6/06
Elastic Energy Conservation
About this lab
Coherent bodies are held together, atom by atom or molecule by
molecule, by fundamental forces, mainly electrostatic. (In certain special materials,
magnetism plays a significant role.)
If stress (stretching, compression, twisting etc.) is applied to a macroscopic body
consisting of very large numbers of such small entities, the separation between neighbors
changes very slightly in response. The applied external stress has done work is distorting
the system. If the stress is not too large, this work can be equated to an “elastic” potential
energy, which can be reclaimed when the stress removed and the system relaxes toward its
original dimensions. The system frequently oscillates around its original equilibrium
position. If (as is usual) there is some energy loss during the oscillations, then the
oscillation amplitude damps out. An example is the flexing and recovery of an airplane
wing due to sudden forces from a gust of wind.
But, if the applied stress was too large, bonds are broken – the system is distorted beyond
its elastic limit, and the energy is no longer recoverable. This may occur partially, in
limited portions of the system.
The external force necessary to increase distortion around an equilibrium condition
increases linearly with small distortion. The work done, starting from zero distortion, is
the integral of the work times the distortion distance. It is, thus, quadratic – and the
stored elastic potential energy varies as the square of the distortion. This differs from the
stored gravitational energy arising from a small variation in height near the earth's surface,
which varies only in proportion to the height change.
Elastic deformation and potential energy is thus not at all fundamental, but arises from
fundamental electric forces. Nevertheless, the elastic characteristics of macroscopic
systems are of great practical importance in design of buildings, automobiles, medical
instrumentation, stringed instruments etc.
References: Cutnell and Johnson, 6th: Section 5.7, Section 6.5, Sections 10.3, .7, .8 .
Apparatus: Catapult launcher, track and loop, projectile (payload1), “hot wheels” car
(payload2), meter stick, post and mounting rod, balance, Vernier “Graphical Analysis”
software, yellow electrical tape (projectile range marker).
6/06
Elastic Energy Conservation
Experimental outline
Launch vehicle and stage
Figure 1
Top view of launch platform and (orange) launch vehicle (0.00790 kg) in
release position, showing the four spring latch positions.
Measure the spring stretch (distances vehicle moves from release position to latched
position) x1 to x4 for the four latch positions (four launch energy options).
The elastic potential energies E1 to E4 will be ½ k x 2 . Corresponding kinetic energy
is shared between the launch vehicle and the payload in proportion to their masses,
since they will have the same velocity, just before separation on vehicle arrest.
6/06
Elastic Energy Conservation
Figure 2
Bottom view of launch vehicle and platform. Leave assembled, remove
rubber band spring and determine spring constant k. Reassemble.
Part A: Experimental parameters
Measure and record all masses. Measure the loop radius R loop = D loop / 2 .
Measure four values x i of the distance between release position of launch vehicle
and the four vehicle latch points: x 1 to x 4 (meters). Assume that the rubber band is
unstretched when the launch vehicle is at the payload release point. You can thus
take the elastic PE i release for each notch position to be ½ k x i 2. (We will assume
there is negligible elastic potential energy remaining at release, though there is a little
tension in the band at the release position. However, the band goes slack at slightly
shorter length.)
6/06
Elastic Energy Conservation
Remove the rubber band. Characterize elastic properties of the doubled launcher rubber
band (i.e. determine spring constant k for doubled rubber band). Hang various weights
and measure corresponding lengths from support rod. Enter these in the Graphical
Analsis file in your experiment folder and plot force (Newtons) vs. length (meters). Fit a
straight line to to linear portion. The slope is the elastic constant k.
6/06
Elastic Energy Conservation
Figure 3
Determination of elastic constant k of elastic launch spring. Note
doubling, as in launch apparatus assembly.
6/06
Elastic Energy Conservation
Enter into Graphical Analysis program the distance (meters) from horizontal rod vs.
force (Newtons) = mg. Fit linear portion.
Calculate four elastic potential energies PE elastic 1 to PE elastic 4 , for the four latch
lengths. PE elastic = ½ k x 2 .
http://physics.syr.edu/courses/mra/Mar97/spring/osc36.html
Figure 4
Quadratic dependence of stored elastic energy on stretch from rest
position. The force increases linearly with distance from the equilibrium length.
More work is required per additional millimeter of stretch The integral of the area
under the elastic force curve is quadratic, and the elastic potential energy of an ideal
spring is:
PE elastic = ½ k x 2
where x is the spring length less the unstretched spring length.
