PHY 5246: Theoretical Dynamics, Fall 2015 Assignment # 9, Solutions

1 Graded Problems

Problem 1 (1.a) In order to find the equation of motion of the triangle, we need to write the Lagrangian, with generalized coor- dinate θ . The potential energy is going to be based on y { } the location of the center of mass, and we find that via l/2 l/2 ρ l/2 0 − x yCM (θ =0) = 2 dx dyy m 0 √3(l/2 x) Z Z− − 2 m l/2 1 l 2 CM = dx 3 x −m 1 √3 0 2 2 − ! 2 l 2 l Z 2 8 3 1 l l/2 θ = x √3l2 2 3 2 − ! 0

3 4 l l √3l = = , − 2 −√3l2 8 −2√3 where the density is ρ. This is as expected by symmetry considerations and by the fact that the CM has to be at the geometrical center of the triangle, i.e. at one third of it’s height, 1 √3 l l = . −3 2 −2√3 Note the negative sign is because the triangle lies below the x-axis. So, for generalized θ, ρ yCM = cos θ, −2√3 and the potential energy is l U = mgyCM = mg cos θ. − 2√3 To calculate the kinetic energy, 1 T = T (about 0) = I θ˙2, rot 2 3 we need first to find the principal moment of inertia about the axis of rotation, which is an axis perpendicular to the plane of the triangle, through 0. This is the only moment of inertia we need, since ˙ ~ω = θˆe3 and therefore, 1 1 1 T = ~ωT ˆI ~ω = θ˙2I = θ˙2I . 2 · · 2 33 2 3 l/2 0 2 2 I3 = 2ρ dy(x + y ) 0 √3(l/2 x) Z Z− − l/2 0 l/2 0 = 2ρ dxx2 dy +2ρ dx dyy2 0 √3(l/2 x) 0 √3(l/2 x) Z Z− − Z Z− − l/2 l l/2 1 l 3 = 2ρ dxx2√3 x +2ρ 3√3 x Z0 2 − ! Z0 3 2 − ! 1 l l3 1 l4 1 l4 = 2ρ√3 + 3 2 8 − 3 16 4 16! m 1 l4 ml2 = 2√3 = . 1 √3 2 3 16 6 2 2 l

Therefore 1 ml2 T = θ˙2, 2 6 ! and the Lagrangian is 1 ml2 l L = T V = θ˙2 + mg cos θ. − 2 6 ! 2√3 The equation of motion for our generalized coordinate is

d ∂L ∂L ml2 l =0 θ¨ + mg sin θ =0 dt ∂θ˙ − ∂θ → 6 2√3 g θ¨ + √3 sin θ =0. ⇒ l Under small oscillations about θ = 0 this is the equation for a simple harmonic oscillator with frequency g ω = √3 . r l (1.b) For this situation, the θ = 0 position of the CM is y (θ =0)= l . Therefore CM − √3 y l l yCM = cos θ U = mg cos θ. −√3 ⇒ − √3

We need to calculate the moment of inertia I3 with re- l/2 l/2 spect to the new center of rotation (and axis through − x it).

l/2 √3x − 2 2 I3 = 2ρ dx dx(x + y ) 0 √3/2l Z Z− l/2 √3x l/2 √3x 2 − − 2 CM = 2ρ dxx 3 dy +2ρ dx 3 dyy θ 0 √ l 0 √ l Z Z− 2 Z Z− 2 l/2 l l/2 1 l3 = 2ρ√3 dxx2 x + + dx 3 x3 + √3 0 2 0 3 8 l "Z − ! Z − !# − 2 1 l4 1 1 l4 l4 = 2ρ√3 + l4 + −4 16 3 − 4 16 16! m l4 3+2+6 5 = 2√3 − = ml2. 1 √3 2 16 6 12 2 2 l Thus the kinetic energy and Lagrangian are 1 5 T = ml2 θ˙2 2 12  1 5 l L = T V = ml2 θ˙2 + mg cos θ, − 2 12  √3 and the equation of motion is d ∂L ∂L 5 l =0 ml2θ¨ + mg sin θ =0 dt ∂θ˙ − ∂θ → 12 √3 g 12 1 θ¨ + sin θ =0. ⇒ l 5 √3 Thus the frequency of the small oscillation is

