Dynamics of Rigid Bodies

Transcription

1 CHAPTER Dynamics of Rigid Bodies -. The calculation will be simplified if we use spherical coordinates: x rsin cos φ y rsin sin φ z rcos ( z y Using the definition of the moment of inertia, x we have or, ij r ij xk xixj k ρ δ dv ( ρ r z dv ( cos ( cos ρ r r r dr d dφ ( R + π 4 ( cos ( cos ρ r dr d dφ R 4 πρ (4

2 4 CHAPTER The mass of the sphere is Therefore, M 4 π ρ ( R MR (6 Since the sphere is symmetrical around the origin, the diagonal elements of {} are equal: A typical off-diagonal element is ρ xy dv MR (7 ρ r sin sin φ cos φ r dr d cos This vanishes because the integral with respect to φ is zero. n the same way, we can show that all terms except the diagonal terms vanish. Therefore, the secular equation is dφ (8 (9 From (9 and (7, we have MR ( -. a Moments of inertia with respect to the x axes: i x x R CM h x x x x t is easily seen that for i j. Then the diagonal elements become the principal moments, which we now calculate. i ij The computation can be simplified by noting that because of the symmetry, ii. Then, + ρ ( x + x + x dv (

3 DYNAMCS OF RGD BODES which, in cylindrical coordinates, can be written as where ρ + r π h Rz h dφ dz ( r z rd ( M M ρ V πr h Performing the integration and substituting for ρ, we find is given by from which M R + h ( 4 ρ φ ( ρ x + x dv r rdr d dz b Moments of inertia with respect to the x i axes: MR (6 Because of the symmetry of the body, the center of mass lies on the x axis. The coordinates of the center of mass are (,, z, where ( (4 Then, using Eq. (.49, z x dv dv 4 h (7 n the present case, ij ij M a δij aa i j a a and a ( 4 h, so that Mh M R + h Mh M R + h MR (8 -. The equation of an ellipsoid is x x x + + ( a b c

4 6 CHAPTER which can be written in normalized form if we make the following substitutions: Then, Eq. ( reduces to x aξ, x bη, x cζ ( This is the equation of a sphere in the (ξ,η,ζ system. f we denote by dv the volume element in the (ξ,η,ζ system, we notice that the volume of the ellipsoid is ξ + η + ζ ( x i V dv dx dx dx abc dξ dη dζ 4 abc dτ π abc because dτ is just the volume of a sphere of unit radius. system and by dτ the volume element in the The rotational inertia with respect to the x -axis passing through the center of mass of the ellipsoid (we assume the ellipsoid to be homogeneous, is given by M ( x + x dv V ( M abc a ξ + b η d τ ( V n order to evaluate this integral, consider the following equivalent integral in which z r cos : ( sin φ a z dv a z r dr r d d π π R 4 a dφ cos sin d r dr a π (4 Therefore, and 4π a 4 ( a ξ + b η dτ π ( a + b (7 (6 M( a + b (8 Since the same analysis can be applied for any axis, the other moments of inertia are

5 DYNAMCS OF RGD BODES 7 M b + c M a + c (9-4. The linear density of the rod is ρ For the origin at one end of the rod, the moment of inertia is m m m ρ x dx f all of the mass were concentrated at the point which is at a distance a from the origin, the moment of inertia would be ( ( Equating ( and (, we find ma ( a (4 This is the radius of gyration. -. z J a M z a Q a The solid ball receives an impulse J; that is, a force F(t is applied during a short interval of time τ so that The equations of motion are ( t J F dt ( dp dt F ( dl dt r F (

6 8 CHAPTER which, for this case, yield ( t p F dt J (4 ( t L r F dt r J ( Since p(t and L(t, after the application of the impulse, we have ω p MVCM J; L ω r J ( z a J (6 ω so that and V CM J M J ω ω ( z a (8 ω (7 where M a. The velocity of any point a on the ball is given by Eq. (.: For the point of contact Q, this becomes v V + ω r (9 α CM J vq VCM ωa J α J ( z a M a Then, for rolling without slipping, v Q, and we have so that ( a ( z a ( 7 z a ( b Many billiard tricks are performed by striking the ball at different heights and at different angles in order to impart slipping and spinning motion ( English. For the table not to introduce spurious effects, the rail must be at such a height that the ball will be reflected upon collision. Consider the case in which the ball is incident normally on the rail, as in the diagram. We have the following relationships:

