Prof. O. B. Wright, Autumn 007 Mechanics Lecture 9 More on rigid bodies, coupled vibrations Principal axes of the inertia tensor If the symmetry axes of a uniform symmetric body coincide with the coordinate axes, the products of inertia (Ixy etc.) are zero, and Ixx 0 0 I1 0 0 I 0 I 0 0 I 0. Example yy 0 0 I zz 0 0 I This choice of axes is special so we call them the principal axes. In fact in this example of a cylinder we are free to choose the directions of x and y for the principal axes provided that z is parallel to the cylinder axis. What about a body that is not symmetric? It turns out that we can always find principal axes for which I1 0 0 I 0 I 0. 0 0 I Remember that the values of I1, I and I are always positive (or some can be zero when point masses are used). In the principal axes, L1 I1 0 0 1 I11 L L 0 I 0 I. L 0 0 I I L is not parallel to in general. But if the body is spinning about a principal axis then L is parallel to. Can you see why from the above equation? 1
Prof. O. B. Wright, Autumn 007 Visualizing the inertia tensor Consider the equation I ijrr i j 1, where i, j x r y z, where r1 x, r y, r z. I x I y I z I xy I yz I zx 1 xx yy zz xy yz zx This is the equation for a surface in D, in fact an ellipsoid positioned at some angle to the axes. If we choose principal axes, the equation reduces to I x I y I z 1 1 This is an ellipsoid with axes parallel to x, y, z. With principal axes It can be shown* that if the vector cuts the ellipsoid at a point, then the direction of L is normal to the surface, as in this diagram shown for the principal axes. xy section of ellipsoid Remember L I means L is not parallel. *See Physical Properties of Crystals, J. F. Nye, Oxford University Press.
Prof. O. B. Wright, Autumn 007 Euler equations So far we have considered a rigid body that is spinning about a fixed axis. If the body is not symmetric then we have seen that torques must be exerted to keep the axis fixed (see last lecture). But what if the body is not symmetric and is spinning freely with no torques acting on it? In that case L is constant but the angular velocity vector will change with time because will change with time. However, if the body is spinning without external forces then it must spin with the centre of mass fixed (see Lecture ). It I is convenient to put the origin at the centre of mass because of this. The principal axes of the body are fixed with respect to the body, so we cannot choose a fixed set of axes that are principal axes. But we can define a set of axes that x, y, z rotate together with the body at the instant in time considered (see Lecture 8 for a similar treatment when the rotation is about a fixed axis). And let us assume that the axes point along the principal axes of the body. The x, y, z frame is a rest x, y, z frame in which no inertial forces occur. Therefore I1 0 0 I 0 I 0 in the x, y, z frame. 0 0 I Also L x I1 0 0 1 I11 L L 0 I 0 I in this frame. Alternatively, y L z 0 0 I I L = I1 1i + I j + I k, where i, j, k, and 1,, change with time. Note that the angular velocity of a body is a quantity that does not change direction in space or magnitude when we change the axes. (Please do not think that =0 in the frame of x, y, z!) Here we have chosen to express in the x, y, z axes. (Perhaps a better notation would have been L = I1 1i + I j + I k, but we miss off the for simplicity here.) At time t, we can use the equations we derived in Lecture 8:
Prof. O. B. Wright, Autumn 007 d a =a ω a for a vector a = a i + a j + a k in the dt x y z x, y, z frame. A similar argument can be used for the angular momentum vector L = L i + L j + L k L L i + L j+ L k where dl/ dt dl / dt 0. Therefore x y z x y z dl = dt 0 = LωL (1) where rotating L L L x i + L y j + L z k is the rate of change of L measured in the i, j, k x, y, z frame. L x I1 0 01 I11 L L Iω 0 I 0 I since y L z 0 0 I I x, y, z frame. I1 0 0 I 0 I 0 const. in the 0 0 I Expressing Eq. (1) in the x, y, z frame, we obtain I11 1 I11 0 L ω L I I. I I We can write the components of this equation as follows: I ( I I ) 1 1 I ( I I ) Euler equations 1 1 I ( I I ) 1 1 These equations determine the motion of the body seen in a frame fixed to the body 4
Prof. O. B. Wright, Autumn 007 (i.e. x, y, z ). The angular velocity 1 is calculated in the x, y, z frame. (To calculate it from the components of the angular velocity in the x,y,z frame we just need to use different basis vectors at a different angle: i.e. ω = i + j+ k ω = i + j + k, where I have chosen the subscript 0 to 0 x0 y0 z0 x y z refer to the x, y, z frame. This notation arises because we wanted to avoid using the for in the frame for simplicity.) x, y, z It is not always easy to use the Euler equations because we need to know (t). In fact the equations can be solved by expressing in terms of the angles that describe the orientation of the body. Stability of rotational motion Let us consider one application of the Euler equations. Suppose a body with I1, I and I all different is spinning about its rd principal axis. So, initially, 1= =0 and 0. The right hand side of all three Euler equations is zero. That means that (and so) must remain constant. The same argument holds if the body is spinning about the 1st or the nd principal axis. i.e. A body spinning about a principal axis will remain spinning about this axis. We can also see that if the body is spinning so that two or more values of 1, and are non-zero (i.e. when the angular velocity does not point along a principal axis), then does not remain constant. e.g. if 1 and 0, then from the rd Euler equation we can see that 0 (since I1 I). Let us see if a body with I1, I and I all different that is spinning about a principal axis is stable or not. Suppose that the body is spinning with 1=constant and= =0. Then let us assume that there is a small perturbation (e.g. we knock the body a bit) so that and 0 but and 1. 5
