The Rotating Spring. ω Rotating spring. Spring at rest. x L. x' L' Jacob Gade Koefoed

Transcription

1 The Rotating Spring A spring of mass m, which is uniformly distributed along the spring, spring constant k and unstretched length L is rotated about an ais perpendicular to the spring at a constant angular velocity (see gure). It is assumed that the spring obeys Hooke's law. Spring at rest m Rotating spring L k ' L' We dene a function δ() such that = +δ() where is the coordinate describing a position of a point on the unstretched spring and is the coordinate describing the position of the same point on the stretched spring. That is, δ() is the length that the piece of the spring before has stretched. In the following, you may neglect any eects due to gravity. a) We consider the case, where the system has been rotating for a suciently long time, so that the system has reached an equilibrium conguration. Show that, in this case, the function δ() must satisfy the dierential equation d 2 δ d 2 = β2 (δ() + ) (.1) and nd an epression for β in terms of m, L, k and. (Hint: Consider innitesimal pieces of the spring.) The complete solution to the dierential equation found in a) (as you can check) is δ() = A sin β + B cos β (.2) b) Determine the coecients A and B and show that the total length, L, of the rotating spring can be written as L = κl, and nd the dimensionless constant κ in terms of and, where is the natural angular frequency of an ideal, massless spring of the same length with a point mass m at the end, given by = k/m. c) In terms of the given variables, nd the position cm of the center of mass of the rotating spring and check that the centrifugal force, taken to act at the center of mass, does indeed equal the stress in the spring at = (which is equal to = ). If you are using a non-rotating coordinate system, check that the stress at = provides the necessary centripetal acceleration of the center of mass. d) As approaches some critical value c, the spring becomes innitely long. Find the ratio γ = c / when this happens. Show that the same situation occurs when considering a rotating point mass at the end of an ideal, massless spring and nd γ for this situation. Eplain why the two values of γ dier and eplain why the model of the rotating spring predicts this (obviously impossible) result. e) Finally, show that if the spring has not reached the equilibrium position, δ() must satisfy the partial dierential equation 1 2 δ η t 2 = 2 δ 2 + β2 (δ() + ) (.3) and nd η in terms of the given variables. 1

2 Rotating Spring - Solution Part a We chop the unstretched spring into innitely many small pieces of length d and mass dm = µd, where µ = m/l is the mass per unit length, when the spring is at rest. We model each small piece as a point mass dm at the end of a short spring with unstretched length d. We consider a small piece of spring a distance from the ais of rotation. When the spring is rotating, this distance becomes = + δ(). There are three forces acting on the piece: A spring force pulling towards the ais of rotation, a centrifugal force pulling away from the ais of rotation and nally another spring force pulling away from the ais of rotation. In our rotating frame all parts of the spring must be at rest. Fs () d'=d+dδ() dm dδ(+d) Fs (+d) k' +d Fc() Figure 1: Forces acting on the small segment of the spring. The small segment in consideration must have an innitely large spring constant, since any nite etension should require an innitely large force. In fact, if we apply a force of magnitude F to the whole spring, which causes the spring to stretch some fraction α then same force F should also stretch the small spring segment by a factor of α. We see that we may write F = αkl = αk d where k is the spring constant of the small segment d. So we nd that k = kl/d which means, that if the small spring segment stretches an amount dδ the magnitude of the spring force (or the stress in the spring) is k dδ. The centrifugal force on the segment is F c = dm 2 = µ 2 ( + δ())d, and so, we may write Newtons second law as (taking the direction away from the ais of rotation to be positive) k dδ( + d) k dδ() + µ 2 ( + δ())d = (.4) kl kl dδ( + d) d d dδ() = µ2 ( + δ())d (.5) Where dδ() is the etension of the small spring segment at, and dδ( + d) is the etension of the piece of spring at + d. Rearranging, we obtain dδ d dδ +d d = µ2 ( + δ())d = m2 ( + δ())d (.6) kl kl2 dividing on both sides by d we get a second derivative on the left hand side. Thus, we obtain where β 2 = m 2 /kl 2. d 2 δ d 2 = β2 (δ() + ) (.7) Part b Given the solution the the dierential equation, δ() = A sin β + B cos β, we wish to determine the constants A and B. To do this, we rst note that the stretch of the spring before the point = is zero. 2