7 DYNAMCS OF RGD BODES 9 y V CM x Linear Momentum Before Collision p p x y MV CM After Collision p + MV x p y CM Angular Momentum L L L x y z * L x L Ly y L z * The relation between L and V depends on whether or not slipping occurs. y CM Then, we have p p J MV ( x CM L L ω J z a (4 y so that from which CM ω MV z a ( ω Ma ω a ω z a (6 MV MV V CM CM CM f we assume that the ball rolls without slipping before it contacts the rail, then V CM ωa, and we obtain the same result as before, namely, or, z a a (7 7 z a (8 Thus, the height of the rail must be at a height of ( a above the center of the ball. -6. Let us compare the moments of inertia for the two spheres for axes through the centers of each. For the solid sphere, we have s MR (see Problem - (

8 6 CHAPTER For the hollow sphere, R sin or, using 4πσ R M, we have π ( sin π 4 4 σ dφ R R sin d h πσr sin 8 πσr π h d MR ( Let us now roll each ball down an inclined plane. [Refer to Example 7.9.] The kinetic energy is + ( T M y where y is the measure of the distance along the plane. The potential energy is sin U Mg y α (4 where is the length of the plane and α is the angle of inclination of the plane. Now, y R, so that the Lagrangian can be expressed as L M y + y + Mgysin α ( R where the constant term in U has been suppressed. The equation of motion for y is obtained in the usual way and we find gmr sin α y (6 MR + Therefore, the sphere with the smaller moment of inertia (the solid sphere will have the greater acceleration down the plane.

9 DYNAMCS OF RGD BODES 6-7. x R d r φ The force between the force center and the disk is, from the figure F kr ( Only the component along x does any work, so that the effective force is This corresponds to a potential U k x. The kinetic energy of the disk is F kr sin φ kx. x where we use the result 4 T Mx + Mx ( MR for a disk and dx R d. Lagrange s equations give us Mx kx + ( This is simple harmonic motion about x with an angular frequency of oscillations k ω (4 m -8. x h x w d x We let x be the vertical axis in the fixed system. This would be the axis (i.e., the hinge line of the door if it were properly hung (no self-rotation, as indicated in the diagram. The mass of the door is Mρwhd. The moment of inertia of the door around the x axis is h w m dh w ddw whd where the door is considered to be a thin plate, i.e., d w,h. Mw (

10 6 CHAPTER The initial position of the self-closing door can be expressed as a two-step transformation, starting with the position in the diagram above. The first rotation is around the x -axis through an angle and the second rotation is around the x -axis through an angle ψ: x w x x x x x w w h x x x x x x ψ x x x ψ x x The x -axes are the fixed-system axes and the -axes are the body system (or rotating axes which are attached to the door. Here, the Euler angle φ is zero. The rotation matrix that transforms the fixed axes into the body axes ( x x x i i is just Eq. (.99 with φ and since this rotation is performed clockwise rather than counterclockwise as in the derivation of Eq. (.99: cos ψ cos sin ψ sin sin ψ λ sin ψ cos cos ψ sin cos ψ ( sin cos The procedure is to find the torque acting on the door expressed in the fixed coordinate system and then to obtain the x component, i.e., the component in the body system. Notice that when the door is released from rest at some initial angle ψ, the rotation is in the direction to decrease ψ. According to Eq. (.9, where ω ω since φ. ω N ψ ( n the body ( x i system the coordinates of the center of mass of the door are i R w h where we have set the thickness equal to zero. n the fixed ( x i system, these coordinates are t obtained by applying the inverse transformation to R; but, so that λ λ λ (4 w sin ψ t R λ R wcos cos ψ hsin + ( wsin cos ψ + hcos Now, the gravitational force acting on the door is downward, and in the is x i coordinate system F Mg e (6

