CHAPTER 19 THE CYCLOID

Transcription

1 CHAPTER 9 THE CYCLOID 9. Introduction FIGURE IXX. P.5 θ A.5 3 P ψ Let us set up a coordinate sstem O, and a horizontal straight line = a. We imagine a circle of diameter a between the -ais and the line = a, and initiall the lowest point on the circle, P, coincides with the origin of coordinates O. We now allow the circle to roll counterclockwise without slipping on the line = a, so that the centre of the circle moves to the right. As the circle rolls on the line, the point P describes a curve, which is known as a ccloid. When the circle has rolled through an angle θ, the centre of the circle has moved to the right b a horizontal distance aθ, while the horizontal distance of the point P from the centre of the circle is a sin θ, and the vertical distance of the point P below the centre of the circle is a cos θ. Thus the coordinates of the point P are = a( θ + sin θ ), 9.. and = a( cos θ ). 9.. These are the parametric equations of the ccloid. Equation 9.. can also be written = a sin. θ 9..3

2 Eercise: When the -coordinate of P is.5a, what (to four significant figures) is its - coordinate? Solution: We have to find θ b solution of θ + sin θ =. 5. B Newton-Raphson iteration (see Chapter of the Stellar Atmospheres notes in this series) or otherwise, we find θ = radians, and hence =.936a. 9. Tangent to the Ccloid The slope of the tangent to the ccloid at P is d/d, which is equal to ( d / dθ)/( d / dθ ), and these can be obtained from equations 9.. and 9... Eercise: Show that the slope of the tangent at P is tan θ. That is to sa, the tangent at P makes an angle θ with the horizontal. Having done that, now consider the following: Let A be the lowest point of the circle. The angle ψ that AP makes with the horizontal is given b tan ψ =. aθ Eercise: Show that ψ = θ. Therefore the line AP is the tangent to the ccloid at P; or the tangent at P is the line AP. 9.3 The Intrinsic Equation to the Ccloid An element ds of arc length, in terms of d and d, is given b the theorem of Pthagoras: ds = ( d) + ( d) ) /, or, since and are given b the parametric equations 9.. and 9.., b / d d ds = θ. + d And of course we have just shown that the intrinsic coordinate ψ dθ dθ (i.e. the angle that the tangent to the ccloid makes with the horizontal) is equal to θ. Eercise: Integrate ds (with initial condition s =, θ = ) to show that the intrinsic equation to the ccloid is s = 4asin ψ Also, eliminate ψ (or θ) from equations 9.3. and 9.. to show that the following relation holds between arc length and height on the ccloid: s = 4a. 9.3.

3 3 9.4 Variations In sections 9.,,3, we imagined that the ccloid was generated b a circle that was rolling counterclockwise along the line = a. We can also imagine variations such as the circle rolling clockwise along =, or we can start with P at the top of the circle rather than at the bottom. I summarise in this section four variations. The distinction between ψ and θ is as follows. The angle that the tangent to the ccloid makes with the positivel-directed -ais is ψ; that is to sa, d / d = tan ψ. The circle rolls through an angle θ. There is a simple relation between ψ and θ, which is different for each case. In each figure, and are plotted in units of a. The vertical height between vertices and cusps is a, the horizontal distance between a cusp and the net verte is πa, and the arc length between a cusp and the net verte is 4a. I. Circle rolls counterclockwise along = a. P starts at the bottom. The cusps are up. A verte is at the origin. Figure IXX.. = a( θ + sin θ ) 9.4. = asin θ 9.4. s = 4asin θ s = 8a ψ = θ FIGURE IXX

4 4 II. Circle rolls clockwise along =. P starts at the bottom. The cusps are down. A cusp is at the origin. Figure IXX.3. = a( θ sin θ ) = asin θ s = 4a( cosθ) s = 8a( s) ψ = 9 o θ FIGURE IXX

