1/30. Rigid Body Rotations. Dave Frank

Transcription

1 . 1/3 Rigid Body Rotations Dave Frank

2 A Point Particle and Fundamental Quantities z 2/3 m v ω r y x Angular Velocity v = dr dt = ω r Kinetic Energy K = 1 2 mv2 Momentum p = mv

3 Rigid Bodies We treat a rigid body as a system of particles. Each particle on the rigid body rotates with the same angular velocity ω. The angular velocity vector ω of a rigid body whose center of mass stays constant is directed along its axis of rotation. 3/3

4 Angular Momentum The angular momentum of a point particle is defined to be L = r p = r mv = mr (ω r) 4/3 The angular momentum of a rigid body is the vector sum of the individual momenta of it s particles. L = m i r i (ω r i ) Computing these cross products yields the expression L =[ω x m i (yi 2 + zi 2 ) ω y m i x i y i ω z m i x i z i ]î+ [ ω x m i x i y i + ω y m i (x 2 i + zi 2 ) ω z m i y i z i ]ĵ+ [ ω x m i x i z i ω y m i y i z i + ω z m i (x 2 i + yi 2 )]ˆκ

5 Moments of Inertia The sums in our previous expression for L have fundamental importance. m i (yi 2 + zi 2 ) = moment of inertia about the x-axis = I x. m i (x 2 i + zi 2 ) = moment of inertia about the y-axis = I y. m i (x 2 i + yi 2 ) = moment of inertia about the z-axis = I z. m i x i y i = xy-product of inertia = I xy. m i x i z i = xz-product of inertia = I xz. m i y i z i = yz-product of inertia = I yz. 5/3

6 We can now write our expression for angular momentum as L = (ω x I x ω y I xy ω z I xz )î+ ( ωi xy + ω y I y ω z I yz )ĵ+ ( ωi xz ω y I yz + ω z I z )ˆκ 6/3

7 Rotational Kinetic Energy The rotational form of kinetic energy is obtained from it s linear form. K = 1 2 mv2 = 1 2 mv v = 1 m(ω r) v 2 7/3 The total kinetic energy of a rigid body is the sum of the individual kinetic energies of its constituent particles. K = 1 m i (ω r i ) v i 2 Computing the cross product and then the dot product yields K = 1 2 ω m i r i (ω r i ) = 1 2 ω L

8 The Inertia Tensor Our expressions for L and K are complex and cumbersome. We need a more compact form of L and K. This is accomplished using what is known as the inertia tensor. 8/3

9 Recall the previous expression for L. L = (ω x I x ω y I xy ω z I xz )î+ ( ωi xy + ω y I y ω z I yz )ĵ+ ( ωi xz ω y I yz + ω z I z )ˆκ 9/3 In components L x = ω x I x ω y I xy ω z I xz L y = ω x I xy + ω y I y ω z I yz L z = ω x I xz ω y I yz + ω z I z This system of equations can be written in matrix form as L x I x I xy I xz ω x L y = I xy I y I yz ω y L z I xz I yz I z ω z or L = Iω where I is the inertia tensor.

10 We can also express K in terms of I K = 1 2 ω L = 1 2 ω (Iω) 1/3

11 Principal Axes and Diagonalizing the Inertia Tensor Every rigid body has 3 principal axes. Rotations about these axes are stable. (ω is constant in time.) Choosing the principal axes to be our coordinate axes makes the object s products of inertia zero. The inertia tensor then contains only the object s moments of inertia along it s main diagonal. This process is called diagonalizing the inertia tensor. We begin by letting the z-axis be the body symmetry axis. 11/3

12 ( x, y, z) z (x, y, z) 12/3 y x Every point (x,y,z) has a corresponding point ( x, y,z) which is a reflection across the z-axis. Therefore, in the sum of the xz and yz-products of inertia, we have I xz = m i x i z i = I yz = m i y i z i =

13 We now show that the xy-product of inertia is also zero. y ( x, y, z) (x, y, z) 13/3 x ( x, y, z) (x, y, z) Therefore, we can conclude that in the total sum: I xy = m i x i y i =.

14 With this coordinate system orientation, the inertia tensor becomes I x I = I y I z 14/3 Consider again, an object s rotational kinetic energy. K = 1 2 ω L = 1 2 ω (Iω) K = 1 ω x I x ω x ω y I y ω y 2 ω z I z ω z As you can see, this greatly simplifies our computations, which give K = 1 2 (I xω 2 x + I y ω 2 y + I z ω 2 z)

15 Rotating Coordinate Systems y y v P 15/3 r x x r must be the same as measured from both coordinate systems. r = xî + yĵ = x î + y ĵ

16 If we take r = xî + yĵ and differentiate both sides from the rotating frame of reference we have ( ) ( dr dx = dt dt î + dy ) dt ĵ rotating Now we take the same derivative with from the fixed reference frame. ( ) dr = dx dt dt î + xdî dt + dy dt ĵ + ydĵ dt Grouping: ( ) dr dt fixed fixed Substituting the previous result yields ( ) ( ) dr dr = dt dt fixed = ( dx dt î + dy ) ( dt ĵ + x dî ) dt + ydĵ dt rotating + ( x dî ) dt + ydĵ dt (1) (2) 16/3

