Sect 5.7, Part II Now, quantitative predictions under the initial conditions  = 0, θ = 0. Assumption: The rotational kinetic energy about the z axis is very large compared to the maximum potential energy: (½)I 3 (ω 3 ) 2 >> Mg  Under this assumption, the gravitational torques (causing precession & nutation) are small perturbations on the rotation of the top about its figure axis. This situation  “Fast Top” conditions. We’ll be able to get closed form expressions for the nutation frequency & amplitude + the precession frequency!

Nutation amplitude  u 0 - u 1. u 0 = initial value u 0 = cosθ 0 = u 2, a root of f(u), as just discussed. u 1 = cosθ 1 = the other physical root of f(u). Initial energy: E´ = E – (½)I 3 (ω 3 ) 2 = Mg cosθ 0 Note that α  [2E - I 3 (ω 3 ) 2 ]/I 1, β  (2Mg )/I 1, a  (p ψ )/I 1, b  (p  )/I 1  These initial conditions are equivalent to α = βu 0 = βu 2 = β(b/a) (B) u 2 = f(u) = βu 3 - (α + a 2 )u 2 + (2ab - β)u + (α - b 2 ) (A) (B) & (A) together (factors into linear & quadratic parts!)  u 2 = f(u) = (u 0 - u)[β(1 - u 2 ) - a 2 (u 0 - u)] (C)

u 2 = f(u) = (u 0 - u)[β(1 - u 2 ) - a 2 (u 0 - u)] (C)  The roots of f(u) other than u 0 (  u 1 ) are given by: β[1 - (u 1 ) 2 ] - a 2 (u 0 - u 1 ) = 0 (1) Define: x  (u 0 - u), x 1  (u 0 - u 1 )  (1) becomes (using the definitions of various quantities): (x 1 ) 2 + (a 2 /β)x 1 - sin 2 θ 0 = 0 (2) Or: (β/a 2 )(x 1 ) 2 + x 1 – (β/a 2 )sin 2 θ 0 = 0 (2´) Note that: (β/a 2 ) = [2(I 1 /I 3 )Mg ]/[I 3 (ω 3 ) 2 ]  Unless I 3 << I 1 (which would correspond to a weird, “cigar shaped” top!) (β/a 2 ) << 1  The 1 st term in (2) is much smaller than the 2 nd & 3 rd terms.  To a good approximation The NUTATION AMPLITUDE is given by x 1 = (u 0 - u 1 )  (β/a 2 )sin 2 θ 0 = 2I 1 Mg (I 3 ) -2 (ω 3 ) -2 sin 2 θ 0  Nutation Amplitude

Nutation amplitude: x 1 = (u 0 - u 1 )  (β/a 2 )sin 2 θ 0 = 2I 1 Mg (I 3 ) -2 (ω 3 ) -2 sin 2 θ 0 Note: x 1 decreases as (ω 3 ) -2 Recall also that ω 3 = constant.  The faster the top is initially spun about its own z axis, the LOWER the nutation amplitude. Now, find the nutation frequency for this same fast top: We had: u 2 = f(u) = (u 0 - u)[β(1 - u 2 ) - a 2 (u 0 - u)] (C) Use: x  (u 0 - u), x 1  (u 0 - u 1 ) The nutation amplitude x 1 is small  In (C): (1 - u 2 )  [1 - (u 0 ) 2 ] = sin 2 θ 0  (C) becomes: x 2 = f(x)  a 2 x(x 1 - x) To solve, let y = [x – (½)x 1 ]  y 2 = a 2 [(¼)x 1 - y 2 ] Differentiate & get a simple harmonic oscillator equation: y + ay = 0 (D)

y + ay = 0  y = Acos(at) (D) Or: x = y + (½)x 1 = A cos(at) + (½)x 1 Initial conditions: x = 0 at t = 0  A = - (½)x 1  The time dependence of the nutation motion: x = (u 0 - u) = (½)x 1 [1 - cos(at)] (E) where, the nutation amplitude is: x 1 = (u 0 - u 1 )  (β/a 2 )sin 2 θ 0 = 2I 1 Mg (I 3 ) -2 (ω 3 ) -2 sin 2 θ 0 and the nutation frequency: a  (p ψ )/I 1 = ( I 3 /I 1 )ω 3  The nutation amplitude decreases with increasing ω 3 But the nutation frequency increases with increasing ω 3  The faster the top is initially spun about its own axis (at constant ω 3 ), the faster it nutates and the smaller the nutation amplitude!

