Homework 9. Figure 1: Definition of a point by colatitude, ϕ, longitude, θ, and radius, ρ. = 2ωv 0 sin(90 o +ϕ)ê θ (3)

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1 Theoretical Dynamics November 1, 010 Instructor: Dr. Thomas Cohen Homework 9 Submitted by: Vivek Saxena Goldstein. The colatitude 1, ϕ, is defined in the following figure taken from Wikipedia ( org/wiki/file:spherical_coordinates_(colatitude,_longitude).svg). Figure 1: Definition of a point by colatitude, ϕ, longitude, θ, and radius, ρ. Suppose the projectile is launched horizontally, along the ê ϕ direction, at point P. The angular velocity of earth s rotation is ω = ωẑ, and the projectile s initial velocity is v = v 0 ê ϕ. The Coriolis acceleration is a cor = (ω v) (1) = ωv 0 ẑ ê ϕ () = ωv 0 sin(90 o +ϕ)ê θ () = ωv 0 cosϕê θ () Since the acceleration is perpendicular to the initial velocity and points toward the right of the initial trajectory (in the northern hemisphere), the angular deviation is to the right in the northern hemisphere. In time t, the displacement due to the Coriolis acceleration (in the direction perpendicular to the initial velocity) is while the projectile covers a horizontal distance of s per = 1 (ωv 0cosϕ)( t) (5) s hor = v 0 t (6) 1 In the Goldstein problem, θ and ϕ are interchanged. On the surface of a sphere, a horizontal direction is one which has no radial component, i.e. it is of the form c 1ê θ +c ê ϕ. 9-1

2 So the angular deviation in time t is So the rate of angular deviation is δ = s per s hor = ωcosϕ t (7) ωcosϕ () which in Goldstein s notation for the colatitude, is ωcosθ, as required. This resulti s first order in ω because we have dropped the second order term ω (ω r) in computing the deflection. Goldstein. Let (r,θ) denote the polar coordinates of the bug at time t, where r is measured from the center of the wheel. The bug slips when the frictional force just equals the radial force acting on it. The radial force acting on the bug is The tangential force on the bug is F r = m( r r θ ) = mr θ (the bug crawls at constant speed) (9) F θ = m(r θ +ṙ θ) = mṙ θ (the wheel rotates at constant angular speed) (10) The static friction force has a maximum value given by f = µn (11) where µ is the coefficient of static friction. If the bug crawls on the side of the spoke, the normal force which enters this expression is equal to the tangential force F θ for the bug not to fall off the spoke. In this case, slipping takes place when µ(mṙ θ) = mr θ (1) or equivalently r = µṙ θ = (0.0) (0.5cm/s).0 rad/s = 0.1cm (1) But if the bug crawls on the top of the spoke, there are two normal reactions involved: the one that balances the gravitational force due to the bug s mass, and the other which equals the tangential force. In this case, the frictional force is modified to f = µ m g +Fθ (1) so that the condition for slipping becomes mr θ = µm g +ṙ θ (15) whcih gives r = µ g +ṙ θ = 0. (90cm/s ) +(0.5cm/s) (.0rad/s) θ (.0rad/s).66cm (16) This result is intuitively obvious: if the bug crawls along the top of the spoke instead of the side, it can go much farther out before it starts to slip. 9 -

3 Goldstein.5 The net force on the ball due to the rotation of the carousel is F = m( r r θ )ê r +m(r θ +ṙ θ)ê θ (17) As the ball is to remain stationary in the radial direction, ṙ = r = 0, so the force acting on the ball 6s after the carousel starts to move is F = mr θ ê r +mr θê θ (1) = (.0kg)(7.0m)(0.0 πrad/s 6s) ê r +(.0kg)(7.0m)(0.0 πrad/s )ê θ = 11.9Nê r +.6Nê θ (19) This is the force that the girl must provide to keep the ball moving in the circular path. Thus, F girl = 11.9Nê r +.6Nê θ (0) So, the girl must exert a force equal to 1.5N directed at an angle α given by 1.6 α = tan 11.9 = tan 1 (0.09) = 1.6 o (1) The direction of the force that the girl must provide is shown in the figure below. It should be directed Figure : Figure for problem.5 at an angle of 1.6 o measured clockwise from the line joining the ball to the center, or equivalently at an angle of o measured counterclockwise from the radius vector of the ball. 9 -