12 g ω = . s5√3 l Problem 2 (a)

y y y = l cos θ CM √2

θ

l

The cube rotates about the side perpendicular to the plane of the figure, through 0. We are therefore referring to the case in which the frame has axes parallel to the side of the cube and origin at one corner. In figure (1), the cube is rotating about the back side, along the y-axis. The 2 2 corresponding moment of inertia is I22 = 3 ml (also see p. 409-410 in Goldstein). We can find the angular velocity of the cube when one face strikes the plane, by imposing conservation of energy:

l Ei = Ui = mg √2

l 1 2 2 2 Ef = Uf + Tf = mg + ml ωf 2 2 3  l l 1 2 2 Ei = Ef mg = mg + ml ω . ⇒ √2 2 3 f

1 1 1 2 3 g g = lωf ωf = (√2 1) . ⇒ √2 − 2! 3 ⇒ s2 − l z

O y

x

Figure 1:

(b) If the cube can slide, there is no more fixed point. So, we have to calculate the kinetic energy as

CM CM T = Ttrans + Trot 1 2 1 1 2 2 = mvCM + ml ωf , 2 2 6  where: l l vCM =y ˙CM = sin θθ˙ = θ,˙ −√2 −2 circ and when θ = 45 , the cube hits the plane. With ω(θ = 45◦)= ωf , we have

1 l2 1 5 T = m ω2 + ml2ω2 = lω2, 2 4 f 12 f 24 f Or 1 1 5 2 12 g g = lωf ωf = (√2 1) . √2 − 2! 24 ⇒ s 5 − l We could also have calculated the kinetic energy as

CM CM T = Ttrans + Trot 1 2 1 CM 2 1 1 1 2 2 = mvCM + I22 ωf = + l ωf 2 2 2 2 6 1 = l2ω2 = T origin. 3 f rot

2 ˙ l CM ml Now use v = Rθ to find vCM = √2 ωf , and on p. 422 of Goldstein we can find I22 = 6 . We thus obtain the same answer as above. Problem 3 (Goldstein 5.17)

Take M as the mass of the cone, (x1′ , x2′ , x3′ ) as the principal axes going through the CM, which 3 is at (0, 0, a = 4 h) in the body system. x3′ is the cone axis, x2′ is perpendicular to the plane going through x3′ and the segment OA, and x1′ is orthogonal to the plane of x2′ and x3′ . As stated in the problem, R is the radius, h is the height and α is the half-angle of the cone. Due to the condition of no-slipping, the angular velocity of the cone, ~ω, is directed along the instantaneous axis of rotation which is the line of contact between the surface of the cone and the surface on which the cone is rolling (i.e. along the segment OA, see figure).

x3

x′1

O a x2 α θ CM x R ′3

x1 x′2 A ~ω

(a) The kinetic energy (T ) can be calculated either using the center of mass as origin and taking as reference frame the system of principal axes described above, or using a system of axes parallel to that one but with origin at the apex of the cone O, which is a fixed point for the body. In the first case, the kinetic energy is given by the sum of two terms: the kinetic energy of the center of mass (i.e. the kinetic energy of a pointlike object of mass M located at the CM of the cone and moving as CM moves) and the kinetic energy of the body as rotating about the center of mass (expressed in terms of the principal moments of inertia and the component of the angular velocity in the principal axes frame or body frame),