Now, the consider the precessional motion. The angular velocity of precession is, in general:  = (b - a cosθ)sin -2 θ  a(u 0 - u)sin -2 θ  axsin -2 θ In the same fast top approximation as for the nutation discussion (small x  sin 2 θ  sin 2 θ 0 ):   axsin -2 θ 0 (1) Solution we just got for nutational motion is: x = (u 0 - u) = (½)x 1 [1 - cos(at)] (2) Put (2) into (1), using x 1  (β/a 2 )sin 2 θ 0    (½)(β/a)[1 - cos(at)] (3)

  (½)(β/a)[1 - cos(at)] (3) From (3):  (  precession rate) is not constant, but varies sinusoidally with time! Precession Frequency = a = (I 3 /I 1 )ω 3 (the same frequency as for the nutation). –As a side note, we can easily integrate (3) to get  (t)! Define: The average precession frequency ( = time average of (3)):  = (½)(β/a) = Mg (I 3 ω 3 ) -1 (4) NOTE! Typo in the text!!   (ω 3 ) -1 The precession rate decreases with increasing ω 3 ! Goldstein’s Eq. (5.74), p 217 is WRONG!

Now, we have a complete picture of the fast top motion under the initial conditions that  = 0, θ = 0, θ = θ 0. Immediately after release at θ = θ 0 : the z-axis of the top begins to fall under gravity. As it falls, gravity produces a torque about the θ rotation axis.  The top begins to precess with precession velocity   x = cosθ 0 - cosθ. The z- axis begins moving sideways around the vertical. This fall also results in the nutational motion of the z-axis, as well as precession. As the top is initially spun faster & faster (as ω 3 increases) the frequency of nutation increases, the amplitude of nutation decreases, and the frequency of precession decreases.

All of this neglects frictional and damping effects, of course! In real tops, these effects can be very important! In fact, for a sufficiently fast top, damping effects rapidly damp out the nutational motion.  The top appears to precess uniformly.  a “pseudo-regular” precession.  Elementary discussions of top motion often neglect the nutational motion.

Regular Precession Another (related) interesting problem: What are the initial conditions required for the top to undergo a true regular precession (that is precession without nutation)? This case would require θ = constant = θ 0 for all time. θ = θ 0 = θ 1 = θ 2. (1) Lets analyze this in terms of function f(u). u = cosθ We had: u 2 = f(u) = βu 3 - (α + a 2 )u 2 + (2ab - β)u + (α - b 2 ) (2) (1) requires f(u) in (2) to have a double root at u 0 = cosθ 0. Figure: In this case (2)  f(u 0 ) = 0 = (u 2 )| 0 Also: (df/du)| 0 = 0

f(u 0 ) = 0 = (u 2 )| 0 from (2):  0 = β(u 0 ) 3 - (α + a 2 )(u 0 ) 2 + (2ab - β)u 0 + (α - b 2 ) (3) Can show this implies: (α - βu 0 ) = (b -au 0 ) 2 /[1- (u 0 ) 2 ] (4) (df/du)| 0 = 0  (½)β =[a(b - au 0 ) - u 0 (α - βu 0 )][1- (u 0 ) 2 ] -1 (5) Recall that the precession rate  = (b - a cosθ)sin -2 θ With θ = θ 0 and u 0 = cosθ 0 this becomes:  = (b - au 0 )[1- (u 0 ) 2 ] -½ (6) Combining (4), (5), (6): We get a quadratic equation for the precession rate  : (½)β = a  - (  ) 2 cosθ 0 (7) Put in the constants: I 1 a = I 3 (  cosθ + ψ) = I 3 ω 3, β = (2Mg )/I 1  (7) becomes: Mg =  (I 3 ω 3 - I 1  cosθ 0 ) (8) Or: In terms of ψ: Mg =  [I 3 ψ - (I 1 - I 3 )  cosθ 0 ] (9)

Conditions for regular precession (no nutation): Mg =  (I 3 ω 3 - I 1  cosθ 0 ) (8) Or: Mg =  [I 3 ψ - (I 1 - I 3 )  cosθ 0 ] (9) General, initial conditions for heavy top require specification of initial values of : θ,θ, , ,ψ,ψ,(or ω 3 ). ,ψ are cyclic  Their initial values are somewhat irrelevant. However, if we want regular precession (no nutation) then the initial values must also satisfy Eqs. (8) or (9). Can still almost arbitrarily choose θ 0 & ω 3 but then  is determined by (8). Note that (8) & (9) are quadratic for . We want real  !  We must have (I 3 ω 3 ) 2 > 4Mg I 1 cosθ 0 (10)

