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the ideal linear spring , which is quite well approximated by a coil spring , is an important concept in the education of a physicist .
simple problems in mechanics often involve a spring of stiffness is also called force constant or spring constant . ] @xmath0 and negligible mass , with a weight of mass @xmath1 hanging at the free end . in the static case
the extension of the spring is then @xmath2 .
a real spring has mass which is evenly distributed along the unloaded spring .
it is then natural to ask how such a spring deforms under its own weight in a force field .
here we investigate this problem for two different force fields .
first in constant gravity , then in the centrifugal force field due to rotation of the reference frame with constant angular velocity .
we also investigate two different models for the finite mass springs .
first a discrete model where the spring is assumed to consist of point masses connected by weightless springs
. then the continuum limit of this model is considered .
it is pointed out that the deformation in this limit should be given by the function that minimizes the potential energy .
results are then easily obtained from the variational principle .
various authors have considered similar problems involving springs of finite mass .
statics of a slinky has been studied by mak @xcite .
several authors have been interested in the effects on dynamics of the finite spring mass .
rotating springs with @xcite and without @xcite finite mass , have also attracted attention .
the approach and most results presented here are , however , either new or , at least hard to find in the literature .
the discrete model could be taught at the elementary level .
the continuum model should be useful in the teaching of variational principles as simple examples of their use .
consider a light ( negligible mass ) spring of natural length @xmath3 and stiffness @xmath0 with one end fixed .
a particle of mass @xmath1 hangs at the free end .
the equation for static equilibrium is , @xmath4 along the spring in the downwards direction . here
@xmath5 is the extension of the spring .
the length of the loaded spring is thus , @xmath6 where @xmath7 is the relative extension of the spring , @xmath8 a dimensionless quantity .
assume now that the same light spring instead is rotating , in a smooth horizontal plane , with the fixed end on the rotation axis .
a particle of mass @xmath1 at the other end will remain at a fixed distance from the rotation axis if , @xmath9 here @xmath10 is the constant angular velocity of the rotation and @xmath5 is the extension of the spring .
the term on the left hand side is mass times centripetal acceleration .
if it is moved to the other side it plays the role of the centrifugal force in the rotating system in which the spring and mass are at rest .    for the dimensionless relative extension of the spring , defined by @xmath11 , we find , @xmath12 in the rotating case . to compare this with the gravitational result we chose the angular velocity , @xmath13 , so that the centripetal acceleration at the end of the unloaded spring is @xmath14 .
one then finds that , @xmath15 the relative extension is longer than in the gravitational case since the centrifugal force grows with distance from the rotation axis .
below we will compare these elementary results for a light spring with the corresponding results for a heavy spring .
consider a chain of @xmath16 particles , each of mass @xmath17 , connected by @xmath16 light springs of natural ( neutral , unloaded , uncompressed ) length @xmath18 , and stiffness ( spring constant ) @xmath19 , see fig .
we must first establish the properties of the resulting finite mass spring of natural length @xmath20 and total mass @xmath21 .
the stiffness or spring constant @xmath0 must be determined by the formula for the effective spring constant of springs in series .
this means that , @xmath22 is the stiffness of the full chain .
the first spring has a free end and we attach this end to a fixed point and let the chain hang vertically from this point . introduce a downward directed @xmath23-axis along the chain with origin at the point of suspension .
the equilibrium equations for the @xmath16 particles are then , @xmath24 we see that the upward force on the @xmath16th particle is the same as the downward spring force on the @xmath25th particle . in this way
one can easily work one self upwards and find that this system is equivalent to the set of equations , @xmath26 here we have introduced @xmath27 , the extension of the last spring in the chain .
this , perhaps obvious , result says that each spring will carry the weight of all the particles below it , including the particle at its lower end .
