Solution Set Seven. 1 Goldstein Components of Torque Along Principal Axes Components of Torque Along Cartesian Axes...

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1 : Soution Set Seven Northwestern University, Cassica Mechanics Cassica Mechanics, Third Ed.- Godstein November 8, 25 Contents Godstein Components of Torque Aong Principa Axes Components of Torque Aong Cartesian Axes Probem #2: Rotating Pate. 4 3 Godstein Godstein Probem #5: Beads on Rods The Hamitonian Hamitonian Equations Probem #6: Massfu Puey. 9

2 Godstein 5.8. A dumbbe is formed by connecting two sma spherica masses of mass m with a massess rod of ength 2b. The rod is attached to an axe in such a way that it makes a constant ange φ with the axe. The dumbbe rotates about the axe at a rate ω, as shown in Figure. The anguar momentum about this axis was determined in the ast homework as sin 2φ cos ωt L = mωb 2 sin 2φ sin ωt. () 2 sin 2 φ The right handed coordinate system is set up so that ω points aong ẑ, with the origin at the point the barbe crosses ω. Figure : Depiction of the obique dumbbe system examined in probem #.. Components of Torque Aong Principa Axes. The principa axes of a body are the set of orthonorma eigenvectors found from diagonaizing the inertia tensor Î. The components of this tensor are isted in the previous homework; using Mathematica, the principa moments (eigenvaues) are I = 2mb 2 I 2 = 2mb 2 I 3 =, (2) and the corresponding eigenvectors are cot φ sec ωt tan ωt tan φ cos ωt w = w 2 = w 3 = tan φ sin ωt, (3) which are not yet orthonormaized. Note that because w 3 corresponds to a nondegenerate eigenvaue, it can be chosen as a principa axis, and the other two can be made orthogona to it. The remaining two eigenvectors, since they are degenerate, are both orthogona to w 3, but not to each other. Therefore another vector can be chosen as a principa axis, for simpicity w 2 is chosen, so that the ony eigenvector that must be orthonormaized is w. Before doing so, the other two vectors must be normaized. The norm of a vector w i is w i w i (note that assuming < φ < π/2 means that a trig functions are positive definite), so the first two principa axes are ê 2 = sec ωt w 2 = w 2 cos ωt = sin ωt cos ωt ê 3 = sin φ cos ωt sec φ w 3 = w 3 cos φ = sin φ sin ωt. (4) cos φ To find the orthonorma vector w 2, the Gram-Schmidt procedure is used, with the projection operator defined as ê 2 = w 2 projê (w 2 ) projê3 (w 2 ), (5) proj a (b) = a b a a a. (6) Page 2 of 9

3 Note that w is aready orthogona to w 3, so projê3 (w ) =. Therefore the ast orthogona vector is cot φ cos ωt cot φ cos ωt cos φ cos ωt e = cot φ sin ωt ê = cot φ sin ωt = cos φ sin ωt. (7) csc 2 φ sin φ Now that the three principa axes have been determined, the torque projections aong each of these axes can be found by knowing the anguar veocity projected onto the appropriate axis. The components of ω in the coordinate basis are (,, ω), so the transformed anguar veocity in the principa axis basis is (ω ê ) sin φ ω = projê (ω) + projê2 (ω) + projê3 (ω) = (ω ê 2 ) = ω. (8) (ω ê 3 ) cos φ With these components, the torque can be found usinng Euer s equations, given by Godstein Equation Since none of the components of ω are expicity time dependent, the first term in each torque vanishes. Additionay, two of the components depend on the vaue of ω 2, which in this case is zero, so the ony projection that is nonzero is aong ê 2. Therefore the torque aong the principa axes is τ = ω 3 ω (I 3 I ) = ω 2 sin φ cos φ(2mb 2 ) = mω 2 b 2 sin(2φ)ê 2. (9) The torque aong the coordinate axis, from this, is just this mutipied through by ê 2, which has components in the coordinate basis: sin ωt τ = τ ê + τ 2ê2 + τ 3ê3 = mω 2 b 2 sin 2φ cos ωt. ().2 Components of Torque Aong Cartesian Axes. The components of the torque in the coordinate basis can be computed directy, without knowing the principa moments. If the anguar momentum is known (it was cacuated in the ast homework), the net torque can be found by Godstein Equation 5.37, b 2 mω 2 sin 2φ sin ωt sin ωt τ = ω L = b 2 mω 2 sin 2φ cos ωt = mω 2 b 2 sin 2φ cos ωt, () which agrees with the resut from the previous section. Page 3 of 9