11 DYNAMCS OF RGD BODES 6 There the torque on the door, expressed in the fixed system, is N R F so that in the body system we have Thus, e e e Mg w sin w cos cos + h sin w sin cos + h cos ψ ψ ψ wcos cos ψ + hsin Mg w sin ψ (7 wcos cos ψ + hsin cos ψ + wcos sin ψ N λn Mg h sin sin ψ (8 w sin sin ψ and substituting this expression into Eq. (, we have where we have used Eq. ( for N Mgwsin sin ψ (9 ψ ψ Mgw sin sin Mw (. Solving for ψ, This equation can be integrated by first multiplying by ψ : ψ g ψ sin sin ψ ( w g dt w ψψdt ψ sin sin ψψ g sin cos ψ ( w where the integration constant is zero since cos ψ when ψ. Thus, g ψ ± sin cos ψ ( w We must choose the negative sign for the radical since ψ < when cos ψ > from ψ 9 to ψ, π T. ntegrating again, dψ g sin dt cos ψ w (4

12 64 CHAPTER where T sec. Rewriting Eq. (4, Using Eq. (E. 7a, Appendix E, we find π dψ g T sin cos ψ ( w π Γ π 4 cos ψ dψ (6 Γ 4 From Eqs. (E. and (E., Γ.96 4 Γ 4 4 Γ.64 4 (7 And from Eqs. (E. and (E.4, Γ.99 4 Γ 4 4 Γ. 4 (8 Therefore, π dψ π.64.6 cos ψ. (9 Returning to Eq. ( and solving for sin, nserting the values for g, w( m, and T( sec, we find or, w sin (.6 ( gt sin.8. (

13 DYNAMCS OF RGD BODES 6-9. y C C Q P R O a x The diagram shows the slab rotated through an angle from its equilibrium position. At equilibrium the contact point is Q and after rotation the contact point is P. At equilibrium the position of the center of mass of the slab is C and after rotation the position is C. Because we are considering only small departures from, we can write Therefore, the coordinates of C are (see enlarged diagram below so that QP R ( r OA+ AC ( a x R+ sin R cos a y R cos R sin + + ( C C R A Q R P Consequently, O

14 66 CHAPTER a x R+ cos Rcos + R sin a cos + R sin a y R+ sin + R cos + Rsin from which a sin + R cos The kinetic energy is a x + y + R 4 (4 T M( x + y + ( where is the moment of inertia of the slab with respect to an axis passing through the center of mass and parallel to the z-axis: Therefore, where The potential energy is where and where Eq. ( has been used for y. The Lagrangian is The Lagrange equation for is M( + a (6 ( T f ( (7 a f M + R 4 + (8 U Mgy f (9 a f ( Mg R+ cos + R sin ( L f + f ( (

15 DYNAMCS OF RGD BODES 67 Now, d dt d L L dt ( L f L f + ( f L f + a M + R + + MR 4 ( ( f ( Combining, we find a sin cos sin (4 MR + Mg R + R R a a M + R + + MR Mg R+ sin R cos Rsin 4 ( For the case of small oscillations, and The system is stable for oscillations around only if, so that Eq. ( reduces to a Mg R + (6 Ma + 4 a Mg R ω > Ma 4 This condition is satisfied if R a >, i.e., Then, the frequency is + (7 a R > (8 Simplifying, we have ω a Mg R Ma + M + 4 ( a (9

16 68 CHAPTER ω a g R ( + 4a ( According to Eqs. (9 and (, the potential energy is a U( Mg R+ cos + R sin This function has the following forms for R> a and R< a : ( U( U( R > a Mg R a + R < a π/ π/ To verify that a stable condition exists only for R> a, we need to evaluate U a Mg sin + R cos U at : ( and U a Mg cos + R cos R sin ( so that U a Mg R (4 U a > if R > ( -. z R m When the mass m is at one pole, the z component of the angular momentum of the system is L ω MR ω z ( After the mass has moved a distance vt R along a great circle on the surface of the sphere, the z component of the angular momentum of the system is