Condition (I 3 ω 3 ) 2 > 4Mg I 1 cosθ 0 (10)  If θ 0 > (½)π (the top is mounted so that its CM is BELOW the fixed point!): Any value of ω 3 can lead to uniform precession. If θ 0 < (½)π: There is a minimum value of ω 3 for uniform precession. This is given by making (10) an equality. ω 3 ´ = (2/I 3 ) [Mg I 1 cosθ 0 ] ½ Back to Conditions for regular precession (no nutation): Mg =  (I 3 ω 3 - I 1  cosθ 0 ) (8) Or: Mg =  [I 3 ψ - (I 1 - I 3 )  cosθ 0 ] (9) Quadratic Eqtn  2 solutions for .  “Fast Precession” & “Slow Precession” (8) & (9)   can never be zero (for non-zero ψ or ω 3 )  We must always give the top a shove to cause it to uniformly precess.

Conditions for regular precession (no nutation): Mg =  (I 3 ω 3 - I 1  cosθ 0 ) (8) Or: Mg =  [I 3 ψ - (I 1 - I 3 )  cosθ 0 ] (9) Look at fast & slow precession solutions: Slow precession:  cosθ 0 << a = (I 3 ω 3 )/I 1 (8)    (Mg )(I 3 ω 3 ) -1 Fast precession: Mg <<  (I 3 ω 3 - I 1  cosθ 0 ) (8)    (I 3 ω 3 )(I 1 cosθ 0 ) -1

Final Example Another (related) interesting problem: Special case: u = cosθ = 1 (θ = 0) is a root of f(u) We must treat this as special case, since many previous expressions have had [1- u 2 ] in the denominator! Suppose the initial conditions have the top z axis initially vertical: θ 0 = 0. Recall: Angular momentum about the vertical axis p ψ = I 1 a = const. Angular momentum about the z-axis p  = I 1 b = const. In this case, p ψ = p  since the 2 axes are initially in same direction.  a = b Also, the energy E´ = E – (½)I 3 (ω 3 ) 2 = constant. From the general energy equation of earlier, we find, in this special case: E´ = Mg  From the definitions of α & β, we find, under these conditions, α = β. the top is vertical!

In general, we had (u = cosθ) u 2 = f(u) = (1-u 2 )(α - βu) - (b - au) 2 Under the conditions just described (a = b, α = β)  u 2 = f(u) = (1-u 2 )β(1 - u) - a 2 (1 - u) 2 Or: u 2 = f(u) = (1 - u) 2 [β(1+u) - a 2 ] (1) Obviously, u = 1 is a (double) root of f(u). The 3 rd root is clearly: u 3 = a 2 β (2) Consider: Fast Top Conditions:  a 2 β -1 > 2  u 3 > 1 Slow Top Conditions:  a 2 β -1 < 2  u 3 < 1

u = 1 is a root of f(u) the other is u 3 = a 2 β Fast Top Conditions:  (a 2 )(β) -1 > 2  u 3 > 1. Earlier discussion: Physical roots have to be in the range -1 < u < 1.  u = 1 is the only physically allowed root!  The top spins vertically forever! See figure:

u = 1 is a root of f(u) the other is u 3 = a 2 β Slow Top Conditions:  (a 2 )(β) -1 < 2  u 3 < 1. Earlier discussion: Physical roots have to be in range -1 < u < 1.  u 3 is a physically allowed root.  The top nutates between u 3 < u < 1. See figure:

It is thus clear that there is a critical frequency (ω 3 ) c above which ONLY vertical rotation is possible. This is given by (using definitions of the constants) a 2 β -1 = 2= (I 3 ) 2 (I 1 ) -1 [(ω 3 ) c ] 2 [2Mg ] -1 Giving: [(ω 3 ) c ] 2 = 4Mg I 1 (I 3 ) -2 READ the interesting discussion in the text on p 221 about effects of friction on real top motion. READ the interesting discussion in the text on p 222 about applications of all of this to gyroscope motion & applications of gyroscopes to modern (navigational & other) technology.