4 2 Probem #2: Rotating Pate. Consider a rectanguar pate of sides a and b rotating about a diagona axis (orner to corner) with anguar veocity around that axis of ω. Aign a Cartesian coordinate system so the rectange ies in the x z pane, with the origin such that the rectange extends to z = ±a/2 and x = ±b/2, as shown in Figure 2. The components inertia tensor for this pate are given by a b I ij = ρ(r) [(x 2 + y 2 + z 2 )δ ij r i r j ] d 3 r = σ [(x 2 + z 2 )δ ij r i r j ] dx dz, (2) V with r k {x, y, z}, because a point in the body aways has a zero y component of its position. This aso means a cross terms with a y component vanish. The diagona eements are I xx = σ I yy = σ I zz = σ a/2 b/2 a/2 b/2 a/2 b/2 a/2 b/2 a/2 b/2 a/2 b/2 and the ony potentiay non-zero off-diagona terms are I xz = I zx = σ [(x 2 + z 2 ) x 2 ] dx dz = a2 M 2 (3) [(x 2 + z 2 )] dx dz = 2 M ( a 2 + b 2) (4) [(x 2 + z 2 ) z 2 ] dx dz = b2 M 2, (5) a/2 b/2 a/2 b/2 [ xz] dx dz =. (6) Note that because the inertia tensor aong these axes is diagona, they must be a set of principa axes. The torque required to maintain the rotation is given by τ = dl dt = L ω. (7) The anguar veocity vector points in a direction in the x z pane proportiona to the ratio of side engths of the rectange. The corners are a distance a 2 + b 2 from each other and the components are b in the x direction and a in the z direction. This gives the unit vector pointing aong ω, ˆω = a 2 + b (bî + aˆk), (8) 2 given the vector has a magnitude ω, the compete vector is ω = ω ˆω. The anguar momentum of this rotation is given by L = Î ω = M a ω 2 b a 2 + b 2 = M a ω 2 b, (9) 2 a 2 + b 2 b 2 2 a a 2 + b 2 ab 2 where Î is the inertia tensor. The torque can be now be found directy from Godstein Equation 5.37 τ = dl dt + ω L = M ω ω î ĵ ˆk 2 a 2 + b 2 a 2 + b 2 b a a 2 b ab 2, (2) working out the determinant and noting that the ony surviving term is in the y direction, the torque is τ = M ω 2 2 a 2 + b 2 (a3 b ab 3 )ĵ = Mω2 ab b 2 a 2 2 a 2 + b 2 ĵ. (2) Page 4 of 9

5 3 Godstein 5.6. Three equa mass points (mass m) are ocated at (a,, ), (, a, 2a), and (, 2a, a) for the coordinates (x, y, z). The components of the inertia tensor Î about the origin are given by I jk = 3 m (i) [r(i) 2 δ jk r j(i) r k(i) ], (22) i= where i is an index that corresponds to the i th mass, whie j, k {x, y, x}. The Cartesian distance for each mass is {a 2, 5a 2, 5a 2 }. The first component is I = m 3 [r(i) 2 δ jk x (i) x (i) ] (23) i= = m[(a 2 a a) + (5a 2 ) + (5a 2 )], (24) and the other components foow. Therefore the inertia tensor is given by I = ma 2 6 4, (25) 4 6 the principa moments are the eigenvaues of this matrix, and the principa axes are the normaized eigenvectors, assuming they are a orthogona. Using Mathematica the principa moments are I = ma 2 I 2 = ma 2 I 3 = 2ma 2, (26) and the principa axes are ê = ê 2 = ê 3 =, (27) 2 2 which are in fact orthogona. Figure 2: Depiction of the coordinate system and rectange orientation for probem #2. Figure 3: Depiction of the coordinate system for the Foucaut penduum in Probem #4. Page 5 of 9

