1 CHAPTER 19 THE CYCLOID 19.1 Introduction FIGURE IXX.1 2 P 1.5 y 1 2θ 0.5 ψ 0 0 0.5 1 1.5 2 A 2.5 3 P x FIGURE XIX.1 Let us set up a coordinate system O xy , and a horizontal straight line y = 2 a. We imagine a circle of diameter 2 a between the x-axis and the line y = 2 a, and initially 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 y = 2 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 cycloid. When the circle has rolled through an angle 2 θ, the centre of the circle has moved to the right by a horizontal distance 2 aθ, while the horizontal distance of the point P from the centre of the circle is a sin 2θ, and the vertical distance of the point P below the centre of the circle is a cos 2θ. Thus the coordinates of the point P are x = a(2θ + sin 2θ) , 19.1.1 and y = a(1 − cos 2θ). 19.1.2 These are the parametric equations of the cycloid. Equation 19.1.2 can also be written y= 2 a sin2 θ . 19.1.3 2 Exercise : When the x-coordinate of P is 2.500 a, what (to four significant figures) is its y- coordinate? Solution : We have to find 2 θ by solution of 2θ + sin 2θ = 2.5. By Newton-Raphson iteration (see Chapter 1 of the Stellar Atmospheres notes in this series) or otherwise, we find 2 θ = 0.931 599 201 radians, and hence y = 0.9316 a. 19.2 Tangent to the Cycloid The slope of the tangent to the cycloid at P is dy /dx , which is equal to (dy /dθ) / (dx /dθ), and these can be obtained from equations 19.1.1 and 19.1.2. Exercise : Show that the slope of the tangent at P is tan θ. That is to say, 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 by y tanψ = . x− 2 aθ Exercise : Show that ψ = θ. Therefore the line AP is the tangent to the cycloid at P; or the tangent at P is the line AP. 19.3 The Intrinsic Equation to the Cycloid An element ds of arc length, in terms of dx and dy , is given by the theorem of Pythagoras: 2/1 ds = (()()dx 2 + dy 2 ) , or, since x and y are given by the parametric equations 19.1.1 and 19.1.2, by 2 2 2/1 dx dy ds = + dθ. And of course we have just shown that the intrinsic coordinate ψ dθ dθ (i.e. the angle that the tangent to the cycloid makes with the horizontal) is equal to θ. Exercise : Integrate ds (with initial condition s = 0, θ = 0) to show that the intrinsic equation to the cycloid is s = 4a sin ψ . 19.3.1 Also, eliminate ψ (or θ) from equations 19.3.1 and 19.1.2 to show that the following relation holds between arc length and height on the cycloid: s2 = 4 ay . 19.3.2 3 19.4 Variations In sections 19.1,2,3, we imagined that the cycloid was generated by a circle that was rolling counterclockwise along the line y = 2 a. We can also imagine variations such as the circle rolling clockwise along y = 0, 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 cycloid makes with the positively-directed x-axis is ψ; that is to say, dy /dx = tan ψ. The circle rolls through an angle 2 θ. There is a simple relation between ψ and θ, which is different for each case. In each figure, x and y are plotted in units of a. The vertical height between vertices and cusps is 2a, the horizontal distance between a cusp and the next vertex is πa, and the arc length between a cusp and the next vertex is 4 a. I. Circle rolls counterclockwise along y = 2 a. P starts at the bottom. The cusps are up. A vertex is at the origin. Figure XIX.2. x = a(2θ + sin 2θ) 19.4.1 y= 2 a sin 2 θ 19.4.2 s = 4a sin θ 19.4.3 s2 = 8ay 19.4.4 ψ = θ . 