17 Now, recall that v = dr dt = ω r Therefore, we can treat î and ĵ as rotating position vectors and say that dî dt = ω î and dĵ dt = ω ĵ Recalling equation (2) and making these substitutions ( ) ( ) ( dr dr = + x dî ) dt dt dt + ydĵ dt ( ) dr dt fixed fixed ( ) dr = dt ( ) dr = dt ( ) dr = dt rotating rotating rotating rotating + [x(ω î) + y(ω ĵ)] + ω (xî + yĵ) + ω r 17/3

18 This gives us the general operator on any vector A ( ) ( ) da da = + ω A dt dt fixed rotating 18/3

19 Euler s Equations The rotational motion of a rigid body is completely described by Newton s second law in rotational form. dl τ = dt = d(iω) dt To avoid a changing inertia tensor, we must adopt a coordinate system that rotates with the rigid body. Using our previous operator, we have ( ) d(iω) d(iω) τ = = + ω (Iω) dt dt rotating But since the inertia tensor is now constant in time ( ) dω τ = I + ω (Iω) dt rotating 19/3

20 And again using our vector operator, we see that ( ) ( ) dω dω = + ω ω dt fixed dt rotating ( ) ( ) dω dω = dt dt And we now have fixed rotating τ = I dω dt + ω (Iω) Or, in matrix form τ x I x ω x τ y = I y ω y + τ z I z ω z ω x ω y ω z I x I y I z Computing the cross product and matrix multiplication yields τ x ω x I x ω y ω z I z ω y ω z I y τ y = τ z + ω y I y ω z I z ω x ω z I x ω x ω z I z ω x ω y I y ω x ω y I x ω x ω y ω z 2/3

21 Or, in components we have τ x = ω y ω z I z ω y ω z I y + ω x I x τ y = ω x ω z I x ω x ω z I z + ω y I y τ z = ω x ω y I y ω x ω y I x + ω z I z 21/3 These three equations are known as Euler s equations of motion for a rigid body.

22 Force Free Rotations and Precession We wish to consider in detail the case when the net torque on a rotating object is zero. This is true for any object freely falling in a gravitational field. In this case, Euler s equations become 22/3 ω x = I y I z ω y ω z I x ω y = I z I x ω x ω z I y ω z = I x I y ω x ω y I z This is a nonlinear system of three first order equations which is easily solved using ODE45. But let s first consider the nature of such rotations given some initial conditions.

23 Initial Conditions and Equilibrium Solutions The equilibrium points are obtained by setting ω x = ω y = ω z =. = ω y ω z = ω x ω z 23/3 = ω x ω y When solving this system simultaneously, we get any multiple of one of the following a b c, In other words, any point on the coordinate axes is an equilibrium point. Therefore, any solution with an initial condition of one of these forms will remain constant for all time. That is, if ω is directed along one of the three principal axes, it will remain constant.,

24 Let us now consider an initial condition when ω is in some arbitrary direction. L z ω z 24/3 y x In the fixed coordinate system: dl τ = = dt = d(iω) dt In the rotating coordinate system: dω τ = = I dt + ω (Iω)

25 Kinetic Energy and the Ellipsoid Since we are considering force free motion, we expect, from the conservation of energy, that K will also remain constant in time. We can verify this by considering our previous expression for K. 25/3 Taking the time derivative of K K = 1 2 (I xω 2 x + I y ω 2 y + I z ω 2 z) dk dt = 1 2 (2I xω x ω x + 2I y ω y ω y + 2I z ω z ω z ) Substituting in the differential equations for ω x, ω y, and ω z dk dt = 1 ( ) I y I z I z I x I x I y 2I x ω x ω y ω z + 2I y ω y ω x ω z + 2I z ω z ω x ω y 2 I x I y I z dk dt = ω xω y ω z [(I y I z ) + (I z I x ) + (I x I y )] =

26 So, given an initial condition, (ω x, ω y, ω z ) the object s kinetic energy is given by K = 1 2 (I xω 2 x + I y ω 2 y + I z ω 2 z) = constant This gives the equation of an ellipsoid. This surface represents a surface of constant energy called the inertial ellipsoid. Solutions must stay on the ellipsoid for all time. The equilibrium points occur where the axes intersect the ellipsoid. 26/3

27 Example Let I x = 2, I y = 1, and I z = 3. Our system of equations now becomes: ω x = ω y ω z ω y = ω x ω z 27/3 ω z = 1 3 ω xω y And our equation for kinetic energy becomes K = 1 2 (2ω2 x + ω 2 y + 3ω 2 z) If we let K = 12, this becomes the equation of our inertial ellipsoid 24 = 2ω 2 x + ω 2 y + 3ω 2 z If we pick off any initial condition that satisfies this equation, the resulting solution trajectory must remain on the surface of the ellipsoid for all time.

28 This is observed using ODE45 to plot several solution trajectories 28/ wz wy wx 4

29 Examining a solution near a stable equilibrium point 2 wx 29/ t 1 wy wz t wy wx wz t

30 Now consider a solution near an unstable equilibrium point 5 wx 3/ t 1 wy 2 wz t wy wx wz t