4 Problem 1 Part (a) d R z dθ = = sinθ cosθ 0 cosθ sinθ 0 cosθ sinθ 0 sinθ cosθ R z(θ) () } {{ } M z () = R z (θ)m z () = M z R z (5) The solution to this equation is R z (θ) = e θmz (6) where the multiplicative constant is fixed by the requirement that R z (0) = I, the identity matrix. Part (b) tr( R z (θ)) = tr cosθ sinθ 0 sinθ cosθ (7) = cosθ +1 () so cosθ = 1 tr( R z (θ)) 1 (9) Problem Part (a) = R z (φ) R y (θ) (0) cosφ sinφ 0 cosθ 0 sinθ = sinφ cosφ (1) sinθ 0 cosθ = R z(φ) R y(θ) cosθcosφ sinφ sinθcosφ cosθsinφ cosφ sinθsinφ sinθ 0 cosθ 9 - ()

5 So, Mz R n = = cosθcosφ sinφ sinθcosφ cosθsinφ cosφ sinθsinφ sinθ 0 cosθ cosθcosφ cosθsinφ sinθ sinφ cosφ 0 sinθcosφ sinθsinφ cosθ R n 0 cosθ sinθsinφ cosθ 0 sinθcosφ sinθsinφ sinθcosφ M z () () Now, So, from Eqs. (-7), cosθ M z = sinθcosφ M x = sinθsinφ M y = 0 cosθ 0 cos θ sinθcosφ 0 sinθcosφ sinθsinφ sinθsinφ 0 0 (5) (6) (7) Mz R n = cosθ M z +sinθcosφ M x sinθsinφ M y () where ˆn = (sinθcosφ, sinθsinφ,cosθ) T and M = (M x,m y,m z ). = ˆn M (9) Note: The minus sign appearing in equation () is simply due to the sign convention adopted for R z (φ) in equation (1). If we instead took R z (φ) = cosφ sinφ 0 sinφ cosφ (0) which corresponds to a clockwise rotation about the z-axis through an angle φ in the x y plane, then we would get Mz R n = cosθ M z +sinθcosφ M x +sinθsinφ M y = ˆn M (1) 9-5

6 with ˆn = (sinθcosφ,sinθsinφ,cosθ) T. Therefore, as required to be proved. Part (b) ˆn M = Mz R n () As is an orthogonal matrix, R ˆn R n = 1. Therefore, ( ( R ˆn exp( Φ ) R M z ) n = ( Φ ) M z ) k R ˆn R k! n k=0 () ( Φ) k = ( R ˆn Mz R k! n )...( R ˆn Mz R n ) k=0 k times () ( Φ) k [ = k M] k! (using Eqn. ()) (5) = k=0 k=0 1 [ Φˆn k M] (6) k! = exp( Φˆn M) (7) Therefore, R = ( exp( Φ M z )) R n () Part (c) From the result of problem 1, exp( Φ M z ) = and from equation () in part (a) of this problem, = cosφ sinφ 0 sinφ cosφ cosθcosφ sinφ sinθcosφ cosθsinφ cosφ sinθsinφ sinθ 0 cosθ (9) (50) 9-6

7 Substituting into equation () we get cosθcosφ sinφ sinθcosφ cosφ sinφ 0 R = cosθsinφ cosφ sinθsinφ sinφ cosφ 0 sinθ 0 cosθ exp( Φ M z) cosθcosφ cosθsinφ sinθ sinφ cosφ 0 sinθcosφ sinθsinφ cosθ R n (51) That is, R = ( ( ) ] cos φ cos θcosφ + sin θ + cosφsin φ sin θsin[φ]sin[ Φ + cosθsinφ sinθ( cosθcosφ( 1 + cosφ) + sinφsinφ) [ ] sin θsin[φ]sin Φ ( ) cosθsinφ cos φcosφ + cos θcosφ + sin θ sin φ sinθ(cosθ( 1 + cosφ)sinφ + cosφsinφ) sinθ(cosθcosφ( 1 + cosφ) + sinφsinφ) sinθ(cosθ( 1 + cosφ)sinφ cosφsinφ) cos θ + cosφsin θ ) Problem Part (a) Using the explicit form of R derived above, via Mathematica, we find that sinθcosφ R sinθsinφ = } cosθcos {{ } ˆn So, ˆn is an eigenvector of R with eigenvalue +1. Part (b) sinθcosφ sinθsinφ cosθcos (5) Using Mathematica, we find that so that tr( R) = 1+cosΦ (5) cosφ = 1 tr( R) 1 (5) Please refer to the file, problem.nb, containing calculations for this part. A similar statement holds for the alternate form of ˆn mentioned on page 5, corresponding to the alternate sign convention for φ. 9-7

