cambridge tripose ia-dyn-chapter7

Chapter 7 Rotating Frames 7.1 Angular Velocity A rotating body always has an (instantaneous) axis of rotation. Definition: a frame of reference S/primeis said to have angular velocity !with respect to some fixed frame Sif, in an infinitesimal time δt, all vectors which are fixed in S/primerotate through an angle δθ=ω δt about an axis n= !/ωthrough the origin, where ω=| !|. Consider a vector uwhich is fixed in the rotating frame. In a time δtit rotates through an angle δθabout n; i.e., it moves to u+δu=ucosδθ+ (u . n )n(1−cosδθ)−u×nsinδθ =u+n×uδθ+O(δθ2) =⇒ δu=n×uω δt+O(δt2) =⇒ ˙u=ωn×u= !×u. (This fact is often regarded as “obvious” and can be quoted; it is sometimes taken as the definition of !.) Now let Sbe an inertial frame and S/primebe a frame rotating with angular velocity ! with respect to S. Let {e1,e2,e3}be a basis for Sand{e/prime 1,e/prime 2,e/prime 3}a basis for S/prime. Let abe any vector — not necessarily fixed in either SorS/prime— with components aianda/prime i respectively, i.e., a=a1e1+a2e2+a3e3=a/prime 1e/prime 1+a/prime 2e/prime 2+a/prime 3e/prime 3. Then ˙a=d dt(aiei) = ˙aiei 44 since the {ei}are fixed. Also, ˙a=d dt(a/prime ie/prime i) = ˙a/prime ie/prime i+a/prime i˙e/prime i = ˙a/prime ie/prime i+a/prime i !×e/prime i (because the e/prime iare fixed in S/prime) = ˙a/prime ie/prime i+ !×(a/prime ie/prime i) = ˙a/prime ie/prime i+ !×a. Introduce the notations /parenleftbiggda dt/parenrightbigg S≡˙aiei,/parenleftbiggda dt/parenrightbigg S/prime≡˙a/prime ie/prime i. Note that an observer in S/primewho does not know that it is rotating would measure the rate of change of aas (da/dt)S/prime, because he would not include the contribution of ˙e/prime i; he only notices the rate of change of the components a/prime iin his own frame. We have /parenleftbiggda dt/parenrightbigg S=/parenleftbiggda dt/parenrightbigg S/prime+ !×a foranyvector a. We can apply the same method again: ¨a=d dt(˙a/prime ie/prime i+ !×a) = ¨a/prime ie/prime i+ ˙a/prime i˙e/prime i+˙!×a+ !×˙a = ¨a/prime ie/prime i+ 2 !×(˙a/prime ie/prime i) +˙!×a+ !×( !×a). So /parenleftbiggd2a dt2/parenrightbigg S=/parenleftbiggd2a dt2/parenrightbigg S/prime+ 2 !×/parenleftbiggda dt/parenrightbigg S/prime+˙!×a+ !×( !×a). 7.2 The Equation of Motion in a Rotating Frame Since Sis an inertial frame, NIIholds in that frame: F=m/parenleftbiggd2x dt2/parenrightbigg S. 45 Hence F=m/braceleftbigg/parenleftbiggd2x dt2/parenrightbigg S/prime+ 2 !×/parenleftbiggdx dt/parenrightbigg S/prime+˙!×x+ !×( !×x)/bracerightbigg . (7.1) Example: a particle is suspended by a string in a laboratory on the Earth’s surface, with position vector Rrelative to the centre of the Earth which rotates with (constant) angular velocity !. The particle is hanging at equilibrium in the lab frame. What is the tension in the string? Since (d x/dt)S/prime=0, because the particle is at rest in the lab, T+mg=m !×( !×R), i.e., T=−m{g− !×( !×R)}. Without rotation, the answer would have been just −mg; so we call g− !×( !×R) the apparent gravity . Example: in a fairground ride on Midsummer Common, a large cir- cular drum of radius arotates about its own axis, which is vertical. People pay good money to get pinned to the wall as the floor drops away. What is the minimum safe angular velocity of the drum? In the rotating frame of the drum, the position vector xof a person is fixed, so (d x/dt)S/primeand (d2x/dt2)S/primeboth vanish. The forces on the person are mg, a reaction Nfrom the wall and friction R, so mg+N+R=m !×( !×x) =m( !. x) !−m( !. !)x =−mω2x. (Because !is perpendicular to x.) Resolving vertically, R=−mg, so friction is the only thing holding the person up, while horizontally, N=−mω2x. Recall that |R|/lessorequalslantµ|N|from§1.6.2, where µis the coefficient of static friction; so mg/lessorequalslantµmω2a, i.e., ω/greaterorequalslant/radicalbig g/(µa). Ifωdrops below this, the person slides off the ride. 46 The equation of motion is sometimes rearranged (by engineers and physicists) as m/parenleftbiggd2x dt2/parenrightbigg S/prime=F−2m !×/parenleftbiggdx dt/parenrightbigg S/prime−m˙!×x−m !×( !×x). The various terms on the rhs are then called fictitious forces ; they do not really exist but an observer in S/prime(who does not know that S/primeis rotating) feels an acceleration caused by them just as if they were real. For example, in the fairground example above, the fictitious force is −m !