Course of Lectures in Mechanics Unit-F Dynamics of Rigid Bodies A. K. Kapoor http://0space.org/users/kapoor [email protected]; [email protected] Contents 1 Equations of Motion for a Rigid Body 3 1 Equations of motion in inertial frames . .... 3 § L 1.1 d equation–adifficulty ........................ 3 § dt 1.2 Usingbodyfixedaxes .......................... 4 § 2 Understandingprecession of spinningwheel . ...... 4 § 3 Torque free rotation does not always imply constant ~ω ............ 5 § 4 Principalaxesofarigidbody . 8 § 5 Eulerequations.................................. 10 § 6 EndNotes ..................................... 10 § 2 Examples :: Inertia Tensor and Finding Principal Axes 11 1 LessonOverview ................................. 11 § 1.1 LearningObjectives ........................... 11 § 1.2 Prerequisites ............................... 11 § 2 Computingmomentofinertiatensor . 11 § 2.1 Momentofinertiatensorofacube . 13 § 2.2 MI tensor of a cube relative to a corner . 14 § 2.3 Principal axes of a cube relative to a corner . 15 § 3 L~ is not always parallel to ~ω ........................... 15 § 4 EndNotes ..................................... 17 § 3 Applications of Euler’s Equations 18 1 LessonOverview ................................. 19 § 1.1 LearningGoals .............................. 19 § 1.2 Prerequisites ............................... 19 § 2 Bicyclewheelrevisited ............................. 20 § 3 FreeMotionofaRigidBody .......................... 21 § 3.1 Sphericaltop ............................... 21 § 4 Freesymmetricaltop............................... 22 § 4.1 Rigidrotator ............................... 22 § 1 4.2 Symmetrictop .............................. 22 § 5 Asymmetricaltop................................. 23 § 6 EndNotes ..................................... 24 § 6.1 Whytheearthwobbles ......................... 24 § 6.2 Kowalskeyatop.............................. 25 § Draft Copy for Limited Circulation 2 Lesson 1 Equations of Motion for a Rigid Body §1 Equations of motion in inertial frames The motion of rigid body is described by two sets of equations d2X dL cm = F , = ττ. (1.1) dt2 tot dt Important Recall that the separation of equations of motion of a rigid body takes place only if in the second equation above L is taken to be the angular momentum of the rigid body about the centre of mass of the body. Also, it should be remembered that the forces internal to the system, being of action reaction type do not contribute to the right hand sides of the above equations. The above equations are valid in inertial frames. For several purposes, the above equations are sufficient to get information about the motion of the rigid body. dL §1.1 dt equation – a difficulty Note that the second equation in (1.1) requires writing L w.r.t. the inertial frame. This involves moment of inertia tensor components computed relative to the inertial frame. Since the body is moving the components of the inertia tensor keep changing with time. Computing the moment of inertia tensor requires configuration of the body at time t, which in turn can be determined after equations of motion has been solved . / 3 §1.2 Using body fixed axes The above mentioned difficulty is solved by writing the the rate of change of angular momentum with respect to the frame fixed in the body. For this we use the result dL dL = + ω L (1.2) dt inf dt bf × The time dependence of angular momentum L = Iωin the body frame comes from the time dependence of angular velocity, the inertia tensor relative to the body frame remains constant in time. Thus we write the equation of motion as dL + ω L = ττ. (1.3) dt bf × All reference to the space axes has disappeared and we can drop the suffix bf, and write dL + ω L = ττ. (1.4) dt × In this form the above equation is valid if the body axes are chosen to have the origin at the centre of mass. It turns out that the above equations also hold when one point of the rigid body is fixed and that point is chosen as the origin. §2 Understanding precession of spinning wheel We discuss the dynamics of a spinning wheel. You have already watched this video on the moodle course site. A string is tied to one of the ends, P , of axle of a bicycle wheel and is the other end is supported manually by holding it with hand. The string is kept vertical by, for example tying the other end C of the string to the ceiling. When the end held by hand is released, the wheel falls due to gravity. However, if the wheel is set spinning before being released, it does not fall; it precesses in one direction. The direction of precession