CM CM T = Ttrans + Trot kinetic energy 1 1 T CM = = Mv2 = Ma2 cos2 αθ˙2. trans of the CM 2 CM 2     ˙ In the above equation we have used that vCM = a cos αθ as it rotates about ˆx3. This implies that ω (angular velocity about OA) and θ˙ are related. Thus we also have

vCM = a sin αω. Combining these two relations we find v ω = CM = θ˙ cot α. (1) a sin α The components of ~ω in the body frame are cos2 α ~ω =(ω sin α, 0,ω cos α)=(θ˙ cos α, 0, θ˙ ). sin α

This follows from our choice of x2′ and because of the relation (1). Therefore: 1 1 T CM = I ω2 + I ω2 rot 2 1 1 2 3 3 2 I = I = 3 M R2 + h = 1 2 20 4 I = 3MR2.  n 3 10 2 4 1 3 2 h 2 ˙2 1 3 2 cos α ˙2 = M R + cos αθ + MR 2 θ 2 20 4 ! 2 10 sin α 2 4 CM CM 1 2 2 ˙2 1 3 2 h 2 ˙2 1 3 2 cos α ˙2 T = Ttrans + Trot = Ma cos αθ + M R + cos αθ + MR 2 θ 2 2 20 4 ! 2 10 sin α 9 3 1 3 = Mh2 cos2 αθ˙2 + Mh2 tan2 α + cos2 αθ˙2 + Mh2 cos2 αθ˙2. 32 40  4 20 3 To get this last line we used a = 4 h, R = h tan α, and the next line uses the trigonometric identity, 2 1 tan α = 2 1. cos α − 9 3 3 3 3 T = Mh2 cos2 αθ˙2 Mh2θ˙2 Mh2 cos2 αθ˙2 + Mh2 cos2 αθ˙2 32 40 − 40 4 20 3 = Mh2θ˙2(1+5cos2 α). 40 In the second case, the kinetic energy is purely rotational, since O is a fixed point, and is given by: 1 1 T = T O = J ω2 + J ω2 , rot 2 1 1 2 3 3 where J1 and J3 are the principal moments of inertia with respect to a system of axes parallel to the ones drawn in the picture and with origin in O. They can be found using Steiner’s theorem and they are simply given by: J = I + M(a2 a a )= I + Ma2 , 1 1 − 1 1 1 J = I + M(a2 a a )= I , 3 3 − 3 3 3 where we have used that ~a = (0, 0, a) is the position vector of O in the CM frame. It is then − easy to see that, 1 1 1 1 1 1 T = T O = I ω2 + Ma2ω2 + I ω2 = Ma2 cos2 αθ˙2 + I ω2 + I ω2 = T CM + T CM . rot 2 1 1 2 1 2 3 3 2 2 1 1 2 3 3 trans rot (b) The components of the angular momentum are given by

~ ˆ L = L1ˆe1′ + L2ˆe2′ + L3e′3

= I1ω1ˆe1′ + I2ω2ˆe2′ + I3ω3ˆe3′ .

Therefore we know that L2 = 0, and

2 3 2 h ˙2 L1 = I1ω1 = M R + cos αθ 20 4 ! 3 1 3 = Mh2 cos α θ,˙ 20 cos α − 4  3 cos2 α 3 cos2 α L = I ω = MR2 θ˙ = Mh2 tan2 α θ˙ 3 3 3 10 sin α 10 sin α 3 = Mh2 sin αθ.˙ 10 Problem 4 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 ~ω ~ x3 a symmetric top. Therefore, in absence of forces L will be constant and ~ω will precess around it. α Let us calculate the moments of inertia explicitly:

l/2 l/2 3 2 1 2l l l I1 = I2 = ρ dx dy x = ρ 2 x2 l/2 l/2 3 8 2 O Z− Z− ml2 = x1 12 l/2 l/2 2 2 2 ml I3 = ρ dx dy (x + y )= . l/2 l/2 6 Z− Z− Now at t = 0,

ω sin α ω sin α ~ω = , ,ω cos α , √2 √2 ! and the angular momentum is

ml2 ω sin α ω sin α L~ =(I1ω1,I2ω2,I3ω3)= , , 2ω sin α . 12 √2 √2 !