6 4 Godstein The Foucaut penduum experiment consists in setting a ong penduum in motion at a point on the surface of the rotating Earth with its momentum originay in the vertica pane containing the penduum bob and the point of suspension. Consider a massess penduum of ength suspending a mass m in a uniform downwards gravitationa acceeration. Pick a right handed reference frame on the surface of the earth such that the origin is set to the equiibrium position of the penduum. From here the z axis points verticay (norma to the surface of the earth). In the sma osciation imit, the motion of the mass is confined to the horizonta (x y) pane. In this approximation, z and ż can be set to zero. Using Newton s second aw the vector equation of motion for the penduum mass is m r = mg + T 2mω ṙ, (28) as given by Thornton and Marion Equation.42. In this expression r is the distance vector, and T is the tension vector which aways points towards the pivot. The fina term is the force due to the Coriois effect, with ω as the frequency of the Earth s rotation. The projections of T in the x y pane must aways be negative because the mass wi aways be attracted to the equiibrium position at (x, y) = (, ). The projections in this pane are sma (in the sma ange approximation) so the projection in the z axis is approximatey T. For this sma ange approximation (to first order), the projections into x and y can be expressed as the ratio of the coordinate to the ength of the penduum, because sin θ θ. Therefore the equation of motion for each projection becomes ẍ ÿ = + T x ω x ẋ y m 2 ω y ẏ, (29) g ω z so a that remains is the components of anguar veocity. As shown in Figure 3, ω does not have any component in the y direction (out of the page). Given the co-atitude θ, the anguar veocity vector is ω = ( ω cos θ,, ω sin θ). So the cross product in Equation 28 becomes The three equations of motion then become î ĵ ˆk ωẏ sin θ ω ṙ = ω cos θ ω sin θ ẋ ẏ = ( ωẋ sin θ). (3) ωẏ cos θ ẍ = T x 2( ωẏ sin θ) m (3) ÿ = T y 2(ωẋ sin θ) m (32) z = = g + T 2( ωẏ cos θ), (33) m which in the imit that the Earth s rotationa frequency is negigibe compared to the frequency of the penduum, the term with ω can be dropped in the ast equation. Therefore the ast equation can be soved with T mg. This substitution yieds a pair of couped second order equations ẍ = g x + 2ωẏ sin θ (34) ÿ = g y 2ωẋ sin θ. (35) Page 6 of 9

7 If a coordinate ξ is defined to be x + iy, then ξ = g ξ 2iω ξ sin θ ẍ + iÿ = g (x + iy) 2iω(ẋ + iẏ) sin θ (36) = g x 2iωẋ sin θ g iy 2ω(i2 ẏ) sin θ = [ g ] x + 2ωẏ sin θ + i [ g ] y 2ωẋ sin θ (37) and the couped equations can be merged to one, such that Re[ξ(t)] = x(t) and Im[ξ(t)] = y(t). The soutions to the differentia equation in ξ (given by Mathematica ) are ( ) ( )] ξ(t) = e [c iω sin θt exp t g2 2 ω2 sin 2 θ + c 2 exp t g2 2 ω2 sin 2 θ, (38) note that for ω = (no Earth rotation) the equation of motion reduces to the simpe penduum resut, with frequency ω = g/. It has been previousy stated that the rotationa frequency of the Earth is negigibe compared to the frequency of the penduum, the soution can be rewritten using Euer trig identities ξ(t) = e iω sin θt [A cos(ω t + δ ) + ib sin(ω t + δ 2 )]. (39) Assume the penduum s momentum is originay in the vertica pane of the penduum mass and pivot point, so ẋ = v and ẏ =, and it is in the equiibrium position at t =. Consider the first derivative, ẋ() + iẏ() = [( iω sin θ) {A cos(δ ) + ib sin(δ 2 )} + ω { A sin(δ ) + ib cos(δ 2 )}], (4) which impies B =. This is because the eft side is rea ξ() = v, which is entirey in x, and equating rea and imaginary parts means that either B = or the two imaginary parts exacty cance. This can not be the case because the assumption ω ω has been made. The constraints given by this and the initia conditions give = A cos(δ ) (4) v = ( iω sin θ) {A cos(δ )} + ω { A sin(δ )}. (42) The first constraint is that δ is an odd integer mutipe of π/2, so that A = v/ω. Therefore the soution is ξ(t) = e iω sin θt v ( cos ω t + π ) (43) ω 2 = v ( cos ω t + π ) [cos (ω sin θt) i sin (ω sin θt)], (44) ω 2 so the equations of motion in the origina coordinates are x(t) = v ( cos ω t + π ω 2 y(t) = v ( cos ω t + π ω 2 ) cos (ω sin θt) (45) ) sin (ω sin θt). (46) This gives the ange of precession of the penduum pane to be Θ = Ωt = ω sin θt, thus the precessiona frequency is Ω = ω sin θ, and the rotationa frequency of Earth is 2π radians per day. Knowing the reationship between atitude (θ) and coatitude (θ c ) is θ + θ c = (π/2), and sin([π/2] α) = cos(α), the precessiona frequency is Ω = 2π cos θ c per day. Therefore in one day the pane of osciation rotates 2π cos θ c radians, the direction of rotation is based on the sign of the cosine. In the northern hemisphere θ c < π/2, so cosine is positive and the pane of osciation rotates countercockwise. Contrariy, in the southern hemisphere π/2 < θ c π, so the cosine is negative and the penduum precesses cockwise. Page 7 of 9

