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Spiral Physics
Applications
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Linear Oscillations
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Copyright © 1994-2000
Paul D’Alessandris
Spiral Physics
Rochester, NY 14623
All rights reserved. No part of this book may be reproduced or transmitted in any
form or by any means, electronic or mechanical, including photocopying, recording,
or any information storage and retrieval system without permission in writing from
the author.
This project was supported, in part, by the National Science Foundation. Opinions expressed are those of
the author and not necessarily those of the Foundation.
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Model Specifics
Linear Oscillations
In this section, we will investigate a new model that can be viewed as an extension, or
application, of the models previously studied. This application concerns the motion, of either
a particle or a rigid body, after it is displaced from its equilibrium position. (The equilibrium
position is the position at which the total force acting on the object is zero.) The model
hypothesizes the existence of a force that acts to return the object to its equilibrium position.
This force is termed the restoring force. Furthermore, the magnitude of this restoring force
will be restricted to be linearly dependent on the magnitude of the initial displacement from
equilibrium. This means that the farther the object is displaced from equilibrium, the larger
the magnitude of the force acting to return it to equilibrium1. When these two criteria are met,
we will show that the object’s position will oscillate about the location of its original
equilibrium position.
1
This is the first non-constant force we have examined. Not only does the restoring force change
magnitude as the distance of the object from its equilibrium position changes, it also changes direction
as the object oscillates from one side of its equilibrium position to the other.
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Linear Oscillations
Concepts and Principles
Imagine an object at rest at its equilibrium position, for example, a pendulum hanging straight
downward, an object dangling from an elastic cord, or a diving board perfectly horizontal.
When these objects are displaced from equilibrium, and released, they return toward their
equilibrium position. However, their trip back toward equilibrium involves repeated
oscillations about the location of equilibrium. If the initial disturbance is small enough, all of
these systems exhibit mathematically similar behavior. This chapter explores the similarities
in their oscillatory behavior.
We will analyze in detail the motion of a common oscillator, a mass attached to the end of a
spring, from both a work-energy perspective and a force perspective. We will investigate the
resulting motion without including the effects of friction (undamped). You should then be
able to generalize these results for other oscillatory systems. Before we begin, however, we
must learn more about springs.
The Force Exerted by a Spring
The force that a spring exerts on an object depends on the amount by which the spring is
compressed (or stretched). The more the spring is compressed, the larger the force it exerts on
the object. A common model is that the force exerted by a spring is directly proportional to
the amount of deformation of the spring, whether compression or extension. Deformation (s)
is defined to be the difference between the current length of the spring (L) and the equilibrium
length (L0):
s = L − L0
Thus, if a spring is compressed, L < L0, and s is negative. If
the spring is stretched, L > L0, and s is positive. If we define
the positive coordinate direction to point in the same
direction as positive deformation (stretch), then
k
Fspring = −ks
+x
with the proportionality constant, k, referred to as the spring
constant.
The negative sign assures that the force points in the correct direction. Notice that when the
spring is stretched, s is positive, and the spring exerts a force in the negative direction. When
the spring is compressed, s is negative, and the spring exerts a force in the positive direction.
In either case, the force acts to return the object to equilibrium.
4
Elastic Potential Energy
When an object interacts with an spring, or other elastic material, a common model is that the
spring reacts linearly, i.e., with a force directly proportional to the deformation of the spring.
It is possible to calculate the work done by the spring in general, and to rewrite the workenergy relation in such a way as to incorporate the effects of this work from the start. This is
referred to as constructing a potential energy function for the work done by the spring.
Imagine a spring of spring constant k, initially deformed by a distance si. It changes its
deformation, ultimately resulting in deformation sf. To calculate the work done by the spring
on the object causing the deformation:
W = F ∆r cos φ
Wdone by spring = ks r f − ri cos φ
Choosing a coordinate system in which s and r are interchangeable (the origin is located at the
point where the spring is undeformed and the positive direction for r is the same direction as
positive deformation (stretch)) results in:
Wdone by spring = ks s f − s i cos180
Wdone by spring = − ks (s f − s i )
(The direction of the force of the spring is always opposite to its deformation.) Now we come
to a problem. Which s (si or sf) do we use in the relation “-k s”?