17 DYNAMCS OF RGD BODES 69 Lz MR + mr sin φ ( where φ is the new angular velocity. Since there is no external force acting on the system, angular momentum must be conserved. Therefore, equating ( and (, we have Substituting MR ω φ ( MR + mr sin vt R and integrating over the time interval during which the mass travels from one pole to the other, we have Making the substitutions, we can rewrite (4 as φ t πr V MR ω dt (4 MR + mr sin ( vt R t φ vt R u, dt R v du ( π MR ω sin MR + mr u R du v Rω v π du + β sin u (6 where β m M and where we have used the fact that the integrand is symmetric around u π to write φ as twice the value of the integral over half the range. Using the identity we express (6 as u ( sin cos u (7 Rω φ v π du + β β cosu or, changing the variable to x u, Rω π dx φ v (9 + β β cosx Now, we can use Eq. (E., Appendix E, to obtain (8

18 7 CHAPTER ( + β tan( x Rω φ tan v + β + β π where T πrω M v M+ m M ω T ( M+ m πr v is the time required for the particle to move from one pole to the other. f m, ( becomes Therefore, the angle of retardation is or, φ ( m ωt ( ( m ( m α φ φ ( α M ω T M+ m ( -. a No sliding: P P From energy conservation, we have mg C.M. + + ω ( where v is the velocity of the center of mass when one face strikes the plane; v is related to ω by CM C.M. v CM ω ( is the moment of inertia of the cube with respect to the axis which is perpendicular to one face and passes the center: Then, ( becomes 6 m (

19 DYNAMCS OF RGD BODES 7 from which, we have ( m m ω ω m ω mg + 6 g ω ( (4 ( b Sliding without friction: n this case there is no external force along the horizontal direction; therefore, the cube slides so that the center of mass falls directly downward along a vertical line. P P While the cube is falling, the distance between the center of mass and the plane is given by y cos (6 Therefore, the velocity of center of mass when one face strikes the plane is From conservation of energy, we have y sin ω (7 π 4 π 4 from which we have mg mg + m ω + m ω (8 6 g ω ( (9 -. According to the definition of the principal moments of inertia, since ρ + ρ x + x dv + x + x dv j k i k i j ρ x + x dv+ ρ x dv + i j k i ρ x dv ( i we have ρ xi dv >

20 7 CHAPTER j + k i ( -. We get the elements of the inertia tensor from Eq..a: m x + x α α, α, α mb + 4mb + mb mb Likewise 6 mb and mb m x x α α, α, α 4mb m b mb Likewise mb and 4mb Thus the inertia tensor is {} mb The principal moments of inertia are gotten by solving mb λ 6 λ 4 4 λ Expanding the determinant gives a cubic equation in λ: Solving numerically gives λ λ λ λ. λ 4. λ 9.6 Thus the principal moments of inertia are mb 4. mb 9.6 mb To find the principal axes, we substitute into (see example.:

21 DYNAMCS OF RGD BODES 7 For i, we have ( λ λ ω ω + ω i i i i ω + 6 λ ω + 4ω i i i i ω + 4ω + λ ω i i i i ω ω + ω ω 6ω + 4ω ω 4ω + ω Solving the first for ω and substituting into the second gives Substituting into the third now gives or So, the principal axis associated with ω ω ω ω ω : ω : ω :: is x y z ( + Proceeding in the same way gives the other two principal axes: i :.8 x+.9 y.z i :.4 x+.77 y+.6z We note that the principal axes are mutually orthogonal, as they must be. -4. z y x Let the surface of the hemisphere lie in the x-y plane as shown. The mass density is given by M M M ρ V πb πb First, we calculate the center of mass of the hemisphere. By symmetry

22 74 CHAPTER x CM y CM Using spherical coordinates z CM zdv M ρ v φ we have ( z rcos, dv r sin dr d d π π b φ sin cos r dr φ r ρ zcm d d M 4 ( π b b πb 4 8 We now calculate the inertia tensor with respect to axes passing through the center of mass: z z y By symmetry, Also, by symmetry. We calculate using Eq..49: x 8 b. Thus the axes shown are the principal axes. J M v 8 where J the moment of inertia with respect to the original axes J ρ y + z dv v r sin sin φ+ r cos r sin dr d dφ v ( b π M π 4 r dr ( sin sin φ cos d φ sin d πb + r φ π Mb ( π sin + π cos sin d π Thus, from ( Also, from Eq..49 Mb Mb Mb Mb