8 5 Probem #5: Beads on Rods. A bead of mass m is free to side on a frictioness straight rod, which ies in a horizonta pane. The rod is spun with a constant anguar veocity ω about a vertica axis through the midpoint of the rod. 5. The Hamitonian. Consider the poar coordinates r and θ = ωt, with the origin set in the horizonta pane of the rod, at the axis of rotation. The ocation of the bead is then (r, ωt), with speed (ṙ, ω). The kinetic energy of this system is given by T = 2 m(ṙ2 + r 2 θ2 ) = 2 m(ṙ2 + r 2 ω 2 ) = L, (47) which is aso the Lagrangian and tota energy of the system because there is no potentia. So the conjugate momentum for r is p = L q = ṙ 2 m(ṙ2 + r 2 ω 2 ) = mṙ. (48) The Hamitonian is then H = qp L = ṙp 2 mṙ2 2 mr2 ω 2 = p m p ( p ) 2 2 m m 2 mr2 ω 2 (49) = 2m p2 2 mω2 r 2 = 2 mṙ2 2 mω2 r 2 T, (5) so the Hamitonian is not equa to the tota energy of the system. 5.2 Hamitonian Equations. For this system, Hamiton s equations are q = H p ṙ = p m and ṗ = H q ṗ = H r = ( mω 2 r ), (5) differentiating the first equation with respect to time and equating to the second yieds which is the equation of motion for r. m r = mω 2 r, (52) Figure 4: Depiction of the coordinate system for the Atwood in machine in Probem #6. Page 8 of 9

9 6 Probem #6: Massfu Puey. Consider an Atwood machine with masses m and m 2 suspended over a puey of mass M and radius R in a uniform downwards gravitationa acceeration g. This is a one dimensiona probem because the motion of both masses and the puey can be entirey represented by one coordinate x. Define the origin of the coordinate system to be the eve of the center of mass of the puey, with positive x the distance downwards, as shown in Figure 4. Therefore if one mass moves in a distance +x, the other moves a distance x. Let the ength of the string be + πr 2, so that when the masses are at their farthest possibe distance one is at x = and one is at x =. The potentia energy of a mass µ at this position is µg, with g >. It is usefu to note the moment of inertia for a uniform disk about an axis through its center, norma to the circuar face, is I = 2 MR2. This information aows the Lagrangian to be written down immediatey, L = 2 m ẋ m ( ẋ) Iω2 [ m gx + m 2 g( x)], (53) where ω is the rotationa frequency of the puey, in this case just ẋ/r. Simpification and substitution yieds L = 2 (m + m 2 )ẋ 2 + [ ] ( ) ẋ MR2 + m 2 g + g(m m 2 )x (54) R L = (m + m ) 2 M ẋ 2 + m 2 g + g(m m 2 )x. (55) Therefore the conjugate momentum for x is p = L ẋ = Additionay, the Hamitonian is given by (m + m M ) ẋ. (56) H = ẋp L = p ( m + m M)p ( m + m 2 2 ( M) 2 m + m M) 2 p2 m 2 g g(m m 2 )x (57) = 2 p2 ( m + m M) m 2g g(m m 2 )x, (58) after substituting in Equation 56, note this is the tota energy. Hamiton s equations of motion are q = ẋ = H p = p m + m M ṗ = H q = [ g(m m 2 )], (59) which can be reduced for a second order differentia equation in x. By differentiating q with respect to time and substituting in ṗ, this becomes ẍ = g(m m 2 ) m + m M. (6) This answer makes sense because it is either monotonicay decreasing or increasing, or zero depending on m /m 2, with constant acceeration. This is due to the heavier mass moving the system due to gravity, but fought by the other mass moving upward as we as the inertia of the puey. The sign of the answer makes sense, for if m 2 > m, m wi acceerate upwards, which is the negative direction. Page 9 of 9