Prof. O. B. Wright, Autumn 007 The Euler equations tell us what should happen: I ( I I ) const. 1 1 1 I ( I I ) I ( I I ) 1 1 1 1 ( I I )( I I ) I ( I I ) I 1 1 1 1 1 I ( I1 I)( I1 I) 1 II (We can derive a similar equation for ). This is the equation for simple harmonic motion of provided that I1 is the largest or smallest moment of inertia (compared to I and I). This means that the rotation is stable, i.e. the system just oscillates about a single dominant value of, provided that we rotate about one of two axes. If on the other hand we rotate about the principal axis that has the intermediate value of moment of inertia, then the sign of the square bracket in the above equation is negative, so that the value of rapidly moves away from zero, and the motion is unstable. An illustration of which is the largest and smallest I component for a rectangular block this is shown in the figure above. Coupled vibrations Consider the coupled system of masses and springs below: mx kx k( x x ) a a b a mx kx k( x x ) b b b a 6
Prof. O. B. Wright, Autumn 007 Add and subtract these two equations to get: d m ( x ) ( ) a xb k xa xb dt d m ( x ) ( ) a xb k xa xb dt These equations are uncoupled equations with variables x a +x b and in x a -x b. The solutions are x a x b x 1 (t) Acos( 1 t 1 ), x a x b x (t) B cos( t ), 1 k / m k / m Solving for the positions of the masses: x a 1 Acos( 1t 1 ) 1 Bcos( t ) 1 (x 1 x ) x b 1 Acos( 1t 1 ) 1 Bcos( t ) 1 (x 1 x ) We call x 1 and x the normal coordinates of the system. If B=0 the masses oscillate in mode 1 with SHM at a single frequency 1, with x a = x b. If A=0 the masses oscillate in mode with SHM at a single frequency 1, with x a = -x b. If a single mode is present all masses pass through their equilibrium positions simultaneously. Each mode has its own 'shape'. Mode 1 and mode are called the normal modes of the oscillating system. The total number of modes (here ) is equal to the number of degrees of freedom (the number of coordinates) of the system. Here, x 1 corresponds to the motion of the centre of mass x 1 /=(x a +x b )/. x is the compression of the central spring, or, equivalently, the relative displacement of the two masses. If neither A or B are zero the motion is a combination of oscillations at both 1 and. Since we have derived the general solution, any motion can be expressed as a linear combination (a superposition) of the motion in the modes 1 and : 7
Prof. O. B. Wright, Autumn 007 xa 1 cos( 1t 1) 1 cos( t) xa xa A B x cos( t ) cos( t ) x x b 1 1 b 1 b This is a consequence of the principle of superposition that applies because the differential equation describing the motion is linear. Beats Beats arise where the motion of a part of an oscillating system is a superposition of two SHM at different frequencies. For example, the two SHM can be the two normal modes of system with degrees of freedom. For simplicity let us calculate the superposition of two harmonic oscillations with the same phase (=0) and amplitude: x 1 Acos 1 t, x Acos t x x 1 x Acos 1 t Acos t We can rewrite this in an interesting form. x Acos 1 t cos 1 t Now define the average angular frequency by av =( 1 + )/ and the 'modulation' angular frequency by mod =( 1 - )/, so x Acos mod tcos av t This can be thought of as an oscillation at angular frequency av, with an amplitude Acos mod t. If 1 then mod << av. This corresponds to nearly SHM at frequency av with slowly varying amplitude at frequency mod. See example below. 8
Prof. O. B. Wright, Autumn 007 Amod is a maximum twice every modulation cycle (one modulation cycle takes Tmod=/mod). We call the frequency beat=mod, the beat frequency. For the case of the system of two masses and three springs above, we can above the phenomenon of beating by starting the system of with the initial conditions: x A, x 0, x 0, and x 0. The solution is then a b a b x a Acos 1 t Acos t A cos mod tcos av t x b A cos 1 t Acos t Asin mod tsin av t where, in this case, av =1.7 k / m, mod =0.7 k / m. When the amplitude Acos mod t=0 the energy is nearly all in the mass b. When Asin mod t=0 the energy is nearly all in the mass a. The energy passes from mass a to b and back again to a in the time that cos mod t takes to go from 1 to 0 and back to 1, i.e. in a time / mod. This is equal to the beat period / beat. Matrix method for coupled vibrations We need a more rigorous method to find the normal modes. This can be done using matrices. For the case of the system of two masses and three springs above, 9
Prof. O. B. Wright, Autumn 007 mx kx k( x x ) kx kx a a b a a b mx kx k( x x ) kx kx b b b a a b m 0 xa k k xa 0 m x b k k xb or MxKx Notice that the matrix K is symmetric about its diagonal. This is not a coincidence. It is the result of Newton s rd law. Try x a it a e e b x b it m 0 a k k a 0 mb k k b or a K M b ( ) 0 This has a solution for non-zero a and b only if the matrix (otherwise we could multiply by the inverse matrix ( K M) ( K M) has no inverse 1 and then both a and b would be zero). The condition for no inverse matrix is that the determinant of ( K M) should be zero. That is, e f Det ( K M) =0, where Det eh gf g h. In our example k m k K M k m k 0 k k m So k m k k k,. m m k k m 10
Prof. O. B. Wright, Autumn 007 So we have found the possible vibrational frequencies. There are two of them. Finding the normal modes We need to find the possible values of a b in order to find the possible a e b it x. 1) For 1 k m k k and a 0. k k b, K 1 M K 1 M k k a 0 a b 0 a b. k k b A e A x 1 it, where A can be a complex number. x x A cos( t ) a b 1 1 Normal mode 1: ) For k m k k and a 0. k k b, K 1 M K 1 M k k a 0 a b 0 a b. k k b B e B x it, where B can be a complex number. x x Bcos( t ) a b Normal mode : 11