23 DYNAMCS OF RGD BODES 7 ( J should be obvious physically So J M J ρ x + y dv v ρ r sin dr d dφ Mb v 4 Thus, the principal axes are the primed axes shown in the figure. The principal moments of inertia are 8 Mb Mb -. P g We suspend the pendulum from a point P which is a distance from the center of mass. The rotational inertia with respect to an axis through P is where is R MR + M is the radius of gyration about the center of mass. Then, the Lagrangian of the system L T U Mg ( cos ( Lagrange s equation for gives + Mg sin ( For small oscillation, sin. Then, ( or, Mg + (4 g + R + (

24 76 CHAPTER from which the period of oscillation is π τ π ω R + g (6 f we locate another point P which is a distance from the center of mass such that the period of oscillation is also τ, we can write R + R + (7 g g from which R. Then, the period must be τ π + g (8 or, τ π + g (9 This is the same as the period of a simple pendulum of the length +. Using this method, one does not have to measure the rotational inertia of the pendulum used; nor is one faced with the problem of approximating a simple pendulum physically. On the other hand, it is necessary to locate the two points for which τ is the same. -6. The rotation matrix is cos sin λ sin cos ( The moment of inertia tensor transforms according to That is ( t ( ( A+ B ( A B cos sin cos sin sin cos ( A B ( A B sin cos + C Ι λ Ι λ (

25 DYNAMCS OF RGD BODES 77 or ( ( A+ B cos + ( A B sin ( A+ B sin + ( A B cos cos sin sin cos ( A B cos ( A B sin ( A B sin ( A B cos C ( A+ B cos + ( A B cos sin + ( A+ B sin ( A B sin + ( A B cos ( A B ( A B ( A+ B ( A B + ( A+ B C cos sin sin sin cos cos ( A+ B + ( A B cos sin ( A B cos ( A B sin ( A B sin + ( A B cos ( A+ B ( A B cos sin ( C f π 4, sin cos. Then, A B (4 C ( -7. x x x

26 78 CHAPTER The plate is assumed to have negligible thickness and the mass per unit area is ρ s. Then, the inertia tensor elements are Defining A and B as above, Also, ρ s r x dx dx s ρ x + x dx dx ρ x dx dx A ( s s s ρ r x dx dx ρ x dx dx B ( ( r x dx dx ρs( x x dx dx ρ + ( s becomes Therefore, the inertia tensor has the form A+ B (4 ρs xx dx dx C ( ρs xx dx dx C (6 ρs xx dx dx (7 ρs xx dx dx (8 A C C B (9 A + B {} -8. The new inertia tensor { } is obtained from { } by a similarity transformation [see Eq. (.6]. Since we are concerned only with a rotation around the x -axis, the transformation matrix is just where λ φ, as defined in Eq. (.9. Then, Therefore, the similarity transformation is λ λ ( φ t φ φ φ λ λ ( cos sin A C cos sin sin cos C B sin cos A+ B Carrying out the operations and simplifying, we find

27 DYNAMCS OF RGD BODES 79 { } A C + B C + ( B A A + B cos sin sin cos sin C cos + B A sin Asin + C sin + Bcos ( Making the identifications stipulated in the statement of the problem, we see that A, B C (4 and Therefore A+ B A + B ( A C C B (6 A + B { } n order that and x be principal axes, we require C : x Ccos ( B A sin (7 or, C tan B A (8 from which C tan B A (9 Notice that this result is still valid if A B. Why? (What does A B mean? -9. x π/ π η x The boundary of the plate is given by r ke α. Any point (η, has the components x x η cos η sin (

28 8 CHAPTER The moments of inertia are α π ke ρ A x ηdηd π α ke ρ sin d η dη The integral over can be performed by using Eq. (E.8a, Appendix E, with the result where n the same way, 4 k A ρ P α 4πα e P 6 4 ( + α α π ke ρ B x ηdηd ( ( α π ke ρ cos d η dη (4 Again, we use Eq. (E.8a by writing cos sin, and we find Also 4 ρk B P( + 8α ( α α π ke C ρ x x ηdηd α π ke cos sin d d η ρ η n order to evaluate the integral over in this case we write Eq. (E.8, Appendix E. We find C k 4 ρ P Using the results of problem -7, the entire inertia tensor is now known. cos sin sin and use According to the result of Problem -8, the angle through which the coordinates must be rotated in order to make { } diagonal is Using Eqs. (, (, and (7 for A, B, C, we find C tan B A C B A α (6 (7 (8 (9

