Lecture 31: Dynamics in Non-Inertial Frames • Last time, we found the following relationship between velocities measured in interial and non-inertial frames: = + + v f V vr ×r • To find the acceleration, we take the time derivative: ≈ dv ’ d (V + v + ×r) ’ = f = r a f ∆ ÷ ∆ ÷ dt dt « ◊fixed « ◊fixed dv d ( ×r) = dV + r + dt dt dt ≈ dV ’ ≈ dv ’ ≈ dr ’ = ∆ ÷ + ∆ r ÷ + " ×r + × ∆ ÷ « dt ◊fixed « dt ◊fixed «dt ◊fixed • Since our goal is to relate quantities in the two frames, we need to make some substitutions on the left-hand side – i.e., want to write it in terms of quantities measured in the non-inertial frame • The first term is just the relative acceleration of the frames, which is the same in both frames (up to a minus sign…) • The second term we can write as: dv ’ ≈ dv ’ r = r + × ∆ ÷ ∆ ÷ vr « dt ◊fixed « dt ◊noninertial = + × ar vr • The third term we can modify in the same way: dr ’ ≈ dr ’ × ∆ ÷ = × ∆ ÷ + × ( ×r) «dt ◊fixed «dt ◊noninertial = × + × ( × ) vr r • Putting this together, we have: = "" + + × + × + × + × ( × ) a f R f ar vr vr " r r = "" + + × + × + × ( × ) R f ar 2 vr " r r Dynamics in a non-inertial frame • Now that we understand how acceleration looks in the two frames, we can write Newton’s Law – which is valid only in the inertial frame! = = "" + + × + × + × ( × ) F ma f mR f mar 2m vr m " r m r • From this, we can derive a “Newton-like” law that’s valid in the non-inertial frame: = "" + + × + × + × ( × ) F mR f mar 2m vr m " r m r = = − "" − × − × − × ( × ) Feff mar F mR f 2m vr m " r m r • So there appear to be additional “forces” causing the particle to accelerate in this frame Fictitious Forces • So the set of “fictitious” forces we need to apply a Newton-like law in a non-inertial frame are: − "" − × 1. mR f and m " r : these arise from the linear and angular acceleration of the non-inertial frame 2. −m × ( ×r) : this is the familiar “centrifugal force” • i.e., the “force” you feel pulling you to the right when you make a left-hand turn − × 3. 2m vr : this is the “coriolis force” • note that it only applies to a particle that is moving in a rotating reference frame Example • Problem 10-18 from the text: • A warship fires a shell due south near the Falkland Islands (latitude 50o S). The shell has initial velocity of 800m/s and initial angle of 37o. How big is the correction for the non-inertial nature of the reference frame fixed to the warship? • Step 1 is to find an inertial frame – We’ll choose a frame fixed at the center of the Earth – This is still not exactly inertial (Earth is orbiting the sun), but it’s a much better approximation than the Earth’s surface is • The real force F acting on the shell is simply the force of gravity • Let’s look now at the menu of fictitious forces: − "" − × 1. mR f and m " r : there’s no angular acceleration, and the linear acceleration is the one for uniform circular motion: "" = ( ) mR f m × × R 2. −m ×( ×r) : of the same form as the above term, but much smaller (since r is much less than R). We’ll neglect it. − × 3. 2m vr : to understand the magnitude of this, we need the components of ω in our frame • Let’s set up our (non-inertial) axes as follows: p o o u = ω (−cos50 i + 0j − cos50 k) = z = + + v vxi 0j vz k v R = Rk 37o 50o x = south • Then the cross-products are: ω R i j k × R = ωR −cos50o 0 −sin50o 0 0 1 = −ωR (−cos50o ) j = ωR cos50o j i j k ×( × R) = ω 2 R −cos50o 0 −sin50o 0 cos50o 0 2 o o o = ω R cos50 ( sin50 i + cos50 k⁄) • The other cross product we need is: i j k × = ω − o − o vr cos50 0 sin50 vx 0 vz = ω − 0 + o ( vx sin50 vz cos50 ) j • Let’s consider the first of these fictitious forces: 2 o o o ×( × R) = ω R cos50 ( sin50 i + cos50 k⁄) • This acceleration is constant throughout the motion of the shell – In fact, if one were to measure g by noting the acceleration of a free-falling object, one would find: = + ( ) g go × × R = + ω 2 2 o + ω 2 o o (go R cos 50 )k R cos50 sin50 i – go is the acceleration due to gravity – Note that the water’s surface is perpendicular to g, not to go! • So we’ll consider g to be the “effective gravity” • There is still anomalous motion due to the Coriolis force: = − = − ω − 0 + o ay 2 × vr 2 ( vx sin50 vz cos50 ) = − ω − o 0 + o − o 2 ( vo cos37 sin50 (vo sin37 gt)cos50 ) dv = 2ω (v c s − (v s − gt)c ) = y o 1 2 o 1 2 dt ≈ 2 ’ ’ ( ) = ω − − gt = dy vy t 2 ∆voc1s2t ∆vo s1t ÷c2 ÷ « « 2 ◊ ◊ dt ≈ ≈ 3 ’ ’ = ω 2 − 2 − gt y(t) ∆voc1s2t ∆vo s1t ÷c2 ÷ « « 3 ◊ ◊ • To find the miss distance in y, we evaluate y at the time of landing: 2v s t = o 1 l g 4v3c s s2 ≈ 4v3s3 8v3s3 ’ ’ = ω o 1 2 1 − o 1 − o 1 y(tl ) ∆ 2 ∆ 2 2 ÷c2 ÷ « g « g 3g ◊ ◊ 4ωv3s2 » 1 ÿ = o 1 c s − s c g 2 … 1 2 3 1 2 ⁄Ÿ • Plugging in the numbers, we get: 4(7.4 ×10−5 s-1 )(800m/s)3 (0.6)2 = [ − ] y(tl ) 2 0.61 0.13 (9.8m/s2 ) = 273m.