8 Problem Using the results of the previous parts, R corresponds to a rotation about an angle Φ such that cosφ = 1 tr( R) 1 (55) and the axis of rotation is the eigenvector of R which has an eigenvalue +1. Using Mathematica 5, we find So, tr(r) = 1/. The eigenvalues are The matrix of eigenvectors is R = (56) λ 1 = 1 ( 5+i ) 9 (57) λ = 1 ( 5 i ) 9 (5) λ = 1 (59) V = i 1 i (60) The normalized eigenvector corresponding to the eigenvalue +1 corresponds to the axis of rotation, ˆn, and is given by / 1 ˆn = 0 / (61) 1 The angle of rotation is given by the solution to cosφ = 5 (6) which is about 19 o. 5 Please refer to the file, problem.nb. 9 -

9 In[16]:= Clear "Global` " ; In[17]:= In[1]:= In[19]:= In[150]:= In[151]:= Mz 0, 1, 0, 1, 0, 0, 0, 0, 0 ; Rz x_ Cos x, Sin x, 0, Sin x, Cos x, 0, 0, 0, 1 ; Ry x_ Cos x, 0, Sin x, 0, 1, 0, Sin x, 0, Cos x ; Rn Rz Φ.Ry Θ ; Rn MatrixForm FullSimplify Out[151]//MatrixForm= Cos Θ Cos Φ Sin Φ Cos Φ Sin Θ Cos Θ Sin Φ Cos Φ Sin Θ Sin Φ Sin Θ 0 Cos Θ In[15]:= In[15]:= R Rn.MatrixExp Mz.Transpose Rn ; R MatrixForm FullSimplify Out[15]//MatrixForm= Cos Φ Cos Θ Cos Sin Θ Cos Sin Φ Sin Θ Sin Φ Sin Cos Θ Si Sin Θ Sin Φ Sin Cos Θ Sin Cos Φ Cos Cos Θ Cos Sin Θ Sin Θ Cos Θ Cos Φ 1 Cos Sin Φ Sin Sin Θ Cos Θ 1 Cos Sin Φ Cos Φ In[15]:= Out[15]= In[155]:= n Sin Θ Cos Φ, Sin Θ Sin Φ, Cos Θ Cos Φ Sin Θ, Sin Θ Sin Φ, Cos Θ R.n MatrixForm Simplify Out[155]//MatrixForm= Cos Φ Sin Θ Sin Θ Sin Φ Cos Θ In[156]:= Out[156]= n True In[157]:= Tr R FullSimplify Out[157]= 1 Cos

10 In[1]:= Clear "Global` " ; In[1]:= In[15]:= In[16]:= Rx x_ 1, 0, 0, 0, Cos x, Sin x, 0, Sin x, Cos x ; Rz x_ Cos x, Sin x, 0, Sin x, Cos x, 0, 0, 0, 1 ; Rx Θ MatrixForm Out[16]//MatrixForm= Cos Θ Sin Θ 0 Sin Θ Cos Θ In[17]:= Rz Φ MatrixForm Out[17]//MatrixForm= Cos Φ Sin Φ 0 Sin Φ Cos Φ In[1]:= In[19]:= R Rz Π.Rx Π.Rz Π ; R MatrixForm Out[19]//MatrixForm= In[10]:= Tr R Out[10]= 1 In[16]:= sol Solve Cos Φ 1 Tr R,Φ Solve::ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. Out[16]= Φ ArcCos 5, Φ ArcCos 5 In[15]:= N sol 1, 10 Π Out[15]= 19. In[1]:= Eigenvalues R Out[1]= 1 5 9, 1 5 9, 1 In[1]:= Eigenvectors R ColumnForm Out[1]= 1 1, 9, 1 1, 1 9, 0, 1, 1

11 problem.nb In[15]:= Transpose Eigenvectors R MatrixForm FullSimplify Out[15]//MatrixForm=