The velocity with which ~ω precesses about L~ is (see discussion in class and in the text):

L Ωpr = , I1 where

2 2 2 1/2 2 2 2 1/2 ml ω sin α sin α 2 L = (I1ω1 + I2ω2 + I3ω3) = + + cos α 6 " 8 8 # ml2ω = (1+3cos2 α)1/2. 12 And so the frequency of precession is

(ml2ω/12)(1 + 3cos2 α)1/2 Ω = = ω(1+3cos2 α)1/2. pr ml2/12 2 Non-graded Problems

Problem 5 Take a generic point P and the principal axis going through it (see figure). The principal moments of inertia are

3 2 2 x3 I1 = d rf(~r) ~r x1 V − Z h i 3 2 2 I2 = d rf(~r) ~r x2 V − Z h i 3 2 2 I3 = d rf(~r) ~r x3 , V − V Z h i where f(~r) is the mass density. Therefore

3 2 2 2 Ii + Ij = d rf(~r) 2~r xi xj V − − Z h i 3 2 2 2 x2 = d rf(~r) xi + xj +2xk V P Z h i 3 ~ 2 2 d rf(R) xi + xj ≤ V Z h i 3 2 2 x d rf(~r) ~r xk = Ik. 1 ≤ V − Z h i The equality in the above statement occurs when xk = 0. Here we have assumed that i = j = k for i, j, k =1, 2, 3. 6 6 Problem 6 Using cylindrical coordinates, x = r cos θ z y = r sin θ ( z = z, and we have that r = (R/h)z. First determine the inertia tensor with respect to 0: CM h 2π Rz/h 2 2 I11 = ρ dz dθ dr r(y + z ) 0 0 0 3 Z Z Z h h 2π Rz/h 4 M 2 2 2 2 = 1 dz dθ drR (r sin θ + z ) πR2h 0 0 0 z 3 Z Z Z h 4 h 2 M 1 R 4 2 1 R 2 = 1 2 π 4 z dz +2π z 2 z dz y πR h " 0 4 h 0 2 h # r 3 Z Z M 2 1 2 1 2 3 2 2 θ = 1 2 R hπ R + h = M(R +4h ) 3 πR h 20 5  20 x I22 = I11 by symmetry h 2π Rz/h M 2 2 I33 = 1 2 dz dθ dr r(x + y ) 3 πR h Z0 Z0 Z0 4 5 M R h 3 2 = 1 2 2π 4 = MR . 2 πR h 4h 5 10 By symmetry we can also say that the non diagonal terms are zero (each of the chosen axes is a symmetry axis for the cone). These can also be shown explicitly. For instance: M h 2π Rz/h I12 = 1 2 dz dθ dr rxy =0. − 3 πR h Z0 Z0 Z0 This vanishes because xy = r2 sin θ cos θ and 2π dθ sin θ cos θ =0. Z0 Now in a similar way the other two vanish M h 2π Rz/h I13 = 1 2 dz dθ dr rxz =0, − 3 πR h Z0 Z0 Z0 M h 2π Rz/h I23 = 1 2 dz dθ dr ryz =0 − 3 πR h Z0 Z0 Z0 where we have used 2π xz = r cos θz dθ cos θ =0, ⇒ Z0 2π yz = r sin θz dθ sin θ =0. ⇒ Z0 So, the chosen Cartesian axes are also principal axes with respect to the origin, and 3 2 20 MR 0 0 ˆ 3 2 I0 = 0 20 MR 0 ( 3 2 ) 0 0 10 MR

To obtain IˆCM , we need first to determine (xCM ,yCM , zCM ). By symmetry, we know xCM = yCM = 0, while: 1 M h 2π Rz/h ICM = 1 2 dz dr rz M 3 πR h Z0 Z0 Z0 h 2 3 1 R 2 = 2 2π dz z 2 z πR h Z0 2 h 3 1 R2 h4 3 = 2π = h. πR2h 2 h2 4 4 Finally, we can apply the generalized form of Steiner’s parallel axis theorem, according to which (I ) =(I ) M a2δ a a , CM ij 0 ij − ij − i j 3 h i which in our case ~a = (0, 0, 4 h), so we have 2 3 3 2 2 9 2 (ICM )11 = (I0)11 h M = M(R +4h ) h M − 4  20 − 16 3 h2 = M R2 20 4 !