3 This can also be stated as = at =. Secondly, we require that the stress in the spring, F s = klδ () (where the prime denotes dierentiation with respect to ), equals zero at the far end of the spring. That is, we require that δ() = and δ (L) = (.8) Using the rst of these conditions, we see that The second conditions yields which gives δ() = A sin() + B cos() = B = (.9) δ (L) = Aβ cos(βl) 1 = (.1) The solution is then A = 1 β cos(βl) (.11) δ() = sin(β) β cos(βl) (.12) Remembering that β 2 = m 2 /kl 2 and using 2 = k/m we may rewrite this as δ() = L sin(/( L)) cos(/ ) The total stretch of the spring must be the stretch before = L, so we nd L = δ(l) = δ() = L The total length of the spring must then be L = L + L = L ( tan and we nd κ = ( ) tan (.13) sin(/ ) cos(/ ) L (.14) ) (.15) (.16) Part c From the denition of center of mass, MX cm = i m i i (.17) and using an integral instead of a sum for this continuous distribution of mass, we nd cm = 1 m m L d = 1 L L cos β = Lβ 2 cos βl = 1 cos βl Lβ 2 cos βl = (δ() + ) d = 1 L kl ( 1 m 2 cos βl 1 To determine the stress in the spring at =, we need to compute sin β β cos βl d ) 3

4 δ/l L'/L /L / Figure 2: The left plot shows the stretch of the spring relative to the rest length as a function of the relative position /L on the spring. The right plot shows the total length of the rotating spring relative to the rest length as a function of the angular velocity of rotation relative to the natural angular frequency of the spring. and we see that indeed m 2 cm = klδ (). δ () = β cos() β cos βl 1 = 1 cos βl 1 (.18) Part d Looking at the epression for the total length of the spring L = L ( ) tan (.19) we see that it blows up as the ratio / approaches π/2. That is, in this case γ = π/2. In the case of a point mass at the end of a rotating spring (which has reached equilibrium), the spring force must balance the centrifugal force, so that m 2 (l + l) = k l (.2) where l is the rest length of the spring. Solving this for the etension l we nd l = lm2 k m 2 = l2 2 (.21) 2 and we see that in this case, γ = 1. That is, when the angular velocity of the rotating spring eceeds the natural angular frequency of the spring-mass system the spring becomes innitely long. When considering the simpler situation of the point mass at the end of an ideal spring, it is clear, that when > then, as l becomes larger, the centrifugal force grows faster than the spring force. The error lies in the assumption that Hooke's law holds for these large etensions. In general, at some point, the spring force will deviate greatly from Hooke's law. The reason that the value for γ is greater in the case of a spring with mass, is that the center of mass (at which the centrifugal force can be taken to act) is only at some fraction of the total length, whereas in the case of a point mass it is always at the position l + l. Thus, in the case of the rotating spring, the angular velocity needs to be larger to overcome the spring force. 4

5 Part e To answer this question, we must go back to Newtons second law as used in part a), but this time we will put the acceleration on the right-hand side. k dδ( + d) k dδ(, t) + µ 2 ( + δ(, t))d = µa(, t) d (.22) Now, performing the same steps as in a), we obtain (using that acceleration is the second derivative of position) Finally, we notice that kl 2 δ 2 + µ2 ( + δ(, t)) = m 2 (.23) L t 2 2 t 2 = 2 t 2 (δ() + ) = 2 δ t 2 (.24) (as the coordinate is independent of time) and dividing the dierential equation (as before) by kl, we obtain the result 2 δ 2 + β2 ( + δ(, t)) = mk 2 δ (.25) L 2 t 2 so that 1/η = mk/l 2. If we dene a new function f(, t) such that δ(, t) = f(, t) + δ E (), where δ E () is the equilibrium solution, this partial dierential equation can be separated and solved for f(, t), given an initial shape of the spring δ(, ). The function f(, t) can then be used to construct δ(, t). In particular, if the initial shape is δ(, ) = (that is, the spring is unstretched) and the spring is suddenly rotated with angular velocity (we ignore the eects of angular acceleration) the ratio of the rotating spring to the rest length as a function of time, is given by L L = ( ) tan π 2 2 n= ep( π 2 (n + 1/2) 2 2 /2 + 1 t) (4n 2 + 4n + 1)(4 2 / 2 4π2 n 2 4π 2 n π 2 ) and we see that this indeed equals the result found in b) as t becomes very large. Figure 3: The left plots shows the stretch of the spring before the position /L on the spring. Various curves are plotted showing dierent instants of time measured in units of 1/. The right plots shows the stretch of the whole spring as a function of time. Dierent curves represents dierent values of angular velocity in units of the natural angular frequency of the spring. We can see that the spring reaches the equilibrium position when t 3 which corresponds to roughly half a turn. 5