29 DYNAMCS OF RGD BODES 8 so that Therefore, we also have tan ( α sin + 4α α cos + 4α α 4α + Then, according to the relations specified in Problem -8, ( A A C B ( cos sin + sin Using cos ( ( + cos and sin ( ( cos, we have Now, A ( A+ B + ( A B cos Csin ( 4 ρkp A+ B ( + 4α α 4 A B 4αρk P (4 Thus, or, 4 ρkp 4 α 4 A ( + 4α αρk P ρk P α + 4α + 4α 4 A ρk P Q R ( (6 where + 4α Q α R 4a + (7 Similarly, 4 B ρk P Q+ R (8 and, of course,

30 8 CHAPTER A + B + (9 We can also easily verify, for example, that C. -. We use conservation of energy. When standing upright, the kinetic energy is zero. Thus, the total energy is the potential energy E U mg ( b is the height of the center of mass above the floor. When the rod hits the floor, the potential energy is zero. Thus b E T ω where is the rotational inertia of a uniform rod about an end. For a rod of length b, mass/length σ: Thus By conservation of energy b end σ σ x dx b T U 6 mb ω T b mg mb ω 6 mb ω g b -. Using to denote the matrix whose elements are those of { }, we can write L ω (.4 L ω (.4a t We also have x λ x and x λ x and therefore we can express L and ω as L λ L (.a t substituting these expressions into Eq. (.4, we have t ω λω (.b

31 DYNAMCS OF RGD BODES 8 and multiplying on the left by λ, t λ L λω t or by virture of Eq. (.4a, we identify t λλ L λλ ω t t ( λλ L ω t λλ (.6 -. According to Eq. (.6, Then, so that ( ij λik k λ j kl, { } λ λ tr ii ik k i i i k, k, k, λ λ k i ik i δ ( tr k k kk k { } tr { } ( This relation can be verified for the examples in the text by straightforward calculations. Note: A translational transformation is not a similarity transformation and, in general, tr { } is not invariant under translation. (For example, tr { } will be different for inertia tensors expressed in coordinate system with different origins. -. We have Then, λλ (

32 84 CHAPTER λλ λ λ λλ so that, ( This result is easy to verify for the various examples involving the cube. -4. x a/ a/ x a The area of the triangle is A a 4, so that the density is ρ M 4M A a ( a The rotational inertia with respect to an axis through the point of suspension (the origin is x x ρ x + x dx dx a ρ dx x + x dx ( a x ρa Ma When the triangle is suspended as shown and when, the coordinates of the center of mass are (, x,, where (

33 DYNAMCS OF RGD BODES 8 x M ρ M ρx dx dx a dx x dx ( a x a ( The kinetic energy is T Ma (4 and the potential energy is Mga U ( cos ( Therefore, Mga (6 L Ma + cos where the constant term has been suppressed. The Lagrange equation for is g + sin (7 a and for oscillations with small amplitude, the frequency is ω g (8 a b The rotational inertia for an axis through the point of suspension for this case is x x a x ρ + dx a dx x x The Lagrangian is now Ma Mga ( 4 L Ma + cos (9

34 86 CHAPTER and the equation of motion is g + sin a ( so that the frequency of small oscillations is ω g a ( which is slightly smaller than the previous result. -. x R ρ ρ r x The center of mass of the disk is (, x, where Now, the mass of the disk is so that ρ x x dx dx x dx dx M + lower upper semicircle semicircle ρ R π R π ( r sin rdrd + ( r sin rdrd M π ρr M M ρ πr + ρ πr ρπ R ( x ( 4 R ( 9π The direct calculation of the rotational inertia with respect to an axis through the center of mass is tedious, so we first compute with respect to the x -axis and then use Steiner s theorem.