Since the force of the spring depends linearly on s, it turns out that the correct choice is to use
the average value of the deformation,
s average =
1
(s f + si )
2
Thus,
1
Wdone by spring = − k ( (s f + s i ))(s f − s i )
2
1
2
2
Wdone by spring = − k ( s f − s i )
2
1
1 2
2
Wdone by spring = − ks f + ks i
2
2
The
1 2
ks terms are referred to as the elastic potential energy. Thus, the work done by a
2
spring can be thought of as changing the elastic potential energy in a system.
5
Let’s plug the above result into the work-energy relation:
1
2
mvi
2
1
2
mvi
2
1
2
mvi
2
1
2
mvi
2
1
2
mv f + mgh f
2
1
2
+ mghi + Σ F ∆r cos φ + W spring = mv f + mgh f
2
1
1 2 1
2
2
+ mghi + Σ F ∆r cos φ − ks f + ksi = mv f + mgh f
2
2
2
1 2
1
1
2
2
+ mghi + ks i + Σ F ∆r cos φ = mv f + mgh f + ks f
2
2
2
+ mghi + Σ F ∆r cos φ =
Therefore, this final relation:
1
1 2
1
1
2
2
2
mvi + mghi + ks i + Σ F ∆r cos φ = mv f + mgh f + ks f
2
2
2
2
can be used in place of the standard work-energy relation provided you do not include the
force of the spring a second time by calculating the work done by the spring. Basically, in this
relationship the spring is no longer thought of as a force that does work on objects but rather
as a source of potential energy.
Now we are ready to apply what we know about springs to the investigation of the motion of
a mass attached to the end of a spring.
Spring-Mass Oscillator
One of the simplest systems exhibiting oscillatory
behavior is the spring-mass system pictured at left. A
linear spring of constant k is attached to a mass m. The
equilibrium position of the spring-mass system is
denoted as the origin of a coordinate system.
k
m
+x
Now imagine the block is pulled to the right and let go. Hopefully you can convince yourself
that the block will oscillate back and forth. Let’s apply Newton’s Second Law at the instant
the mass is at its maximum distance from the origin. The only force acting on the mass in the
x-direction is the force of the spring.
ΣF = ma
Fspring = ma
− ks max = ma
6
Because of our choice of coordinate system, the stretch of the spring (s) is exactly equal to the
location of the block (x).
− ks max = ma
− kx max = ma
Since k and m are constants, the maximum magnitude of the acceleration must occur when
the mass is at its maximum position:
− kx max = ma
a max = −
k
x max
m
Thus, the maximum magnitude of the acceleration is directly proportional to the maximum
position of the mass and the quantity
k
. (The negative sign indicates that the direction of the
m
acceleration is always opposite to the sign of the position.)
Now let’s look at the same scenario using the work-energy relation. Let’s examine the motion
between the instant the mass is released from its maximum position and the instant the mass
passes through the equilibrium position.
1
1
1
1 2
2
2
2
mvi + mghi + ks i + Σ F ∆r cos φ = mv f + mgh f + ks f
2
2
2
2
1
1
2
0 + 0 + kx max + 0 = mv 2 + 0 + 0
2
2
1
1 2
2
kx max = mv
2
2
In the equation above, the position is at a maximum value and all other terms in the
relationship are constant. Therefore, the speed as the mass passes through equilibrium must be
at a maximum value.
1
1
2
2
kxmax = mvmax
2
2
k
v max =
x max
m
Thus, the maximum speed of the mass is directly proportional to the maximum position of the
mass and, like the maximum acceleration, depends on the quantity
7
k
.
m
Although we have discovered a great deal about the motion of the spring-mass system, we
have not yet found a way to determine the time it takes for the mass to undergo a single,
complete oscillation. The time for one complete cycle of the motion is termed the period, T,
and is an extremely important parameter for an oscillating system. To determine the period,
you will have to remember some something from a previous (forgotten?) math class.
You should remember that two common oscillating functions are the sine and cosine. For
example, the function,
x(t ) = x max cos(ωt + φ )
or equivalently the sine function, oscillate above and below zero in a manner which may
remind you of the motion of the spring-mass system.2 To refresh your memory,
xmax is the amplitude of the oscillation. The amplitude is the maximum displacement
of the object from equilibrium.
•
φ is the phase angle. The phase angle is used to adjust the function forward
or backward in time. For example, if the particle is at the origin at t = 0 s, φ
must equal +π/2 or -π/2 to ensure that the cosine function evaluates to zero
at t = 0 s. If the particle is at its maximum position at t = 0 s, then the phase
angle must be zero or π to ensure that the cosine function evaluates to +1 or
-1 at t = 0 s.
•
ω is the angular frequency of the oscillation.3
Note that the cosine function repeats itself when its argument increases by 2π. Thus, when
∆(ωt + φ ) = 2π
the function repeats. Since ω and φ are constant,
∆(ωt + φ ) = ω∆t
Therefore, the time interval when
ω∆t = 2π
is the time interval for one complete cycle of the oscillatory motion. The time for one
complete cycle of the motion is termed the period, T. Thus,
T=
2π
ω
Therefore, the physical significance of the angular frequency is that it is inversely
proportional to the period.
2
Sketch a position vs. time graph for the motion of the spring-mass system. Also, plot the function x(t)
= xmax cos (ωt + φ) on a graphing calculator for different values of ω and φ. You should see quite a
similarity between the physical motion of the spring-mass system and the cosine function.
3
Note that ω is not the angular velocity. The block is not rotating; it does not have an angular velocity.
8
Note that the angular frequency must have units of
1
k
. Also note that the quantity
(which
s
m
seems to be popping up everywhere) has units:
N kgm
m = s2 = 1
kg mkg s 2
This suggests that
ω=
k
m
Thus, not only do the maximum acceleration and maximum velocity of the mass depend on
the ratio of spring constant to mass, but so does the period.
In summary, through Newton’s Second Law, the work-energy relation, and a bit of
mathematical trickery, we can readily examine the motion of oscillating objects.
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Linear Oscillations
Analysis Tools
A Vertical Spring-Mass System
An 80 kg bungee jumper is about to step off of a platform high above a raging
river and plummet downward. The elastic bungee cord has an effective spring
constant of 35 N/m and is initially slack, although it begins to stretch the
moment the jumper steps off of the platform.
Let’s see what we can learn about the motion of the bungee jumper by applying Newton’s
Second Law and the work-energy relation.
Place the origin of the coordinate system at the equilibrium location of
the bungee jumper. Note that the initial position of the jumper is not the
equilibrium position. The equilibrium position is where the net force on
the bungee jumper is zero.
k
To find the equilibrium position:
ΣF = ma
m
+x
− Fgravity + Fspring = ma
− mg + ks equilibrium = m(0)
sequilibrium
Note that the force of the spring is directed upward at all times since the
spring will always be stretched downward.
mg
k
80(9.8)
=
35
= 22.4m
s equilibrium =
s equilibrium
s equilibrium
10
Thus, the spring is stretched by 22.4 m at the equilibrium position. Since the bungee jumper starts his
motion 22.4 m above equilibrium, his maximum position, xmax, is 22.4 m. We also know that the
bungee jumper will continue to fall until he reaches a point 22.4 m below equilibrium. Thus, the total
distance he falls is 44.8 m. (Hopefully the raging river is more than 44.8 m below the platform!)
From
k
m
ω=
35
80
ω = 0.661s −1
ω=
Therefore, the time for the jumper to return to the platform (one complete cycle of his motion) is:
T=
2π
ω
2π
0.661
T = 9.51s
T=
We can also determine the maximum acceleration of the jumper.
k
x max
m
35
= − (22.4)
80
= −9.8m / s 2
a max = −
a max
a max
This acceleration occurs when the jumper is at his maximum position, namely the instant he steps off
of the platform. (In addition, he feels a 9.8 m/s2 acceleration upward at the instant he’s at x = -22.4 m;
the bottom of his motion.)
What about the jumper’s maximum velocity? Before you simply use the result derived for the
horizontal spring-mass system, realize that unlike the horizontal case the gravitational potential energy
of the jumper is no longer zero, so that result may be inapplicable.
11
Examining the motion between the instant the jumper steps off of the platform and the instant the
jumper passes through the equilibrium position,
1
1 2
1
1
2
2
2
mvi + mghi + ks i + Σ F ∆r cos φ = mv f + mgh f + ks f
2
2
2
2
1
1
0 + (80)(9.8)(22.4) + 0 + 0 = (80)v 2 + 0 + (35)(22.4) 2
2
2
v max = 14.8m / s
Note, however, that the relationship derived for the horizontal spring-mass system
v max =
k
x max
m
35
(22.4)
80
= 14.8m / s
v max =
v max
gives the same result. This is not a fluke. The results derived for the horizontal spring-mass system are
also valid for the vertical spring-mass system!
Non-Oscillatory Scenarios involving Springs
A 0.15 kg ball is launched vertically upward by means of a spring-loaded
plunger, pulled back 8.0 cm and released. It requires a force of about 10 N to
push the plunger back 8.0 cm.
Although this situation does not involve oscillations, the tools we’ve
developed to deal with oscillations will allow us to analyze it as well.
+r
8 cm
First, we should determine the spring constant of the spring that
launches the ball. If it takes 10 N to push the plunger back 8 cm, the
force the spring is exerting back on the plunger is also 10 N.
Therefore,
Fspring = − ks
10 = − k (−0.08)
k = 125 N / m
12
Applying the work-energy relation between the instant the plunger is released and the instant
the plunger reaches its equilibrium position (and the ball is launched):
1
1 2
1
1
2
2
2
mvi + mghi + ks i + Σ F ∆r cos φ = mv f + mgh f + ks f
2
2
2
2
1
1
0 + (0.15)(9.8)(−0.08) + (125)(−0.08) 2 + 0 = (0.15)v 2 + 0 + 0
2
2
2
0.282 = 0.075v
v = 3.76m / s
The ball leaves the device at 3.76 m/s.
If you’d also like to know the maximum height reached by the ball, you could apply workenergy between the how high the ball the instant the plunger is released and the instant the
ball reaches its maximum height:
1
1 2
1
1
2
2
2
mvi + mghi + ks i + Σ F ∆r cos φ = mv f + mgh f + ks f
2
2
2
2
1
0 + (0.15)(9.8)(−0.08) + (125)(−0.08) 2 + 0 = 0 + (0.15)(9.8)h + 0
2
0.282 = 1.47h
h = 0.19m
The ball reaches a height of 19 cm above the top of the plunger.
The Physical Pendulum
A 0.4 kg meterstick is hung from a pin attached to one end. The meterstick is
displaced by 0.1 radian (5.70) from equilibrium (straight down) and released.
At left is a free-body diagram of the meterstick just before it is released. Since
the oscillatory behavior is occurring in the angular direction, let’s write
Newton’s second law in angular form:
Fpin, Y
Fpin, X
Στ = Iα
τ gravity = Iα
θ
−
FGravity
y
θ
x
13
L
(mg ) sin θ = Iα
2
A meterstick can be approximated as a thin rod, and the rotational inertia of a thin rod rotated
about one end is:
1
I = mL2
3
So
−
1
L
(mg ) sin θ = mL2α
2
3
Since the angle of displacement is small, we will use a common approximation for small
angles:
sin θ ≈ θ
(For the maximum angle attained by our meterstick, sin (0.1) = 0.0998)
Using this approximation, our equation becomes:
L
1
mgθ = mL2α
2
3
3g
−
θ =α
2L
−
This is exactly the same equation as the spring-mass system,
−
k
x=a
m
if
k
3g
is replaced by
m
2L
and
position (x) is replaced by angular position (θ).
Thus, all of the results for the spring-mass system can be used for the physical pendulum if
we simply replace every occurrence of
k
3g
with
and linear variables are replaced by
m
2L
angular variables.
14
Therefore,4
ω=
k
3g
⇒ω =
2L
m
ω=
3(9.8)
2(1)
ω = 3.83s −1
T=
T=
2π
ω
2π
3.83
T = 1.64 s
Notice that the angular frequency, and hence the period, do not depend on the mass of the
meterstick.
k
3g
x max ⇒ α max = − θ max
m
2L
3(9.8)
=−
(1)
2(1)
a max = −
α max
α max = −1.47rad / s 2
v max =
k
3g
θ max
x max ⇒ ω max =
m
2L
ω max =
3(9.8)
(1)
2(1)
ω max = 0.383rad / s
There are two different ω's in the relationships that follow. One is the angular velocity of the meterstick
(which varies as the meterstick oscillates), the other is the angular frequency of the meterstick (which is
a constant). I know they are the same symbol, but try to keep them straight by examining the context in
which they appear.
4
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Linear Oscillations
Activities
16
Sketch the indicated motion graphs for one complete cycle of the motions described below. Set the origin
of the coordinate system at the equilibrium position of the mass.
a. A mass is attached to a vertically-oriented spring. The mass is pulled a short distance below its
equilibrium position and released. Begin the graph the instant the mass is released. Assume air resistance
is so small that it can be ignored.
r
v
t
t
b. A pendulum is displaced from its equilibrium position and released. Begin the graph the instant the
pendulum is released. Assume air resistance is so small that it can be ignored.
θ
ω
t
t
17
Sketch the indicated motion graphs for one complete cycle of the motions described below. Set the origin
of the coordinate system at the equilibrium position of the mass.
a. A mass is attached to a vertically-oriented spring. The mass is pulled a short distance below its
equilibrium position and released. Begin the graph the instant the mass is released. Assume air resistance
is so large that it can not be ignored.
r
v
t
t
b. A pendulum is displaced from its equilibrium position and released. Begin the graph the instant the
pendulum is released. Assume air resistance is so large that it can not be ignored.
θ
ω
t
t
18
For each of the scenarios described below, indicate the amount of kinetic energy, gravitational potential
energy, and elastic energy in the system at each of the events listed. Use a consistent scale throughout both
motions.
a. A 75 kg bungee jumper steps off a platform high above a raging river and plummets downward. The
elastic bungee cord has an effective spring constant of 50 N/m and is initially slack, although it begins to
stretch the moment the jumper steps off of the platform. Set the lowest point of the jumper as the zero-point
of gravitational potential energy
KE
GE
EE
When the jumper first steps off the
platform.
KE
GE
EE
When the jumper has fallen exactly
halfway.
KE
GE
EE
When the jumper reaches his
lowest point.
b. A 75 kg bungee jumper steps off a platform high above a raging river and plummets downward. The
elastic bungee cord has an effective spring constant of 50 N/m and is initially slack, although it begins to
stretch the moment the jumper steps off of the platform. Set the platform as the zero-point of gravitational
potential energy
KE
GE
EE
When the jumper first steps off the
platform.
KE
GE
EE
When the jumper has fallen exactly
halfway.
19
KE
GE
EE
When the jumper reaches his
lowest point.
For each of the scenarios described below, indicate the amount of kinetic energy, gravitational potential
energy, and elastic energy in the system at each of the events listed. Use a consistent scale throughout each
motion. Set the initial position of the object as the zero-point of gravitational potential energy
a. A 0.27 kg toy car is held at rest against a 84 N/m compressed
spring. When released, the car travels around a 0.30 m high loop.
The car's speed at the top of the loop is 2.2 m/s. Assume friction is
so small that it can be ignored.
KE
GE
EE
When the car is first released.
KE
GE
EE
When the car is at the bottom of the
loop.
0.3 m
KE
GE
EE
When the car is at the top of the
loop.
b. A 0.10 kg pinball is launched into a pinball machine by means of a plunger, pulled back 8.0 cm and
released. The surface of the pinball machine is inclined at 130 from horizontal. Assume friction between the
ball and the surface of the pinball machine during the launch of the ball is so small it can be ignored.
KE
GE
EE
When the plunger is first released.
KE
GE
EE
When the ball leaves the plunger.
20
KE
GE
EE
When the ball reaches its maximum
height.
Six crates of different mass are hanging at rest from six springs. Each spring has a stiffness (k), a natural
length (L0), and a current length (L). The natural length of the spring is its length before the crate is hung
from it. The current length of the spring is its length when the crate is hung from it.
k
k
A
B
C
D
E
F
L0
L
20 N/m
0.4 m
0.8 m
20 N/m
0.3 m
0.6 m
10 N/m
0.4 m
0.8 m
10 N/m
0.3 m
1.2 m
30 N/m
0.4 m
0.6 m
30 N/m
0.2 m
0.4 m
a. Rank the force of each spring acting on each crate.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
b. Rank the mass of each crate.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
21
Six crates of different mass (m) are attached to springs of different stiffness (k). The masses are held in
place such that none of the springs are initially stretched. All springs are initially the same length. The
masses are released and the springs stretch.
k
m
A
B
C
D
E
F
m
k
5 kg
20 N/m
20 kg
5 N/m
10 kg
10 N/m
15 kg
20 N/m
5 kg
5 N/m
15 kg
10 N/m
a. Rank the maximum elongation of the spring in each system.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
b. Rank the maximum velocity of the crate in each system.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
c. Rank the maximum acceleration of the crate in each system.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your rankings:
22
Six identical mass crates are at rest on six springs. Each spring has a natural length (L0) and a current length
(L). The natural length of the spring is its length before the crate is placed on top of it. The current length of
the spring is its length when the crate is on top of it.
L0
A
B
C
D
E
F
L
0.8 m
0.4 m
0.6 m
0.3 m
0.8 m
0.2 m
1.2 m
0.3 m
0.6 m
0.4 m
0.4 m
0.2 m
a. Rank the force of the spring acting on each crate.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
b. Rank the stiffness of each spring.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
23
Six crates of different mass (m) are at rest on six springs. Each spring has a natural length (L0) and a current
length (L). The natural length of the spring is its length before the crate is placed on top of it. The current
length of the spring is its length when the crate is on top of it.
m
A
B
C
D
E
F
m
L0
L
10 kg
0.8 m
0.4 m
20 kg
0.6 m
0.3 m
10 kg
0.8 m
0.2 m
5 kg
1.2 m
0.3 m
5 kg
0.6 m
0.4 m
20 kg
0.4 m
0.2 m
a. Rank the stiffness of each spring.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
b. Rank the elastic potential energy stored in each spring.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
24
A pendulum of mass m is released from rest from an angle θ from vertical. All pendulums are the same
length.
θ
m
m
A
B
C
D
E
F
θ
2 kg
60°
1 kg
60°
4 kg
30°
8 kg
15°
2 kg
30°
3 kg
45°
Rank the maximum speed of each pendulum.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
25
A pendulum of mass m is moving at velocity v as it passes through the vertical. All pendulums are the same
length.
v
m
A
B
C
D
E
F
m
v
2 kg
4 m/s
1 kg
4 m/s
4 kg
2 m/s
8 kg
1 m/s
2 kg
2 m/s
3 kg
3 m/s
Rank the maximum angle from vertical reached by the pendulum.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
26
A 75 kg bungee jumper is about to step off of a platform high above a raging river and plummet downward.
The elastic bungee cord has an effective spring constant of 50 N/m and is initially slack, although it begins
to stretch the moment the jumper steps off of the platform.
a. How far does the bungee jumper fall?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
b. What is the maximum speed of the bungee jumper?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
27
A 75 kg bungee jumper is about to step off of a platform 65 m above a raging river and plummet
downward. He hopes to get just the top of his head wet. The elastic bungee cord acts as a linear spring and
is initially slack, although it begins to stretch the moment the jumper steps off of the platform.
a. What is the spring constant of the bungee cord?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
b. How long does it take the jumper to reach the bottom of his motion?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
28
A 45 kg bungee jumper is about to step off of a platform 65 m above a raging river and plummet
downward. She hopes to get just the top of her head wet. The elastic bungee cord acts as a linear spring
and is initially slack. The cord does not begin to stretch until the jumper has fallen 10 m.