5 5 III. Circle rolls clockwise along =. P starts at the top. The cusps are down. A verte is at =. Figure IXX.4. = a( θ + sin θ ) 9.4. = a cos θ 9.4. s = 4asin θ s = 8a(a ) ψ = 8 o θ FIGURE IXX

6 6 IV. Circle rolls counterclockwise along = a. P starts at the top. The cusps are up. A cusp is at =. Figure IXX.5. = a( θ sin θ ) = a cos θ s = 4a( cosθ) s 8as + 8a(a ) = ψ = 9 o + θ FIGURE IXX

7 9.5 Motion on a Ccloid, Cusps Up 7 We shall imagine either a particle sliding down the inside of a smooth ccloidal bowl, or a bead sliding down a smooth ccloidal wire, figure IXX.6. FIGURE IXX.6.5 R.5 mg We shall work in intrinsic coordinates to obtain the tangential and normal equations of motion. (For a reminder of the use of intrinsic coordinates, the were used briefl, for eample, in Section 7.3) These equations are, respectivel: & s& = g sin ψ 9.5. and mv ρ = R mgcos ψ Here R is the normal (and onl) reaction of the bowl or wire on the particle and ρ is the radius of curvature. The radius of curvature is ds/dψ, which, from equation 9.3., (or equations and 9.4.5) is ρ = 4a cos ψ

8 8 From equations 9.3. and 9.5. we see that the tangential equation of motion can be written, without approimation: g & s = 4a s This is simple harmonic motion of period 4π ag, independent of the amplitude of the motion. This is the isochronous propert of the ccloid. Likewise, if the particle is released from rest, it will reach the bottom of the ccloid in a time π ag, whatever the starting position. Let us see if we can find the value of R where the generating angle is ψ. Let us suppose that the particle is released from rest at a height above the -ais (generating angle = ψ ); what is its speed v when it has reached a height (generating angle ψ)? Clearl this is given b mv = mg ( ), and, following equation 9.3., and recalling that θ = ψ, this is = ga(cosψ cos ψ ) v On substituting this and equation into equation 9.5., we find for R: R mg = ( + cosψ cos ψ ) cosψ 9.6 Motion on a Ccloid, Cusps Down We imagine a particle sliding down the outside of an inverted smooth ccloidal bowl, or a bead sliding down a smooth ccloidal wire. We shall suppose that, at time t =, the particle was at the top of the ccloid and was projected forward with a horizontal velocit v. See figure IXX.7. This time, the equations of motion are & s& = g sin ψ 9.6. and mv ρ = mg cos ψ R B arguments similar to those made in Section9.5, we find that g & s = 4a s

9 9 FIGURE IXX.7.5 R mg The general solution to this is pt s = Ae + Be pt, where p = g ( a) With the initial condition given (at t =, s=, s& = v ), we can find A and B and hence: s a pt pt = v ( e e ) g Again proceeding as in Section 9.5, we find for R: R m = ( 4ga cos ψ v ) a cosψ So what happens?

10 If the constraint is two-sided (bead sliding on a wire) R becomes zero when cos ψ=v/ ( ga ), and thereafter R is in the opposite direction. If the constraint is one-sided (particle sliding down the outside of a smooth ccloidal bowl):. If v > 4ga, the particle loses contact at the moment of projection.. If v < 4ga, the particle loses contact as soon as cos ψ = v / ( ga ). If v is ver small (i.e. ver much smaller than ga ), this will happen when ψ = 45 o ; for faster initial speeds, contact is lost sooner. Eample. A particle is projected horizontall with speed v = m s from the verte of the smooth ccloidal hill = a( θ + sin θ ) = a cos θ, where a = m. Assuming that g = 9.8 m s, how long does it take to get halfwa down the hill (i.e. to = a)? We have to use equation With the numerical data given, this is s =.56548t.56548t.45754( e e ). We can find s from equation 9.4., which gives us s =.8847 m t. If we let ξ = e, we now have to solve = ξ / ξ, or ξ ξ =. From this, ξ = and hence t =.9 s. I leave it to the reader to calculate R at this time and indeed to see whether the particle loses contact with the hill before then. Perhaps the fact that I got a positive real root for ξ means that we are all right and the particle is still in contact but I wouldn't be sure of that. I leave it to the reader to investigate further. 9.7 The Brachstochrone Propert of the Ccloid A small point. The word is sometimes spelled brachistochrone, and I have no recommendation one wa or the other. For what it's worth, the onl dictionar within eas reach of m desk has brachiopod and brachcephalic. In an case, the word is derived from Greek, and means shortest time. The famous brachstochrone problem problem is this: A smooth wire, which can be of an desired length, is to connect two points O and P; P is at a lower level than O, but is not verticall below O. The wire is to be bent to a shape, and cut to a length, such that the time taken for a bead to slide down the wire from O to P is least.