6/06
Elastic Energy Conservation
This contrasts with gravitational potential energy near the earth's surface, where
the work required to lift an additional millimeter is independent of height. The
integral of the area under the gravitational force curve is linear, and the
gravitational potential energy (near earth approximation) is:
PE gravitational = mgh.
Figure 5
Determination of force constant k = 65.1 Newtons/meter (slope), for
doubled rubber band.
6/06
Elastic Energy Conservation
Figure 6
Upper Graph: Rubber Band Force vs. Length. See Figure 5.
Lower Graph: Rubber Band Work vs. Length, obtained from upper graph as
shown.
The elastic force
(see Figure 5).
F (upper graph) is proportional to (band length – 0.05) meters
The work ∫ F⋅dx done in stretching the band to 0.09 meters is shown in the
upper graph. Integrals from the upper graph are plotted in the lower. The lower
graph is quadratic in (length – 0.05) (proportional to (length – 0.05) 2, because the
height of the triangular areas also increases in proportion to the stretch.
This is in contrast to the gravitational situation near the earth's surface, where the
gravitational force remains (essentially) constant, resulting in a gravitational
potential which is linearly proportional to the height.
6/06
Part B
Elastic Energy Conservation
Test of theory: Catapult launch for gravitational (ballistic) flight
Study graphic below.
Reassemble catapult. Measure table height h from floor. Choose a latch position and
predict the range (calculate horizontal launch velocity based on elastic potential
energy), for horizontal launch from table edge. (Remember that the payload1 gets
only a fraction of the elastic energy: Mass fraction 1 = [m payload1/(m payload1 + m launch
vehicle] ). Align launcher release point with edge.
Figure 7
Projectile payload ready for launch. Weigh payload (projectile or car)
to determine energy fraction on separation. Notch gives the release position; align
with table edge and measure range on floor from edge of table. Hold down launch
platform to avoid recoil. Take several range measurements and estimate average.
Record, with estimated error. Predict the range from the horizontal launch velocity
(elastic energy prediction), gravitational acceleration g and table height h. x =
distance from notch to nose of orange launch vehicle (nose is at notch at release,
6/06
Elastic Energy Conservation
with negligible remaining elastic energy).
DO NOT FIRE CAR FROM TABLE.
Use a strip of yellow tape on floor for visual reference; move toward initial test impact
point until close enough to impacts that you can easily elate impact position to tape
position.. Test and report range reproducibility.
Repeat for other latch positions as directed.
]
http://www.walter-fendt.de/ph14e/projectile.htm (See Prep.)
Figure 8
Projectile flight for horizontal launch. Horizontal velocity does not
change (no horizontal force). Vertical velocity was initially zero, and increases
downward with time (constant vertical gravitational force m payload 1 g). Applet
gravitation acceleration was set to ~ 1/10 that of earth surface, so fall is slow, fall
time is increased and range (for given launch velocity) is increased.
6/06
Elastic Energy Conservation
Horizontal range = (fall time) * (horizontal velocity).
Part C
Test of theory: Catapult launch for inside loop flight.
Figure 9
A successful inside loop. (Hold down loop and launcher to avoid recoil
vibration.) The velocity at top of the loop was greater than the critical velocity
v critical =
  gR
or, equivalently, the kinetic energy at the top exceeded the critical energy:
KE critical = ½ mgR = ½ m v critical 2 , (see more detailed equation below)
which can be calculated from the payload fraction of elastic kinetic energy at launch
and the change of gravitational potential energy as the car rises to a height of 2R loop
6/06
Elastic Energy Conservation
Figure 10 If the velocity becomes less than the critical velocity, the car will leave
the track. This is not a successful loop, the car has left the track! Watch the top of
the loop!
The critical velocity to fly off the track is the same for an outside loop, but the
criterion is that the velocity at loop top must be greater than the critical velocity
(and correspondingly for energy)
Procedure for loop-the-loop:
Predict, for the four launcher latch positions, whether your car will perform a successful
loop (see pictures). The theory is developed below (Inside Loop Energy Inequality).
You will be dealing with both elastic and gravitational energies, so the energy form will be
particular convenient in predicting whether the car can traverse the inside loop without
falling off. The predictions will omit a consideration which might be significant: friction,
6/06
Elastic Energy Conservation
which can only produce a loss in total organized mechanical energy.
The initial energy input to your car will be elastic. Assuming idealized elastic behavior, this
is a conservative force with elastic potential energy converting to energy of motion
(kinetic energy). This (or a part of it) will then convert to gravitational potential energy as
the car climbs the track with the track exerting force but doing no work on the car,
because the track force is normal (perpendicular) to the motion at all times.
