Lecture 38: Equations of Rigid-Body Motion It s going to be easiest to find the equations of motion for the object in the body frame i.e., the frame where the axes are principal axes In general, we can write: ω= φ+ θ + ψ To write the components of this vector in the body frame we need to do some geometry: ψ x 3 x 1 x 3 θ φ φ x 2 ψ is along the x 3 direction: ψ ψ e = 3 x θ is in the x 1 -x 2 plane: 1 ψ x θ = θ cosψ e 1 θ sinψ e2 2 φ has components in all three θ directions: φ = φ sinθ sinψe + φsinθ cosψe + φcosθ e 1 2 3
Adding up these vectors, we find that the components of ω in the body frame are: ω1 = φ sinθ sinψ + θ cosψ ω2 = φ sinθ cosψ θ sinψ ω = φ cosθ + ψ 3
Euler s Equations for Rigid Body Motion Now that we ve defined the geometry, we can set up the Lagrangian for rigid body motion To make things easy, we ll first consider the case where no external forces act on the body, so: L= T In that case, we can put the origin of both the fixed (meaning inertial) and body frames at the center of mass Furthermore, we can cleverly choose the body frame to coincide with the principal axes of the body, so that: 1 2 T = Iiωi 2 i Principal moment of inertia associated with the ith axis
Since the Eulerian angles fully specify the orientation of the rigid body in the inertial frame, we can take them to be our generalized coordinates The equation of motion for ψ is: T d T = ψ dt ψ which becomes (using the chain rule): T ω d T ω = ω ψ ω ψ i i i i dt i i T One piece that enters everywhere is, which is simply: ωi T = Iiωi ω i
The other relations we need are: ω1 = φ sinθ cosψ θ sinψ = ω2 ψ ω2 ψ ω3 ψ = φ sinθ sinψ θ cosψ = ω = 1 ω1 = ψ ω2 = ψ ω3 = 1 ψ Putting these back into the Lagrange equation gives: d I1ωω 1 2 + I2ω2( ω1) I3ω3 = dt I I ωω I ω = 1 2 1 2 3 3
The choice of which axis to call x 3 was completely arbitrary That means we can find similar relations for the other components of the angular velocity I I ωω I ω = 1 2 1 2 3 3 I I ωω I ω = 3 1 3 1 2 2 I I ωω I ω = 2 3 2 3 1 1 Euler s Equations for force-free motion
Rigid Body Motion With an External Torque If an external torque N is applied, we know that: dl dt = But that s true only if everything is measured in a fixed inertial frame, not in the body frame (where we can take the coordinate axes to be principal axes) To see how we translate between the two coordinate systems, recall that we can write the velocity of a single rotating particle (in an inertial frame) as: If the rotation was happening in a non-inertial frame, we d have to modify this to: dr dr = + ω r dt dt N dr v = = ω r dt inertial non-inertial
That last equation is true for any vector (see Chapter 1). The relationship relevant to us is: dl dl N = = + ω L dt dt inertial where all the terms on the right-hand side can be evaluated in the body frame Remember that in this frame the axes are principal axes Just taking the z component of the above vector equation gives: N = L + ω L ω L = I ω + ω I ω ω Iω body 3 3 1 2 2 1 3 3 1 2 2 2 1 1 = I ω I I ωω 3 3 2 1 1 2
By carrying out the same procedure for the other components, we find: N = I ω I I 2 2 2 3 1 3 1 N = I ω I I ω ω ω ω 1 1 1 2 3 2 3 We can summarize all of this in the single expression: ( Ii I j) ωω i j ( I kωk Nk) εijk = These imply that the motion of any rigid object depends only on the three numbers I 1, I 2, and I 3 and not on any other details of the shape of the object We can define an equivalent ellipsoid for any object k
Motion of a Symmetric Top We ll now apply the equations of motion to a symmetric top Meaning an object with at least two of its principal moments of inertia equal We ll choose I 1 = I 2 (i.e., the z axis is the symmetry axis) and start with the case where no external forces act on the body The equations of motion then become: I I ωω I ω = 1 1 1 2 3 3 I I ωω I ω = 3 1 3 1 1 2 I I ωω I ω = 1 3 2 3 1 1 The first equation tells us that: I ω =, so ω is a constant 3 3 3
We can now rewrite the other two equations as: I3 I1 ω1 = ω3 2 I ω 1 I3 I1 ω2 = ω3 1 I ω 1 The term in parentheses is constant; for simplicity we write it as Ω We re left with the following coupled differential equations: ω = Ωω ω 1 2 =Ωω 2 1
To solve the equations, we multiply the second by i and add it to the first: ω + i ω = Ω ω + iωω 1 2 2 1 ω + i ω +Ωω iω ω = 1 2 2 1 ω + i ω i Ωω iω ω = 2 1 2 2 1 We now define a new variable so the equation becomes: ω + i ω iω ω + iω = 1 2 1 2 η ω + iω 1 2 η Ω i η = That s an equation we know how to solve: η = i( Ω+ t δ ) To make life easy, we ll choose the phase δ to be zero Ae
Writing the equation in terms of the components of ω gives: iωt ω + iω = Ae 1 2 Writing the real and imaginary parts of the above equation separately gives: ω = Acos( Ωt) 1 ω = Asin ( Ωt) 2 To interpret this result, note that the magnitude of the angular velocity vector must be constant: ω 2 2 2 2 2 2 2 2 = ω + ω + ω = A cos ( Ω t) + A sin ( Ω t) + ω 1 2 3 3 = A + ω = 2 2 3 const