PHY 5246: Theoretical Dynamics, Fall Assignment # 10, Solutions. (1.a) N = a. we see that a m ar a = 0 and so N = 0. ω 3 ω 2 = 0 ω 2 + I 1 I 3

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1 PHY 54: Theoretical Dynamics, Fall 015 Assignment # 10, Solutions 1 Graded Problems Problem 1 x 3 a ω First we calculate the moments of inertia: ( ) a I 1 = I = m 4 + b, 1 (1.a) I 3 = ma. b/ α The torque is zero! This can be seen in several ways: for instance, from the definition of the torque and the force of gravity F a = m a g we can see x 1 CM x N = a r a F a = a m a r a g. Here r a is defined with respect to the center of mass since that is the axis about which we want to calculate the torque. However, since the center of mass vector R is defined such that R = am ar a = 0, a m a we see that a m ar a = 0 and so N = 0. (1.b) Euler s equations are { I1 ω 1 (I 1 I 3 )ω ω 3 = 0 I 1 ω (I 3 I 1 )ω 3 ω 1 = 0 I 3 ω 3 = 0 { ω1 I 1 I 3 I 1 ω 3 ω = 0 ω + I 1 I 3 I 1 ω 3 ω 1 = 0 ω 3 = 0 The last equation implies ω 3 is constant, and is just the projection of ω onto the x 3 axisω 3 = ωcosα = ωb b +4a.

2 Now to solve for ω 1 and ω, we can rewrite the first two equations as { ω 1 +Ωω = 0 ω Ωω 1 = 0, where Ω = I 3 I 1 ω 3. I 1 Taking another derivative of the first equation with respect to time and inserting the second equation we find ω 1 +Ω ω = 0 ω 1 +Ω ω 1 = 0, which has solution ω 1 (t) = Acos(ωt+δ), so that means ω (t) = Asin(Ωt+δ). Since the phase is the same we can set δ = 0 and our full solution is { ω1 (t) = Acos(Ωt) ω (t) = Asin(Ωt) ω 3 (t) = ωb b +4a Thus wehavethat ω precesses inaconearoundx 3 withangular frequency x 3 α ω ω x Ω = I 3 I 1 ω 3 I 1 = ma m = 3a b 3a +b ( a 4 + b 1 ma + mb 4 1 ) ωb b +4a ωb b +4a x 1 ω 1 This also tells us what our constant A should be: A = ω 1 (0) = ωsinα = aω b +4a. (1.c) The kinetic energy will be with T = T (CM) (about CM) trans +T rot, T (CM) trans = 1 m(v 0 gt) (about CM) T rot = 1 I 1(ω1 +ω)+ 1 I 3ω3 = 1 ( ) a m 4 + b A + 1 ma 1 ω 3 = 1 ( ) a m 4 + b 4a ω 1 b +4a + 1 ma = 1 ) m a ω (a + b b +4a 3 + b = 1 ma ω (a + 5 ) b +4a b. ω b b +4a

3 Adding these together we find Problem T = 1 m(v 0 gt) + 1 ma ω b +4a For vertical motion, we have θ = 0 so that ω 3 = φ+ ψ and The energy is simply p φ = p ψ = I 3 ω 3. E = 1 I 3ω 3 +Mgh, and since ω 3 is constant we can define the conserved quantity E = Mgh = E 1 I 3ω 3. (a + 5 b ). In order to study the nature of the θ = 0 equilibrium position, we now study how the system (top) behaves when it s displaced by an angle θ. For an arbitrary displacement about θ = 0 we can then write: E = 1 I 1ω I ω + 1 I 3ω 3 +Mghcosθ = 1 I 1 θ + p ψ (1 cosθ) I 1 sin θ +Mghcosθ = Mgh, where I 1, I, and I 3 are the principal moments of inertia relative to the a body system with origin at the (fixed) tip of the top, and we have used that I 1 = I, as well as ω 1 = φsinθsinψ + θcosψ, ω = φsinθcosψ θsinψ. Also, we have expressed the φ component of the angular velocity as a function of the p φ and p ψ constants of the motion, which, for the initial conditions given in this problem, satisfy p φ = p ψ = I 3 ω 3 (see initial discussion): φ = p φ p ψ cosθ I 1 sin θ We can then recast the energy equation in the form, = p ψ(1 cosθ) I 1 sin θ Mgh(1 cosθ) = 1 I 1 θ + p ψ (1 cosθ) I 1 sin θ which is well defined even at θ = 0. Using the change of variables, z = cosθ (so that ż = sinθ θ) and solving the above equation for ż we get: Mgh(1 z) = 1 I 1 sin θ + p ψ (1 cosθ) I 1 sin θ ż ż = 1 I 1 1 z + p ψ (1 z) I 1 (1 z ) ż = (1 z) [ ] MghI1 (1+z) I I1 3 ω 3..,