19.4.5 FIGURE IXX.2 2 y 1 0 2 4 6 8 10 12 x FIGURE XIX.2 4 II. Circle rolls clockwise along y = 0. P starts at the bottom. The cusps are down. A cusp is at the origin. Figure XIX.3. x = a(2θ − sin 2θ) 19.4.6 y= 2 a sin 2 θ 19.4.7 s = 4a 1( − cos θ) 19.4.8 s2 = 8a(y − s) 19.4.9 ψ = 90 o − θ. 19.4.10 FIGURE IXX.3 2 y 1 0 2 4 6 8 10 12 x FIGURE XIX.3 5 III. Circle rolls clockwise along y = 0. P starts at the top. The cusps are down. A vertex is at x = 0. Figure XIX.4. x = a(2θ + sin 2θ) 19.4.11 y = 2a cos 2 θ 19.4.12 s = 4asin θ 19.4.13 s2 = 8a 2( a − y) 19.4.14 ψ = 180 o − θ. 19.4.15 FIGURE IXX.4 2 y 1 0 0 2 4 6 8 10 12 14 x FIGURE XIX.4 6 IV. Circle rolls counterclockwise along y = 2 a. P starts at the top. The cusps are up. A cusp is at x = 0. Figure XIX.5. x = a(2θ − sin 2θ) 19.4.16 y = 2a cos 2 θ 19.4.17 s = 4a 1( − cos θ) 19.4.18 s2 − 8as + 8a 2( a − y) = 0 19.4.19 ψ = 90 o + θ . 19.4.20 FIGURE IXX.5 2 y 1 0 0 2 4 6 8 10 12 14 x FIGURE XIX.5 7 19.5 Motion on a Cycloid, Cusps Up We shall imagine either a particle sliding down the inside of a smooth cycloidal bowl, or a bead sliding down a smooth cycloidal wire, figure XIX.6. FIGURE IXX.6 2 1.5 R y 1 mg 0.5 0 0 0.5 1 1.5 2 2.5 3 x FIGURE XIX.6 We shall work in intrinsic coordinates to obtain the tangential and normal equations of motion. (For a reminder of the use of intrinsic coordinates, they were used briefly, for example, in Section 7.3) These equations are, respectively: s&& = − g sin ψ 19.5.1 mv 2 and =R − mg cosψ . 19.5.2 ρ Here R is the normal (and only) 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 19.3.1, (or equations 19.4.3 and 19.4.5) is ρ = 4a cos ψ . 19.5.3 8 From equations 19.3.1 and 19.5.1 we see that the tangential equation of motion can be written, without approximation: g s&& = − s. 19.5.4 4a This is simple harmonic motion of period 4π a g , independent of the amplitude of the motion. This is the isochronous property of the cycloid. Likewise, if the particle is released from rest, it will reach the bottom of the cycloid in a time π a g , 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 y0 above the x-axis (generating angle = ψ0); what is its speed v when it has reached a height y (generating angle ψ)? Clearly this is given by 1 v 2 2 m= mg(), y0 − y 19.5.5 and, following equation 19.3.2, and recalling that θ = ψ, this is v 2 =2ga (cos 2ψ − cos 2 ψ 0 ). 19.5.6 On substituting this and equation 19.5.3 into equation 19.5.2, we find for R: mg R =(1 + 2 cos 2ψ − cos 2 ψ ). 19.5.7 2 cos ψ 0 19.6 Motion on a Cycloid, Cusps Down We imagine a particle sliding down the outside of an inverted smooth cycloidal bowl, or a bead sliding down a smooth cycloidal wire. We shall suppose that, at time t = 0, the particle was at the top of the cycloid and was projected forward with a horizontal velocity v0. See figure XIX.7. This time, the equations of motion are s&& = g sin ψ 19.6.1 mv 2 and =mgcosψ − R . 19.6.2 ρ By arguments similar to those made in Section19.5, we find that g s&& = s. 19.6.3 4a 9 FIGURE IXX.7 2 1.5 R y 1 mg 0.5 0 0 0.5 1 1.5 2 2.5 3 x FIGURE XIX.7 The general solution to this is s= Aept + Be − pt , 19.6.4 where p= g().2 a 19.6.5 v With the initial condition given (at t = ,0 s= ,0 s& = 0 ), we can find A and B and hence: a s = v ()e pt − e− pt .