solving this system of equations successively for @xmath28 , @xmath29 , and so on , one easily finds that : @xmath30 \delta \\
\ldots &   & \ldots \\
\nonumber   x_{n-1 } & = & ( n-1 ) a + [ \sum_{i=0}^{n-2 } ( n - i ) ] \delta \\
\nonumber   x_n & = & n a + [ \sum_{i=0}^{n-1 } ( n - i ) ] \delta .\end{aligned}\ ] ] we therefore get , @xmath31 \delta = na + \frac{n(2n - n+1)}{2}\delta,\ ] ] as the general result for the @xmath32th particle s position .
we note that when we turn off gravity , by letting @xmath33 , the particles are at their natural positions , @xmath34 .
we can easily perform the sums indicated by using the formula for the sum of an arithmetic series .
the position of the last ( @xmath16th ) particle will give the length of the hanging massive chain .
the result is , @xmath35 here we can use @xmath36 , @xmath37 , @xmath38 , and @xmath39 , to get @xmath40.\ ] ] for a real coil spring with continuously distributed mass we must let @xmath41 . comparing to ( [ q.ext.g ] )
we then find that a heavy spring of mass @xmath1 and stiffness @xmath0 is extended half as much ( @xmath42 ) as the corresponding light spring with a weight of mass @xmath1 at the end .
to find how gravity deforms a spring with continuously distributed mass we will now use that potential energy minimization determines the static equilibrium . to pass to the continuum one can follow a procedure by goldstein @xcite ( it can also be found in sakurai @xcite ) .
he derives the wave equation for longitudinal waves along a one dimensional elastic continuum by first considering the discrete case of a horizontal chain of masses and springs . in the limit of infinitely many particles
the sum which is the lagrangian for the discrete case is replaced by an integral of a lagrangian density .
the euler - lagrange equation of the corresponding action is the wave equation .
here we adapt to the static hanging case by skipping the kinetic energy and instead including the gravitational potential energy .
the potential energy of the discrete system of the previous section is given by , @xmath43 where we assume @xmath44 .
we rewrite this in the form , @xmath45 put @xmath46 , @xmath47 , and note that @xmath48 , to get , @xmath49,\ ] ] here @xmath50 is young s modulus , that goes under this name in the theory of elasticity @xcite . ] and @xmath51 the linear mass density of the un - deformed spring .
we can now pass to the continuum by the identifications : @xmath52 here @xmath53 represents the @xmath23-coordinate of points on the un - deformed spring , and @xmath54 is the @xmath23-coordinate of the same point on the spring deformed by gravity under its own weight .    for the continuum case
the potential energy is now a functional of the function @xmath54 which is the limit of the sum ( [ eq.pot.energy.discrete3 ] ) as @xmath55 , and @xmath56 .
the potential energy functional is thus , @xmath57 = \int_{0}^{\ell } { \mbox{\rm d}}x\ , v(\xi,\xi ' ) = \int_{0}^{\ell } { \mbox{\rm d}}x \left [ \frac{y}{2 }   \left ( \frac{{\mbox{\rm d}}\xi}{{\mbox{\rm d}}x } - 1 \right)^2 - \lambda g\ , \xi(x ) \right],\ ] ] where it is assumed that @xmath58 . here ,
@xmath59 , is a potential energy per unit length .
variation of the functional ( [ eq.pot.energy.cont ] ) in the usual way @xcite gives , @xmath60 assuming that the functional is stationary .
then one notes that @xmath61 is a derivative and takes advantage of integration by parts to get rid of this derivative .
after this step one has , @xmath62_0^{\ell}= 0 , \ ] ] for the variation of @xmath63 . at this point one usually invokes the requirement that @xmath64 and thus arrive at the usual euler - lagrange equation .
here we must have @xmath65 since the top end of the spring is fixed . at the lower end , however , this is not obvious .
the variation @xmath66 therefore results in the equation , @xmath67 since both the function @xmath68 and its value at @xmath69 are arbitrary , apart from the condition @xmath65 , the only way to satisfy this equation is to have both the euler - lagrange equation , @xmath70 and the boundary condition , @xmath71 satisfied .    using the explicit form for @xmath72 of equation ( [ eq.pot.energy.cont ] ) we get , the simple equation , @xmath73 from the euler - lagrange equation . here the constant on the right hand side
can also be written @xmath74 using the definitions above eq .