11 It is not eas to prove that the required curve is a cusps-up ccloid; but it is quite reasonable to speculate or to guess that this might be so. And, having speculated that it might be a ccloid, it is eas to verif that the required curve is indeed a cusps-up ccloid, the bead starting from rest at a cusp of the ccloid. A speculation might go something like this. Generall one would epect that the further P is from O, the longer it will take for the bead to slide from O to P. But, if O and P are connected with a ccloidal wire, the time taken to go from O to P does not increase with distance. (See the isochronous propert of the ccloid discussed in Section 9.5.) Thus, as ou increase the distance between O and P, the time taken to travel b an route other than the ccloidal one must take longer than the ccloidal route. This argument ma not sound like a rigorous proof, though it is enough to arouse our suspicions and to test whether it is correct. Since I am going to deal with a bead sliding downwards under gravit, I am going to find it convenient to set up our coordinate aes such that increases to the right, and increases downwards. In that case, the parametric equations to a cusps-up ccloid, with the origin at a cusp, are = a( θ sin θ ), 9.7. and = a and these are the equations that we shall be testing. sin θ 9.7. The time taken for the bead to travel a distance ds along the wire, while it is moving at speed v is ds/v. In (, ) coordinates, ds is + ' d, where ' = d/d. Also, the speed reached is related (b equating the gain in kinetic energ to the loss of potential energ) to the vertical distance dropped b v = g. Thus the time taken to go from O to P is g P + ' d = O P g O f (, ') d This is least (see Chapter 8 for a discussion of this theorem from the calculus of variations) for a function () that satisfies d d f ' f = We have: f = + ', 9.7.5

12 f / ( + ' ) = 3 /, and f ' = ' ( + ' ) / / It is left for the reader to see whether equations 9.7. and satisf equation You should 4 find that both sides of the equation are equal to / ( 4 a 3/ sin θ). Thus our speculation is confirmed, and a cusps-up ccloid is indeed the curve that offers passage from O to P in the shortest time. 9.8 Contracted and Etended Ccloids As in Section 9., we consider a circle of radius a rolling to the right on the line = a. The point P is initiall below the centre of the circle, but, instead of being on the rim of the circle, its distance from the centre of the circle is r. If r < a, the path described b P will be a contracted ccloid; if r > a, the path is an etended ccloid. (I think there s a case for using this nomenclature the other wa round, but most authors seem to use contracted for r < a and etended for r > a.) It should not take long to be convinced, b arguments similar to those in Section 9., that the parametric equations to a contracted or etended ccloid are = a θ + r sin θ 9.8. and = a r cos θ These are illustrated in figures IXX.8 and IXX.9 for a contracted ccloid with r =.5a and an etended ccloid with r =.5a.

