Sect. 4.9: Rate of Change of a Vector Use the concept of an infinitesimal rotation to describe the time dependence of rigid body motion. An arbitrary vector (or pseudovector) G (could, e.g., be the position r), which is time dependent: G = G(t). Time dependence of G depends on coord system in which the observation is made. If body is rotating &/or translating, G(t) is different when referred to body axes (x, y, z) than when referred to space axes (x, y, z). In infinitesimal time dt, dG = (change in G) differs in the 2 coord systems. Can write: (dG) space = (dG) body + (dG) rot where: (dG) rot  the effect of body rotation on G

If G is fixed (constant) in the body, (dG) body = 0 so (dG) space = (dG) rot In this case, G is rotating with the body & we can use the results of the last section to write (counterclockwise rotation of G): (dG) rot  dΩ  G where dΩ  differential vector angle associated with the rotation (last section discussion)  (dG) space = dΩ  G

If G is not fixed in the body, (dG) body  0 so (dG) space = (dG) body + dΩ  G (1)  In time dt, relation between rates of change of G in 2 systems is obtained by dividing (1) by dt.  (dG/dt) space = (dG/dt) body + ω  G (2) Have defined: Angular velocity ω dt  dΩ Text makes a big deal about emphasizing that (dΩ/dt) = ω is NOT a derivative of a function Ω(t), which the previous section made a big point does not exist (for finite rotations). Instead, (dΩ/dt) = ω = ratio of 2 infinitesimal quantities. Direction of ω  instantaneous axis of rotation direction. More formal derivation of (2): See text p Uses orthogonal transformation formalism.

(dG/dt) space = (dG/dt) body + ω  G (2) (2)  A GENERAL statement about transformation of time derivatives of vectors between body & space axes. Slight change of notation: space  s, body  r (rotating system):  (dG/dt) s = (dG/dt) r + ω  G (2) Since G is arbitrary, take (2) as the definition of an operator equation relating TIME DERIVATIVE OPERATORS in the 2 systems: (d/dt) s = (d/dt) r + ω  (3) Will make repeated use of (3) ( or (2) )!

 (dG/dt) s = (dG/dt) r + ω  G (2) Or: (d/dt) s = (d/dt) r + ω  (3) –Note: This is, of course, for a body rotating without translating! Convenient & useful to express angular velocity ω in terms of Euler angles & their time derivatives. –From Euler angle & infinitesimal rotation discussions: General dΩ associated with vector ω consists of 3 successive infinitesimal rotations through Euler angles ,θ,ψ: Vector angular velocities associated with each: ω   (d  /dt) =  ω θ  (dθ/dt) = θ ω ψ  (dψ/dt) = ψ  ω = ω  + ω θ + ω ψ

ω = ω  + ω θ + ω ψ (4) Practical difficulty with (4): Directions of ω , ω θ, ω ψ are not only not orthogonal, they are not even symmetrical! Recall Euler angle discussion for directions: ω  along space z axis, ω θ along line of nodes, ω ψ along body z axis. Can always get components of ω , ω θ, ω ψ with respect to any set of axes by use of individual orthogonal transformations B, C, D & total A = BCD of Euler angle discussion.

For example components of: ω  (|| z axis) along the body axes: (ω  ) x =  sinθ sinψ, (ω  ) y =  sinθ cosψ, (ω  ) z =  cosθ ω θ (|| ξ axis or line of nodes) along the body axes: (ω θ ) x = θ cosψ, (ω θ ) y = - θ sinψ, (ω θ ) z = 0 ω ψ (|| z axis) along the body axes: (ω ψ ) x = 0, (ω ψ ) y = 0, (ω ψ ) z = ω ψ = ψ Components of the total ω = ω  + ω θ + ω ψ along the body axes (add component by component): ω x =  sinθ sinψ + θ cosψ, ω y =  sinθ cosψ - θ sinψ ω z =  cosθ + ψ Components of these along other axes? See problem 15. Time dependence of Cayley-Klein & Euler parameters? See problem 16.!

Sect. 4.10: Coriolis Effect; “Fictitious” Forces (from Marion) For transformation between time derivatives from space system (s) and rotating system (r) we had: (d/dt) s = (d/dt) r + ω  (3) (3): Basic kinematical law for discussing dynamics of rigid body motion (Ch. 5). Not restricted to rigid bodies! Valid whenever we want to discuss the motion of a particle or particles relative to a rotating coordinate system. Important application: Motion relative to coordinates of the rotating Earth. “The Coriolis Effect” comes in. An example of “fictitious” or non-inertial forces! Next, we will discuss non- inertial forces in general. After the general discussion, then application to Coriolis Effect. My treatment is similar to but much more general than the text’s. It comes mainly from Marion’s text.