gets reversed when if the wheel is set spinning in the opposite direction. How do we understand this? Let the horizontal direction in which the axle points be chosen as X2 axis so that the angular velocity and angular momentum points along the X2. Let the X3 axis be chosen upwards so that the force F = mg due to gravity is along the negative X2 direction. The radius vector ~x of the centre of mass of the wheel is along the positive X axis. So that the torque, ~x F~ is then along the negative X axis. 2 × 1 The change in angular momentum in a short time time ∆t is given by ∆L~ = ~τ∆t and is shown in Fig.1(a). The vector sum of L~ and ∆L~ , drawn in Fig.1(b) is along P D. The implies that after time ∆t the axle has rotated about the X3 axis so that the end C moves to the point D. If the wheel is initially set spinning in the opposite direction, the torque and hence the change in angular momentum ∆L~ remains the same, but the direction of L~ is reversed 4 and the angular momentum after time ∆t will be as shown in Fig.1(c). This means that the end C must move to the point E and hence the axle rotates in the opposite direction. X 3 D X C 2 P E X 1 D L+ ∆L D K ∆ P L L P C C L ∆ L+ L E (a) (b) (c) Fig. 1 Movement of Spinning bicycle wheel [Q/T] := The above analysis has been presented from the point of view of the space axes. Analyze the motion from the point of view of the body frame. §3 Torque free rotation does not always imply constant ~ω )(Short Examples 1 What is the moment of inertia tensor for a rectangular lamina of || sides a and b. Choose the centre of the plate as the origin and X,Y axes parallel to the sides as shown in the figure. 5 Y O α X b a Fig. 2 Figure for the Examples 1 and 2 We divide the plate into small small pieces and find contribution of each piece to the moment of inertia tensor components. So at first divide the lamina into a large number of thin vertical strips parallel to the Y axis. consider one such vertical strip between x and x + dx. We will sum over all such strips by integrating over x from a/2 to a/2. − Next divide the strip at x into small small rectangular pieces by drawing horizontal lines parallel to the Y axis. One such rectangular element is shown as shaded. This element is located at (x,y, 0) has an area of dxdy and its mass will be σdxdy dm where ≡ σ is the mass per unit area. The total mass is M therefore σ = M/ab.The contribution 2 2 to 11 component moment of the inertia tensor is dm(y + z ) and the total value of I11 is obtained by summing (integrating) over all pieces. This is done by integrating over x from a/2 to a/2 over y from b/2 to b/2. So we have − − a/2 b/2 2 Ixx = σ(y + 0)dy dx (∵ z = 0) Z a/2 Z b/2 − n − o a/2 y3 b/2 = σ dx Z 3 a/2 n b/2 o − − a/2 b3 = σ dx Z a/2 12 − b3 a/2 = σ [x] dx 12 a/2 − b3 = σ a 12 Mb2 = (∵ σ = M/(ab)) (1.5) 12 Remember a double integral is computed by evaluated the inner integral over y first Then integrating the result over the outside variable x. In this example the order of integration is unimportant. 6 Ma2 Similarly I = . Also, for this case ( not in general) I = I + I . ( Prove this 22 12 33 xx yy yourself.) Next we compute I12. a/2 b/2 Ixy = σxydy dx − Z a/2 Z b/2 − n − o (1.6) The inner integral, over y, is b/2 y2 b/2 b2 b2 ydy = = = 0. (1.7) Z 2 8 − 8 b/2 b/2 − − Similarly other off diagonal components of the moment of inertia tensor turn out to be zero. This is because of our choice of the coordinate axes is symmetrical w.r.t. the plate. )(Short Examples 2 Find torque acting on a rectangular plate when it rotates with a || constant angular velocity about (a) the Z- axis (b) about its diagonal, so nˆ = 1 (1, 1, 0). √2 The choice of axes is as in the previous example. The moment of inertia tensor is given by b2 0 0 M I = 0 a2 0 (1.8) 12 0 0 (a2 + b2) dω Using dt = 0 in the Euler’s equations dL~ + ~ω L~ = ~τ (1.9) dt × total will give the required torque. Now the components of angular momentum are given by 2 2 Lx b 0 0 ωx b ωx M 2 M 2 Ly = 0 a 0 ωy = a ωy (1.10) 12 12 L 0 0 (a2 + b2) ω (a2 + b2)ω z z z ~ d~ω Note that the components of L are proportional to components of omega~ Hence dt = 0 dL~ means dt = 0. Therefore, The required torque is given by ~τ = ω L.~ (1.11) total × (a) In the first case angular velocity and angular momentum components are given by M ~ω = (0, 0,ω), L~ = (0, 0, (a2 + b2)) (1.12) 12 and required torque vanishes because ~ω and L~ parallel (both being along the Z- axis.) 7 (b) In this case Mω ω = ω(cos α, sin α, 0), L~ (b2 cos α, a2 sin α, 0) (1.13) 12 where α is the angle the diagonal makes with the X- axis.