13 3 FIGURE IXX FIGURE IXX

14 4 9.9 The Ccloidal Pendulum FIGURE IXX..5 P Let us imagine building a wooden construction in the shape of the ccloid = a( θ sin θ ) 9.9. = a cos θ 9.9. shown with the thick line in figure IXX.. Now suspend a pendulum of length 4a from the cusp, and allow it to swing to and fro, partiall wrapping itself against the wooden frame as it does so. If the arc length from the cusp to P is s, then the length of the free string is 4a s, and so the coordinates of the bob at the end of the pendulum are = o a(θ sin θ) + (4a s)cos(8 ψ) = a(θ sin θ) + (4a s) sin θ and = o a cos θ (4a s)sin(8 ψ) = a cos θ (4a s)cosθ

15 5 (You will need to remind ourself of the eact meaning of ψ and also make use of equation 9.4..) Now equation tells us that s = 4a( cosθ), and, on substitution of this in equations and 4, we find (after a ver little algebra and trigonometr) for the parametric equations to the path described b the bob of the pendulum: = a( θ + sin θ) and = a cos θ Thus the path of the pendulum bob (shown as a dashed line in figure IXX.) is a ccloid, and hence its period is independent of its amplitude. (Recall Section 9.5.) Thus the pendulum is isochronous or tautochronous. It is astonishing to learn that Hugens constructed just such a pendulum as long ago as Eamples of Ccloidal Motion in Phsics Several eamples of ccloidal motion in phsics come to mind. One is the nutation of a top, which is described in Section 4. of Chapter. Earth s ais nutates in a similar fashion. Another well known eample is the motion of an electron in crossed electric and magnetic fields. This is described in Chapter 8 of the Electricit and Magnetism section of these notes. In cosmolog, if the mean densit of the Universe is low, the Universe epands indefinitel, but, if the densit is higher than a certain critical densit, the (dimensionless) scale factor R of the Universe epands and contracts with time t according to the following parametric ccloidal equations: Ω R = ( cos θ), 9.. ( Ω ) Ω t = (θ sin θ) / ( Ω ) Here t is epressed in units of the reciprocal of the present Hubble constant, and Ω is the ratio of the present densit of the Universe to the densit required to close the Universe. A less well known eample concerns the propagation of sound in the atmosphere. In the troposphere, which is the lower part of the atmosphere up to about km, the temperature decreases roughl linearl with height. In that case sound travels through the troposphere in a ccloidal path. The speed of sound in a gas is proportional to the square root of the temperature. (If ou are wondering how it depends on pressure P and densit ρ, the answer is that it depends on the ratio P/ρ - and this ratio is proportional to the temperature.) In an case, if the temperature decreases linearl with height, the sound speed v varies with height as v = v c, 9..3

16 6 where c is a constant, equal to about.3 km.now to trace a sound ra through the atmosphere, we have to understand how the direction of propagation changes as the sound passes through laers of air of different temperature. This is governed, as with light, b Snell s law (see figure IXX.): dv v = tan ψ dψ v + dv ψ ψ + dψ v FIGURE IXX. Snell s law states that when sound (or light) enters a slower medium (i.e. one in which the speed of propagation is slower) it is bent towards the normal. I have drawn figure IXX. to represent the situation in the troposphere where the temperature (and hence the sound speed v) decreases with height. That is, dv/d is negative. In other words dv in figure IXX is negative, and equation 9..4 indicates that dψ is positive, as drawn. In case ou do not recognize this differential form of Snell s law, tr integrating it from v to v and from ψ to ψ, and it should assume its more familiar integral form. If ou now eliminate v between equations 9..3 and 4, ou will get a differential relation between and ψ, which, upon integration, becomes where ψ is the ground-level value of ψ. If we introduce cos ψ c =, 9..5 cos ψ a =, c cos ψ 9..6

17 7 equation 9..5 can be convenientl re-written = a(sin ψ sin ψ ) = a(cos ψ cos ψ) Now tan ψ = d/d, and elimination of between this and equation 9..7 will give a differential relation between and ψ, which, upon integration, becomes = a ( ψ ψ ) + sin ψ sin ] [ ψ Equations 9..7 and 9..8 are the parametric equations of the sound path through the troposphere, and describe a ccloid. Problem for a Rain Da: If =. and =.6, what are ψ and ψ? o I make it ψ = 69 7', ψ = 5 o 5'.