You will need some system parameters to make your predictions: m1 (projectile mass),
m2 (car mass), M (launch vehicle mass = 0.00790 kg), h (height of table from floor), R
(loop radius), x i (i = 1,2,3,4) (initial stretch from launch release length) and k (rubber
band spring constant). Most of these should already be available from your projectile
studies.
Measure loop diameter (2 R loop). Predict from the energy inequality equation below, for
each latch position, whether the car (payload2) will fall off the inside top of the loop.
Remember that the payload2 gets only a fraction of the elastic energy:
mass fraction 2 = [m payload2/(m payload2 + m launch vehicle).
Launch from the floor. Test first the minimum x latch which should produce a successful
inside loop.. Record Y/N (agreement with prediction). Test the other latch positions and
record results. Discuss any disagreement with prediction in terms of possible friction
effects.
Figure 11
Loop the loop geometry. Launch on floor.
6/06
Elastic Energy Conservation
Figure 12 Outside Loop: Roller coasting without falling off. A gravity-powered
loop-the loop. If the outside speed at the top is too great, the car will leave the track.
But, if the car is underneath, it will leave the track if the inside speed is too little.
If the velocity is too great, the car flies off, for an outside loop. If it is too small, it falls
off for an inside loop.
The radius of curvature and velocity of the car at the top are the keys to staying on the
track, for either outside or inside travel.
The critical velocity for leaving the track (no track force) at the top of the loop is
developed below, based on Newton's Law: F = ma, where acceleration a is v 2/R
(directed downward at loop top) and F is only mg (directed downward, no track
force).
Canceling the equal gravitational and inertial m factors, the critical velocity for the
payload is
v critical =
  gR
and
KE critical = ½ m v critical 2 = 1/2 mgR at the top, or
KE critical =
5
5
2
mv critical= mgR at the bottom of the loop
2
2
(i.e., the energy needed at the bottom is
6/06
Elastic Energy Conservation
mgh = mg(2R) to get up plus 1/2 mgR to stay on at the top).
The elastic band must deliver this minimum kinetic energy to payload 2:
(Payload2 mass fraction)*(elastic band energy) > (5/2) m 2* g*R loop or
m2
 m 2 M 
* (elastic band energy) > (5/2) m 2* g*R loop or, finally
Inside Loop Energy Inequality:
Elastic band potential energy = ½ k x2 > (5/2) (m 2 + M) gR loop .
The initial question in this result was: What payload2 kinetic energy at the bottom
of the loop is necessary for it to climb a gravitational potential hill of 2 m 2 gR loop
and arrive at the top with at least the critical velocity necessary to avoid falling off.
But, in giving energy to the payload the elastic band must also give energy to the
the orange launch vehicle that carries the payload, in proportion to the launch
vehicle mass M.
So, even though the launch vehicle does not attempt to loop, it must receive energy
just as if it did. (The launch vehicle and the payload travel together at the same
velocity before separation, so their kinetic energies are proportional to their masses.)
The total mass (m 2 + M) matters, not either one separately, and low total mass is
better (from an energy point of view). Note that the launcher construction is pretty
light, considering that it gets repeated shocks when arrested in flight at the release
point.
Satellite launches have a similar problem – energy must go to the launch vehicle.
But the satellite launch might utilize a second stage boost, after separation from the
primary launch vehicle, whereas our payload2 has no additional boost.
The four calculated band elastic potential energies should be successively entered
6/06
Elastic Energy Conservation
numerically into the above inequality. Try first experimentally the smallest (lowest
latch stretch x which satisfies the energy inequality), then all others. Record your
results (success/failure). Can friction explain any discrepancies with your
theoretical predictions? (Remember: Friction always dissipates energy, never
increases it.)
If you wish, you can fool around with http://www.funderstanding.com/k12/coaster/ :
Figure 13 Roller coaster game.
Hill 1 is too high or else the radius of hill 2 is
too tight (small). Either way, the outside velocity at hill 2 may not be more than the
critical maximum velocity (depends on R 2). For the inside loop path, the velocity at
the top may not be less than the critical velocity (depends on R loop).
6/06
Elastic Energy Conservation
Friction and (initial) speed are set to zero (must redo if reset). The cart leaves hill 1
at zero initial speed. Height of hill one is increased until the cart flies off the track
on hill 2, at which time the indicated speed locks. Assuming that lift-off occurred at
the top, the radius of hill 2 can be calculated.