequency of oscillation?
(1) 0.67 Hz
(2) 1.5 Hz
(3) 10 Hz
(4) 15 Hz
time
T=
# cycles
1
f=
T
f = 10 cycles/15 s = 0.67 Hz
# cycles
f =
time
SHM
Position as function of time – sinusoidal function
motion ↔ graph
Note: Amplitude is the greatest displacement of the
object from equilibrium
Frequency – Physical Properties
! Sect 16.3
• How does the frequency of oscillation depend on physical properties of
system?
• Experiment – change masses and springs
• Math
Note: ω , T, f are
Independent of
Amplitude!
Remember: f = 1/T,
ω = 2π/T = 2πf
If you know one, you know the other.
Which is getting bigger and which is getting smaller can be
confusing.
A mass is hung from a spring and set into oscillation. It oscillates with
a given frequency f1. Now a second identical spring is also attached to
the mass (same k, same length). How does the new frequency
compare to the old?
1)The new frequency is double the old
2)The new frequency is sqrt(2) times the old
3)The new frequency is the same as the old
4)The new frequency is 1/sqrt(2) times the old
5)The new frequency is half the old
Double force with two springs. As if single spring of double strength.
(keff = 2k)
Stiffer spring, greater frequency.
1
f1 =
2π
k
m
1
f2 =
2π
2k
=
m
1
2
2π
( )
k
= 2 f1
m
f, T, ω, A:
• Make sure you are clear on definitions and relations between
them
• How would you measure or calculate each?
• Reference text definitions
• Units – text, slides
Frequency f – number of cycles per second – (1/s) or Hz
Period T – time for one complete cycle – (s) – T = 1/f
Angular frequency – also angular speed – 2πf – (rad/s)
Amplitude – maximum distance from equilibrium – (m)
# cycles
f =
time
time
T=
# cycles
Mass-Spring System - Horizontal
Simulation – Horizontal mass and spring
http://loncapa.phy.ohiou.edu/res/ohiou/physlets/shm/shm_mass_spring_novec.html
Mass and Spring – PRS 1
http://loncapa.phy.ohiou.edu/res/ohiou/physlets/shm/shm_mass_spring_prs_a1.html
Mass and Spring – PRS 2
http://loncapa.phy.ohiou.edu/res/ohiou/physlets/shm/shm_mass_spring_prs_v1.html
Full Simulation
http://loncapa.phy.ohiou.edu/res/ohiou/physlets/shm/shm_mass_spring_1.html
Will place links on LON-CAPA
Mass and Spring System
• Force greatest at extremes – where spring stretched or
compressed the most (zero at equilibrium)
• Note: Amplitude is the greatest displacement of the object from
equilibrium, (ie. the “Extremes”)
• F=ma or a = F/m, so where the force is biggest, the
acceleration is biggest.
• Object speeds up as it heads towards equilibrium (F and a in
direction of travel)
• As it goes through equilibrium, F and a change direction, so starts
to slow down.
• Max v at equilibrium, zero v at extremes.
The graph represents displacement of a mass in a mass-spring system. At
what point(s) is the magnitude of the force at a maximum?
(1) A
(6) A and I
(2) C
(7) A, E, and I
(3) E
(8) C and G
(4) G
(9) A,C,G, and
I
(5) I
The magnitude of the force is greatest at the extremes.
maximum acceleration at what point(s)? A, E, and I
minimum force & acceleration at what point(s)? C and G
minimum (zero) velocity at what point(s)? A, E, and I
maximum velocity at what point(s)? C and G
Describe Temporal (Time) Behavior
Compare to Circular motion
• Frequency (1/s) or Hz
• Period (s) = T = 1/f
• x(t),v(t), a(t)
θ = ωt
x = A cos(ωt )
Comparison of Circular Motion and Harmonic Motion http://
loncapa.phy.ohiou.edu/res/ohiou/physlets/shm/shm_circle_1.html
Position as a function of time
x = A cos(ωt )
A = Amplitude
ω = 2πf
USE RADIANS!
ω is angular speed or angular frequency
Velocity as function of time
v = −ωA sin(ωt )
Acceleration as function of time
2
a = −ω A cos(ωt )
Spring-Mass – Review formulas
k
m
ω spring =
f spring
1
=
2π
m
k T
spring = 2π
k
m
Note: T, f, ω are Independent of Amplitude!
ω (lowercase greek omega) is angular frequency or angular speed
xmax = A
x = A cos (ω t )
v = − Aω sin (ω t )
2
a = − Aω cos (ωt )
sin and cos range
(-1 to 1)
vmax = Aω
amax = Aω
2
Suggest making list of equations while reading for reference in class