Non-Inertial Frames General Discussion Inertial Reference Frame: –Any frame in which Newton’s Laws are valid! –Any reference frame moving with uniform (non- accelerated) motion with respect to an “absolute” frame “fixed” with respect to the stars. By definition, Newton’s Laws are only valid in inertial frames! F = ma: Not valid in a non-inertial frame! By definition, the concept of “Force”, as defined in Ch. 1 (& by Newton), is only valid in inertial frames! It’s always possible to find an inertial frame to do the dynamics of a system!

“Fictitious Forces” Some problems (like rigid body rotation!): Using an inertial frame is difficult or complex. Sometimes its easier to use a non-inertial frame! “Fictitious Forces”: If we are careful, we can the treat dynamics of particles in non-inertial frames. –Start in inertial frame, use Newton’s Laws, & make the coordinate transformation to a non-inertial frame. –Suppose, in doing this, we insist that our eqtns look like Newton’s Laws (look like they are in an inertial frame).  Coordinate transformation introduces terms on the “ma” side of F = ma inertial. If we want eqtns in the non-inertial frame to look like Newton’s Laws, these terms are moved to the “F” side & we get: “F” = ma noninertial. where “F” = F + terms from coord transformation  “Fictitious Forces” !

Non-Inertial Frames The most common example of a non-inertial frame = Earth’s surface! Another is rigid body rotation. We usually assume Earth’s surface is inertial, when it is not! –A coord system fixed on the Earth is accelerating (Earth’s rotation + orbital motion) & is thus non-inertial! –For many problems, this is not important. For some, we cannot ignore it! Will do rotating coordinate systems. However, to understand more about “fictitious forces” its simplest to first consider only translational motion. Consider an inertial coord system S: Cartesian coordinate axes (X,Y,Z). Coord system S: Cartesian coordinate axes (X,Y,Z). S is in accelerated translational motion with respect to S. Assume motion is such that axes (X,Y,Z) remain parallel to (X,Y,Z). (No rotation for now!)

Translating Coordinate Systems Consider a point P in space, located at r = (x,y,z) & r = (x,y,z) with respect to O & O. O position = r 0 = (x 0,y 0,z 0 ) in S: Clearly: r = r + r 0  x = x + x 0 y = y + y 0 z = z + z 0

r = r + r 0  Velocity: v = (dr/dt) = (dr/dt) + (dr 0 /dt) Or: v = v + v 0  Acceleration: a = (dv/dt) = (dv/dt) + (dv 0 /dt) a = (d 2 r/dt 2 ) = (d 2 r/dt 2 ) + (d 2 r 0 /dt 2 ) Or: a = a + a 0 Suppose particle, mass m, is at P, acted on by force F.  Newton’s 2 nd Law (only valid in inertial frame S) : ma = F or m(d 2 r/dt 2 ) = F

Newton’s 2 nd Law (in S): ma = F or m(d 2 r/dt 2 ) = F But, a = a + a 0  m(a + a 0 ) = F Or m(d 2 r/dt 2 ) + m(d 2 r 0 /dt 2 ) = F To treat the dynamics of particle in the accelerated frame S, rewrite as ma = F - ma 0 (1) Or m(d 2 r/dt 2 ) = F - m(d 2 r 0 /dt 2 ) (2) (1) & (2)  Mass times acceleration in accelerated frame = Physical force (F) minus ma 0 (or m(d 2 r 0 /dt 2 )). 2nd term comes from coordinate transformation of Newton’s 2 nd Law from inertial frame S to non-inertial frame S.  If we insist on writing ma = F  We must have ma 0 = a (fictitious) “force”, even though its not a physical force!

Newton’s 2 nd Law (inertial frame S): ma = F or m(d 2 r/dt 2 ) = F Newton’s 2 nd Law (accelerated frame S): ma = F or m(d 2 r/dt 2 ) = F, Where F = F - m(d 2 r 0 /dt 2 ) = F - ma 0 ma 0  A non-inertial or “fictitious force”: Comes solely from kinematics of the coord transformation! Note: For non-accelerated (inertial) frames, ma 0 = 0  F = F  Newton’s Laws are the same in all inertial frames.