4 400 r= r=1 r=0. 00 f(z) z 3 z Figure 1: A plot of the function f(z) given below. For simplicity we have set the parameter ξ = 150 and given three values of the ratio r = ω3 /ω c. Note that r = 1 is the critical case. z For clarity, let us rewrite the equation in the following form, ż ξ = (1 z) [(1+z) r] f(z), where r = ω 3 /ω c and with critical frequency ω c given by, ξ = I 1 I 3ω c, ω c I 3 MghI1. The function f(z) is plotted in the figure for three values of r: r < 1, r = 1, and r > 1, corresponding to three different values of ω 3 : ω 3 < ω c, ω 3 = ω c, ω 3 > ω c. The three zeros of the function f(z) (solutions to ż = 0) are values of z such that the motion is stationary (stable or turning point), since they correspond to θ = 0. We can see that the equation has two zeros at z = 1 and a third zero at z = r 1 such that, z > 1 (unphysical) corresponds to r > 1, i.e. ω 3 > ω c ; z < 1 (physical) corresponds to r < 1, i.e. ω 3 < ω c. We can therefore describe the motion of the top as following: for ω 3 ω c the top spins vertically (θ = 0 is the only allowed position; for ω 3 < ω c the top spins nutating between θ = 0 and θ = arccos(r 1).

5 If the top is set to spin vertically (θ = 0) with ω 3 ω c it will be stable, otherwise it will nutate. In the presence of friction, even if the top is started vertically with ω 3 > ω c, friction will eventually reduce its angular velocity until it drops below ω c and the top starts nutating. When friction is very low the top can spin vertically for a long time before nutations set in (case of a sleeping top).

6 Problem 3 x 3 l ω The center of mass in the fixed coordinates is and for this setup we have CM = (l,l,l), ω = 1 (1,1,) ω x 1 4l explicitly: CM L x (a) I 1 = 8 m α (yα +z α ) = m(yα +z α ) α α=1 = 8m(l +4l ) = 40ml I = α m α (x α +z α) = 40ml = ω ˆn, where ˆn = 1 (1,1,). Since ω isconstant (inboththefixed frameandthebody frame since we have ( ) ( ) d ω d ω =. dt fixed dt body From the symmetry of the problem we can tell that I 1 = I I 3 (symmetric top). We can calculate these I 3 = α m α (x α +y α ) = 8m(l +l ) = 1ml I 1 = α m α x α y α = m(l +l l l +l +l l l ) = 0 Where the other off-diagonal elements vanish similarly. Thus, 40ml 0 0 Î = 0 40ml ml Since the angular velocity is constant, this is not a force-free motion, since we know that the angular velocity of a symmetric top in the absence of forces precesses about the fixed direction of the angular momentum. Indeed the angular momentum is not constant in the fixed frame. We have that: L body = I 1 ω 1 ê 1 +I ω ê +I 3 ω 3 ê 3 = 1 ωml (40,40,3) = 8 ml ω(5,5,4) = constant.

7 See the figure for this vector. Therefore; ( ) d L dt fixed = ω L, and we see explicitly that L is not constant in the fixed frame. In that this force tells us that L precesses about the direction of ω. We can also see that L (ê 3 ω) = L ( ω ê 1 +ω 1 ê ) = (I 1 I )ω 1 ω = 0. So, both L and ê 3 precess about the direction of ω, keeping in the same plane with respect to each other and with respect to ω. (b) We can use Euler s equations, observing that in this frame ω 1 = ω = ω 3 = 0, giving N 1 = (I I 3 )ω ω 3 N = (I 3 I 1 )ω 3 ω 1 N 3 = (I 1 I )ω 1 ω From this we find N 1 = (40 1)ml ω = 8ml ω N = (1 40)ml ω = 8ml ω N 3 = 0 N = 8ml ω ( 1,1,0).

8 Problem 4 x 3 α ω In this problem the body axes are the principal axes, and ω can move in the the body fixed frame. It s easy to see that the plane is a symmetric top. Therefore, in absence of forces L will be constant and ω will precess around it. Let us calculate the moments of inertia explicitly: l x 1 O x l/ l/ I 1 = I = ρ dx x l/ l/dy = ρ 1 l 3 l 3 8 = ml 1 I 3 = ρ l/ l/ l/ dx dy (x +y ) = ml l/. Now at t = 0, ω = ( ωsinα, ωsinα ),ωcosα, and the angular momentum is L = (I 1 ω 1,I ω,i 3 ω 3 ) = ml 1 ( ωsinα, ωsinα ),ωsinα. The velocity with which ω precesses about L is (see discussion in class and in the text): where Ω pr = L I 1, L = (I 1 ω1 +I ω +I 3 ω3) 1/ = ml ω = ml ω 1 (1+3cos α) 1/. And so the frequency of precession is Ω pr = (ml ω/1)(1+3cos α) 1/ ml /1 [ sin α 8 + sin α 8 = ω(1+3cos α) 1/. ] 1/ +cos α