(ICM )22 = (ICM )11 by symmetry 3 (I ) = (I ) = MR2, CM 33 0 33 10 and the non diagonal elements are still all zero. Problem 7 Defining ρ as the mass density, we use polar coordinates x = r cos θ y ( y = r sin θ (x,y)

Given a coordinate system (x′,y′) which rotates with the disk (see figure (3)), the location of the CM isx ¯CM = 0 and ρ R r y¯CM = dxdyy +2 dxdyy ρ θ M top bottom x Z Z  2ρ ρ R π R 2π = drr dθr sin θ +2 drr dθr sin θ M (Z0 Z0 Z0 Zπ ) ρ R3 R3 2 ρR3 4R = 2 4 = = . M ( 3 − 3 ) −3 M − 9π Where in the last line we have used Figure 2: πR2 πR2 3 M = ρ +2ρ = ρπR2. 2 2 2 Again referencing Figures (2) and (3) we see that the relationship between the lab coordinates and the coordinates of the center of mass are 4R xCM = Rθ y¯CM sin θ = Rθ 9π sin θ −| | −4R yCM = R y¯CM cos θ = R cos θ. n −| | − 9π ⇓ ˙ 4R ˙ x˙ CM = Rθ 9π cos θθ 4−R ˙ y˙CM = sin θθ. n 9π The kinetic energy of the disk is made of 2 terms:

T = Ttrans + Trot, where Ttrans is the translational kinetic energy of the mass M with coordinates (xCM ,yCM ) and velocity (x ˙ CM , y˙CM ), and Trot is the rotational kinetic energy about the center of mass. Thus 1 1 T = M(x ˙ 2 +y ˙2 )+ I θ˙2. 2 CM CM 2 3

Here I3 is the moment of inertia about the CM, with respect to the z-axis, perpendicular to the plane of the figure. To find I3 we will calculate it with respect to the center of disk (since it is easier!) and use Steiner’s theorem to obtain it with respect to the CM.

R π R 2π 2 2 2 2 I3 = ρ drr dθ(x + y )+2ρ drr dθ(x + y ) 0 0 0 0 π Z Z Z Z R4 R4 3 1 = ρ π +2ρ π = ρR4 π = MR2 4 4 4 2 2 2 1 2 16 R I3 = I3 My¯ = MR M CM |0− CM 2 − 81 π2

1 32 2 = MR 1 2 . 2  − 81π  Then we can calculate the kinetic energy

1 2 2 1 2 T = Ttrans + Trot = M(x ˙ +y ˙ )+ I3 θ˙ 2 CM CM 2 CM

2 2 1 2 ˙2 4 4 1 2 ˙2 32 = MR θ 1 cos θ + sin θ + MR θ 1 2 2 " − 9π  9π  # 4  − 81π  1 2 ˙2 16 8 1 16 = MR θ 1+ 2 cos θ + 2 2  81π − 9π 2 − 81π  1 3 8 = MR2θ˙2 cos θ . 2 2 − 9π  The potential energy must also be defined with respect to the center of mass, 4 U = MgyCM = MgR 1 cos θ ,  − 9π  and the Lagrangian is the sum of these two terms, 1 3 8 4 L = T V = MR2θ˙2 cos θ mgR 1 cos θ . − 2 2 − 9π  −  − 9π 

y′

y y′

θ ρ

x′ 2ρ y¯CM (xCM,yCM) x′ x

Figure 3: