Transcript Springs and Hooke`s Law Lecture

Springs, Hooke’s Law, Work,
and Energy
Chapter 4 in Book
3/27/2016
Dr. Sasho MacKenzie - HK 376
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Objects that behave like springs
•Pole vault pole
•Golf club shaft
•Tennis ball
•Tennis racquet frame and strings
•Hockey stick
•A rubber band
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The Purpose of a Spring
• Springs can be power amplifiers. Springs can
return energy faster than the energy was stored
in the spring.
• Power is the rate at which work is done. P = W/t
• Work done on an object changes the energy of
the object by an amount equal to the work done.
• Therefore, power is the rate at which energy is
changing.
• Power can also be calculated by multiplying the
instantaneous net force acting on an object by
the object’s instantaneous velocity. P = Fv
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Pole Vault Example: Store
Energy
1. The vaulter applies force to the
pole which causes the pole to
bend
2. The more the pole bends, the
more force is required to add
additional bend.
3. The bend results in the storage
of strain energy in the pole.
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Pole Vault: Return Energy
4. The vaulter then maneuvers into
a position in which she does not
supply enough resistance to
maintain the bend in the pole.
5. The pole then releases the stored
strain energy in the form of
kinetic energy to the vaulter.
6. The correct timing of the return
of energy from the pole and the
fact that the energy is returned at
a faster rate, enables the vaulter
to jump higher.
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Hooke’s Law
• If a ‘spring’ is bent, stretched or compressed
from its equilibrium position, then it will exert a
restoring force proportional to the amount it is
bent, stretched or compressed.
• Fs = kx (Hooke’s Law)
• Related to Hooke’s Law, the work required to
bend a spring is also a function of k and x.
• Ws = (½ k)x2 = amount of stored strain energy
– K is the stiffness of the spring. Bigger means Stiffer
– x is the amount the spring is bent from equilibrium
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Based on Hooke’s Law
• A force is required to bend a spring
• The more force applied the bigger the bend
• The more bend, the more strain energy stored
Relationship between F and x
slope of the line = k
Force (N)
Area under curve is
the amount of stored
strain energy
x (m)
Stored strain energy = ½ k x2
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Kinetic Energy
• When the strain energy stored in a spring is
released, it is converted into kinetic energy.
• The kinetic energy of an object is determined
from an object’s mass and velocity. KE = ½ mv2
• A 2 kg object moving at 3 m/s has a kinetic
energy of…KE = ½(2)(3)2 = 9 J
• Mass doesn’t change. This means that when
strain energy is released, either the end of the
spring or an object attached to it undergoes an
increase in velocity.
• The power is amplified.
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The Work-Energy Relationship
• The work done on an object is equal to the
net average force applied to the object
multiplied by the distance over which the
force acted. W = Fd
• This means that no matter how long or how
hard you apply a force to an object, no
work is done if that object doesn’t move.
• Mechanical work is different from
physiological work.
• 100 J of work are done if a 100 N net force
moves an object 1 m.
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Total Energy = KE + PE
• An object’s total energy is calculated by adding
its current kinetic (KE) and potential (PE)
energies.
• Potential energy is dependent upon an object’s
current displacement from some reference
point and a potential average force that will act
on the object if it moves from its current position
to that reference point.
• An example of PE is an object at a certain height
(h) above the ground. PE = mgh
• A 4 kg object 2 m above the ground has
PE = (9.81)(4)(2) = 78.5 J of potential energy.
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Total Energy = KE + PE
• If the object is displaced from it’s current
point to the reference point by that potential
force, then it’s kinetic energy will increase by
the amount of potential energy the object
initially possessed.
• A 4 kg object, initially at rest, that falls 2 m
will have a KE = PE = (9.81)(4)(2) = 78.5 J.
• Since KE = ½ mv2,…v = sqrt(2KE/m)
• Therefore, the object will have a velocity of…
v = sqrt[ 2(78.5)/4] = -6.3 m/s when it lands.
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Potential Energy in a Spring
• A second example of PE is stored strain energy.
• Earlier it was shown that the strain energy
stored in a linear spring is… SE = (½ k)x2, or
½Fx
• In this case, the F stands for the amount of force
required to deform the spring by its current
displacement x.
• The ½ is required for this PE equation because
over the range of displacement, the force will fall
to zero. Thus ½F represents the average force
applied over the displacement.
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Strain PE vs. Gravitational PE
Strain
Force (N)
0
½Fx
Gravitational
Fx
or
mgh
0
Force (N)
“mg”
x (m)
x (m) “h”
• The area under the curve on the left equals the
energy stored in a linear spring, or the amount of
work required to deform the spring.
• The area under the curve on the right equals the
potential energy due to the constant force of gravity
(mg), or the work required to lift an object x m.
• Note that one area is square and the other triangular.
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Pole Vault Example
• Yelena Isinbayeva (1.74m/65kg)
deflects her 4.5m pole to 70% of
it’s full length. The pole has a
bending stiffness of 1000 N/m.
At this point in the vault, her
vertical velocity is 3 m/s and she
is 2.5 m above the ground.
1. How much strain energy is
stored in the pole?
2. What is her potential energy due
to gravity?
3. How much kinetic energy does
she have in the vertical
direction?
4. What will be her peak height in
the vault?
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Arampatzis et al., 2004
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Arampatzis et al., 2004
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Example: Slingshot
• Jimmy loads a 2 kg water balloon into his giant
slingshot which has a stiffness of 400 N/m. Jimmy
stretches the slingshot 1.5 m from equilibrium and
slings the balloon straight up into the air. If no
strain energy is lost to heat or sound, how high will
the balloon fly relative to the equilibrium point?
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A Tennis Serve
•
During a tennis serve there are 3
springs in action
1. Frame
2. Ball
3. Strings
•
Each spring has a specific stiffness
(k) and therefore a different ability to
act as a power amplifier.
•
The frame, the ball, and the strings all
store and release energy differently
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The Strings
Stiff strings
Force
Compliant strings
Load
x
The area under each line, is the amount of strain energy
stored in the strings. You can see that less stiff springs (in this
case strings) can store more strain energy for a given force.
However, the ball also has the ability to store and release
strain energy. Perhaps the ball departs the strings before the
ball has returned to equilibrium.
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Ball leaves before it (the ball)
returns to equilibrium
Force
This graph represents
the ball only
Load
x
This strain energy will
not increase the ball’s
velocity.
Ball leaves at this displacement
This must be considered when the strings, frame and ball
act as springs simultaneously during contact in a serve.
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Frame
Strings
Force
Ball: a player cannot alter
the ball stiffness
x
The frame of a tennis racquet is a poor spring. Therefore, it is
better to have the strings and ball performing most of the power
amplification. Thus the frame should be the stiffest element.
Why would a
good player still
want stiff strings?
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Diving Board
1. How would you determine the actual stiffness
of a diving board?
2. How does the stiffness of a diving board relate
to the strain energy that can be stored
temporarily in the board?
3. Sketch a vertical force-time profile for the
force that acts on a diver’s feet during his
time of contact with the board. Sketch the
corresponding vertical force-time profile for
the force from his feet pressing on the board.
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Dr. Sasho MacKenzie - HK 376
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(Cross, 2000: Sport Eng)
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