a. What is the spring constant of the bungee cord?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
b. What is the maximum speed of the bungee jumper?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
29
A 55 kg bungee jumper is about to step off of a platform 45 m above a raging river and plummet
downward. She hopes to get just the top of her head wet. The elastic bungee cord acts as a linear spring
with spring constant 30 N/m and is initially slack, but does not immediately begin to stretch when she steps
off the platform.
a. How far can she safely fall before the bungee cord must begin to stretch?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
b. How fast is she moving when she hits the river if she falls 15 m before the bungee cord begins to stretch?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
30
A 70 kg skier is zooming down a mountain out of control when he decides to veer off the trail into the
Safety-Spring. The Safety-Spring is a 900 N/m spring into which the skier can collide and be brought to rest
much more gently than if he hit a tree. The skier is traveling at 20 m/s when he pulls off onto a horizontal
path and collides with and sticks to the spring. Assume friction is so small that it can be ignored.
a. What is the maximum force that acts on the skier?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
b. How long does it take for the Safety-Spring to stop the skier’s motion?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
31
A skier is zooming down a mountain out of control when she decides to veer off the trail into the new and
improved Safety-Spring. The new and improved Safety-Spring is a 900 N/m spring into which the skier can
collide and be brought to rest much more gently than if she hit a tree. The skier is traveling at 25 m/s when
she pulls off onto a horizontal path and collides with and sticks to the spring. She slides 13 m before
coming (momentarily) to rest. Assume sliding friction is so small that it can be ignored.
a. What is the skier’s mass?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
b. What is the maximum force that acts on the skier?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
32
A 0.10 kg pinball is launched into a pinball machine by means of a plunger, pulled back 8.0 cm and
released. The surface of the pinball machine is inclined at 130 from horizontal. It requires a force of 12 N
to pull the plunger back 5.0 cm. Assume friction between the ball and the surface of the pinball machine
during the launch of the ball is so small it can be ignored.
a. What is the speed of the ball as it leaves the plunger?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
b. How far does the ball travel after leaving the plunger, assuming friction is small enough to be ignored?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
33
At a UPS distribution center, a 40 kg crate is sliding down an 8° ramp at 3 m/s. At the bottom of the ramp,
8 m away, is a 150 N/m spring designed to bring the crate to rest. The coefficient of friction between the
crate and the ramp is (0.2, 0.1).
a. What is the speed of the crate when it hits the spring?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
b. How far does the spring compress before bringing the crate to rest?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
34
A 0.27 kg toy car is held at rest against a 84 N/m compressed
spring. When released, the car travels around a 0.30 m high loop.
The car's speed at the top of the loop is 2.2 m/s. Assume friction is
so small that it can be ignored.
a. What is the initial compression of the launcher?
Mathematical Analysis
Free-Body Diagram
Event 1:
Event 2:
b. With what speed does the cart enter the loop?
Free-Body Diagram
Mathematical Analysis
Event 1:
Event 2:
35
0.3 m
Stranded on a desert island with only a meterstick and a nail, a physicist builds a pendulum clock by nailing one end
of the meterstick to a tree branch. He then displaces the meterstick by 5° and lets it go. Assume friction in the mount
is so small that it can be ignored.
Free-Body Diagram
Mathematical Analysis
Questions
What is the period of the pendulum clock?
What is the maximum speed of the bottom edge of the meterstick?
If the meterstick is cut in half, how will the period change?
36
Stranded on a desert island with only a 2 m long string, a 0.5 kg ball with a convenient hook in the top, a nail, and a
hammer, a physicist builds a pendulum clock by hanging the ball from the string and mounting the “clock” on a
tree branch. She then displaces the “clock” by 5° and lets it go. Assume friction in the mount is so small that it can
be ignored.
Free-Body Diagram
Mathematical Analysis
Questions
What is the period of the pendulum clock?
What is the maximum velocity of the ball?
37