( [ eq.pot.energy.discrete3 ] ) .
the boundary condition gives , @xmath75 since @xmath76 corresponds to an un - deformed spring this result has the simple physical interpretation that the spring is not stretched ( deformed ) at its lower end where there is no mass below that pulls on it .
the solution of ( [ eq.eul.lgr.grav1 ] ) is trivial to find once the boundary conditions , @xmath58 and @xmath77 , are taken into account .
one finds that the new @xmath23-coordinates of the particles of the spring are given in terms of the unloaded @xmath23-coordinates by , @xmath78 the total length of the hanging spring is then given by , @xmath79 in agreement with the discrete result ( [ eq.length.masive.chain.in.grav.1 ] ) when @xmath41 .
we first consider the discrete case and assume that the first spring of the chain is attached to a fixed point and that the spring rotates , with angular velocity @xmath10 around this point , on a smooth horizontal plane .
we choose @xmath80-axis in the rotating system to be parallel to the spring with origin at the fixed point .
the fictitious centrifugal force on a particle of the spring is then @xmath81 .
the equilibrium equations corresponding to the eqs .
( [ eqs.grav.equil ] ) in the gravitational case are then , @xmath82 where we have introduced @xmath83 . the trick used in the gravitational case to solve for @xmath84 does not work here .
instead one might note that the typical equation of this sequence is the recursive relation , @xmath85 this is a so called difference equation that can be treated by standard methods , but here we chose to proceed directly to the continuum case .
recalling that the potential energy of the centrifugal force is @xmath86 , we can take the result ( [ eq.pot.energy.cont ] ) for the gravitational case and get , @xmath87 = \int_{0}^{\ell } { \mbox{\rm d}}z\ , v(\zeta,\zeta ' ) = \int_{0}^{\ell } { \mbox{\rm d}}z \left [ \frac{y}{2 }   \left ( \frac{{\mbox{\rm d}}\zeta}{{\mbox{\rm d}}z } - 1 \right)^2 - \frac{1}{2}\lambda \omega^2\ , \zeta^2(z ) \right],\ ] ] for the centrifugal case .
the euler - lagrange differential equation is now , @xmath88 if we introduce the notation @xmath89 , the solution obeying the boundary conditions , @xmath90 , is easily found to be , @xmath91 this function and the corresponding one for gravity ( [ eq.sol.grav ] ) are plotted and compared in fig .
[ fig1 ] .
the length of the spring in the centrifugal force field is , @xmath92.\ ] ] for small angular velocity @xmath10 we get the approximate result , @xmath93 a light rotating spring with a mass @xmath1 at the end becomes longer by the amount given in eq .
( [ q.ext.omega ] ) , which gives , @xmath94 for small angular velocity @xmath10 .
( [ eq.zeta.length.approx ] ) shows that if the mass instead is distributed along the spring the extra relative length is reduced to @xmath95 , to first order in @xmath10 .
this concludes our study of the rotating heavy spring .
the study of the deformation of springs under their own weight presented above provides nice illustrations of some general principles of statics .
the most important of these is the fact that static equilibrium often is given simply by minimizing potential energy .
the most well known non - trivial example of this principle found in the literature is the catenary , which usually is presented , together with the brachistochrone , as a basic application of the variational calculus @xcite .
unfortunately these classical examples yield quite difficult differential equations who s solution require considerable mathematical skill .
the examples here , giving the deformation of heavy springs , provide simple but interesting , non - trivial , and easily understood results .
the static equilibrium deformation of a heavy spring due to its own weight is calculated for two cases . 
 first for a spring hanging in a constant gravitational field , then for a spring which is at rest in a rotating system where it is stretched by the centrifugal force . 
 two different models are considered . first a discrete model assuming a finite number of point masses connected by springs of negligible weight 
 . then the continuum limit of this model . in the second case 
 the differential equation for the deformation is obtained by demanding that the potential energy is minimized . in this way 
 a simple application of the variational calculus is obtained .