35 DYNAMCS OF RGD BODES 87 R π R π ρ r rdr d + r rd r d π Then, 4 4 πρ R MR (4 Mx 6 MR M R 8π MR 8π When the disk rolls without slipping, the velocity of the center of mass can be obtained as follows: Thus ( ρ ρ x CM R x R x CM sin y R x CM CM CM cos x R x cos y x sin x + y V R + x R x cos CM CM where V a (6 a R + x R x cos (7 Using (, a can be written as The kinetic energy is 6 8 a R + cos (8 8 π 9 π

36 88 CHAPTER T T + T trans rot Substituting and simplifying yields Mv + (9 The potential energy is 8 cos T MR 9π ( U Mg R+ x cos Thus the Lagrangian is 8 MgR cos 9π ( 8 8 cos g cos 9π 9π ( L MR R -6. Since ω φ lies along the fixed x -axis, the components of ω φ along the body axes ( x i φ are given by the application of the transformation matrix λ [Eqs. (.98 and (.99]: ( ωφ ( ωφ ( ωφ Carrying out the matrix multiplication, we find which is just Eq. (.a. φ φ λ ( φ φ ( ωφ ( φ ( ωφ sin ψ sin ω φ cos ψ sin ( cos The direction of ω coincides with the line of nodes and lies along the x axis. The components of ω along the body axes are therefore obtained by the application of the transformation matrix λ ψ which carries the x i system into the x i system: ( ( ω ( ω ω cos ψ sin ψ ψ λ (

37 DYNAMCS OF RGD BODES 89 which is just Eq. (.b. Finally, since ω ψ lies along the body which is just Eq. (.c. Combining these results, we obtain ω x ( ωψ ( ωψ ( ωψ -axis, no transformation is required: ψ (4 ( ωφ + ( ω + ( ω ψ ( ωφ + ( ω + ( ωψ ( ωφ + ( ω + ( ωψ which is just Eq. (.. φ sin ψ sin + cos ψ φ φ cos ψ sin sin ψ cos + ψ ( -7. x x L α ω x nitially: L ω L Lsin ω ω sin α Thus From Eq. (. Since ω ω cos α, we have L Lxos ω ω cos α tan L L tan α ( ω φ cos + ψ

38 9 CHAPTER φ cos ω cos α ψ ( From Eq. (. ( becomes ψ Ω From (, we may construct the following triangle ω φ cos ω cos α ( tan α from which co s + tan α Substituting into ( gives ω φ sin α + cos α -8. From Fig. -7c we see that ω φ is along the x -axis, ω is along the line of nodes, and ω ψ is along the x -axis. Then, ψ e whee is the unit vector in the x direction. φ φ Projecting the lines of nodes into the x - and x -axes, we obtain ω φ e ( ( e cos e sin ω φ + φ ( ω ψ has components along all three of the x i axes. First, we write ω ψ in terms of a component along the x -axis and a component normal to this axis: ψ ψ ( e sin + e cos ω ( where Then, e e sin φ e cos φ (4 ( e sin sin e sin cos e cos ω ψ ψ φ φ+ (

39 DYNAMCS OF RGD BODES 9 Collecting the various components, we have ω cos φ+ ψ sin sin φ ω sin φ ψ sin cos φ (6 ω ψ cos + φ -9. When the motion is vertical. Then, according to Eqs. (. and (.4, and using Eq. (.9, we see that ( φ ψ Also, when (and, the energy is [see Eq. (.8] Furthermore, referring to Eq. (.6, P + P ( φ ψ Pψ Pφ ω ( E ω + Mgh ( E E ω Mgh ( f we wish to examine the behavior of the system near in order to determine the conditions for stability, we can use the values of P ψ, P φ, and E for in Eq. (.6. Thus, ( cos ω Mgh + + Mgh cos ( sin Changing the variable to z cos and rearranging, Eq. ( becomes ( z z Mgh ( z w + (6 The questions concerning stability can be answered by examining this expression. First, we note that for physically real motion we must have z. Now, suppose that the top is spinning very rapidly, i.e., that ω is large. Then, the term in the square brackets will be negative. n such a case, the only way to maintain the condition z is to have z, i.e.,. Thus, the motion at will be stable as long as or, 4Mgh ω < (7 4Mgh ω < (8