×( !×x) which equals mω2x(radially outwards ) — this is called the “centrifugal force”. As far as the observer in S/primeis concerned, the “centrifugal force” is cancelled by N. The most physically important rotating system is the Earth itself. Our weather patterns are strongly influenced by the Earth’s rotation, and in particular by the “Coriolis force”, i.e., the fictitious force −2m !×(dx/dt)S/prime. Consider the motion of air masses in the atmosphere. Air close to the surface is warmer than air higher up, which produces an upwards force that more or less balances the downwards pull of gravity: hence we can ignore vertical motion as it is essentially static. However, there are significant effects in horizontal directions. As an example, consider a region of low atmospheric pressure p, known as a depression . Weather maps usually include isobars which are con- tours of constant p, and the (real) forces acting on air currents push them in the direction of −∇p(i.e., from high to low pressure). There- fore, we might na¨ ıvely expect air flow to be orthogonal to the isobars (because ∇fis always perpendicular to a curve of constant ffor any function f(x)). However, in the Northern hemisphere, !has a component vertically upwards and − !×(dx/dt)S/primetherefore pushes to the right when viewed from above. So the fictitious Coriolis force causes winds to be de- flected rightwards. (In the Southern hemisphere, winds are deflected to the left.) This deflection continues until the pressure force −∇pis in balance with the Coriolis force; this happens when the wind direction (dx/dt)S/primeis along the contour lines. The result is that winds actually circle anticlockwise around a depression, parallel to the isobars, rather than orthogonal to them. (In reality, other effects such as friction cause the air flow to move slightly inwards as it circulates.) 7.3 The Foucault Pendulum This is just a simple pendulum, with a bob of mass mand a string of length lattached to the point (0 ,0, l). We assume that the displacements are small, i.e., |x| /lessmuchl. We choose our axes such that the x-axis points East and the y-axis North. We note that x2+y2+ (l−z)2=l2, so z=1 2l(x2+y2+z2). 47 Since x, y, z are all small, zis actually very small (second order), and we therefore take it to be zero. So we can use plane polar coordinates ( r, φ) in the ( x, y) plane. Note that θ(the angle of the pendulum from the vertical) is small and that sin θ=r/l; so the components of T= (Tx, Ty, Tz) are Tz=Tcosθ≈T, Tx=−Tsinθcosφ =−T(r/l)(x/r) =−T x/l, Ty=−T y/l. Now our co-ordinate frame is rotating with the Earth at (constant) angular velocity !. Since ω=| !|is small, we may ignore terms of order ω2; hence, from (7.1), T+mg=m/braceleftbigg/parenleftbiggd2x dt2/parenrightbigg S/prime+ 2 !×/parenleftbiggdx dt/parenrightbigg S/prime/bracerightbigg . Since (d /dt)S/primerefers to the rate of change measured in this frame , (dx/dt)S/prime= ( ˙x,˙y,˙z) and (d2x/dt2)S/prime= (¨x,¨y,¨z). We also have != (0, ωcosλ, ωsinλ) where λis the latitude. Hence Tx/m= ¨x+ 2ω( ˙zcosλ−˙ysinλ), Ty/m= ¨y+ 2ω˙xsinλ, Tz/m−g= ¨z−2ω˙xcosλ. Since z≈0, the last of these three equations gives Tz≈mg+O(ω); since T≈Tz, we obtain Tx=−(mg/l )x+O(ωx/l ). As both ωandx/lare small, we can ignore the second-order correction term to Txand obtain ¨x=−g lx+ 2ω˙ysinλ, (7.2) ¨y=−g ly−2ω˙xsinλ. (7.3) Now let ζ=x+ iy. Taking (7.2) + i(7.3), ¨ζ=−g lζ−2iω˙ζsinλ. The auxiliary equation is α2+ 2iωαsinλ+g l= 0, 48 i.e., α=−iωsinλ±/radicalbigg −ω2sin2λ−g l≈ −iωsinλ±i/radicalbig g/l. The general solution is therefore ζ= e−iωtsinλ/parenleftbig Acos(/radicalbig g/l t) +Bsin(/radicalbig g/l t)/parenrightbig . In particular, argζ=−ωtsinλ+ arg/parenleftbig Acos(/radicalbig g/l t) +Bsin(/radicalbig g/l t)/parenrightbig . We note that φ= arg ζ, because the ( x, y)-plane is simply the Argand plane for ζ. The only effect of ωis to cause φto decrease at a constant rate ωsinλ; that is, the direction of swing of the pendulum moves clockwise at constant angular speed ωsinλwhile the pendulum continues to swing to and fro with frequency/radicalbig g/l. (This rotation is slow: its period is 2 π/(ωsinλ), and ω= 2π/(1 day), so at λ= 52◦N the period is around 301 2hours.) 49