40 9 CHAPTER Suppose now that the top is set spinning with but with ω sufficiently small that the condition in Eq. (8 is not met. Any small disturbance away from will then give z a negative value and will continue to increase; i.e., the motion is unstable. n fact, will continue until z reaches a value z that again makes the square brackets equal to zero. This is a turning point for the motion and nutation between z and z z will result. From this discussion it is evident that there exists a critical value for the angular velocity, ω c, such that for ω > ωc the motion is stable and for ω < ωc there is nutation: Mgh ω c (9 f the top is set spinning with ω > ωc and, the motion will be stable. But as friction slows the top, the critical angular velocity will eventually be reached and nutation will set in. This is the case of the sleeping top. -. f we set, Eq. (.6 becomes Re-arranging, this equation can be written as ( Pφ Pψ cos ( cos E V + Mghcos ( Mgh ( E + Pψ + ( PφP ψ Mgh + ( E P cos cos cos φ ( which is cubic in cos. V( has the form shown in the diagram. Two of the roots occur in the region cos, and one root lies outside this range and is therefore imaginary. V( ( + cos

41 DYNAMCS OF RGD BODES 9 -. The moments of inertia of the plate are We also note that cos α + ( cosα + cos α ( cosα sin α Since the plate moves in a force-free manner, the Euler equations are [see Eq. (.4] Substituting ( and ( into (, we find These equations simplify to ωω ω ωω ω ωω ω ( α ω ω ( sin cos α ω ( α ω ω ( cos cos α ω ωω ω ω ω ω ωω tan α ω ω ω ω ( ( ( (4 ( From which we can write ntegrating, we find Now, the initial conditions are ω ωω ωω ωω ωω α (6 cot ω ω ω + ω ω cot α + ω cot α (7

42 94 CHAPTER Therefore, the equations in (7 become ω Ωcosα ω ω Ωsinα (8 From (, we can write ω ω (9 +Ω cos α ω cot α +Ω cos α ( ω ω ω and from (9, we have ω cot ω α. Therefore, ( becomes ω α ( ω cot and using ω Ω sin α ω tan α from (9, we can write ( as ω cot α ω α α tan Ω sin Since ω dω dt, we can express this equation in terms of integrals as ( Using Eq. (E.4c, Appendix E, we find dω cot α dt ω α α ( tan Ω sin Solving for ω, tan tanh ω α t tan sin sin cot α (4 α Ω α Ω α ( t α ( t ω Ωcos tanh Ω sin α ( -. a The exact equation of motion of the physical pendulum is + MgLsin where or or Mk, so we have d dt gl sin k ( ( gl dcos k d

43 DYNAMCS OF RGD BODES 9 so gl d( d ( cos k gl + a k cos where a is a constant determined by the initial conditions. Suppose that at t, and at gl that initial position the angular velocity of the pendulum is zero, we find a cos. So k finally gl k ( cos cos b One could use the conservation of energy to find the angular velocity of the pendulum at any angle, but it is exactly the result we obtained in a, so at, we have gl ω k ( cos cos.7 s -. Cats are known to have a very flexible body that they can manage to twist around to a feet-first descent while falling with conserved zero angular momentum. First they thrust their back legs straight out behind their body and at the same time they tuck their front legs in. Extending their back legs helps to resist spinning, since rotation velocity evidently is inversely proportional to inertia momentum. This allows the cat to twist their body differently to preserve zero angular momentum: the front part of the body twisting more than the back. Tucking the front legs encourages spinning to a downward direction preparing for touchdown and as this happens, cats can easily twist the rear half of their body around to catch up with the front. However, whether or not cats land on their feet depends on several factors, notably the distance they fall, because the twist maneuver takes a certain time, apparently around. sec. Thus the minimum height required for cats falling is about.m. -4. The Euler equation, which describes the rotation of an object about its symmetry axis, say x, is where Nx x ω ω ω N x x y z y z x bω is the component of torque along Ox. Because the object is symmetric about Ox, we have y z, and the above equation becomes b t dω x x x bωx ωx e ω dt x

44 96 CHAPTER