Lecture Notes for PHY 405 Classical Mechanics From Thorton & Marion's Classical Mechanics Prepared by Dr. Joseph M. Hahn Saint Mary's University Department of Astronomy & Physics November 9, 2004 Chapter 10: Motion in a Noninertial Reference Frame Recall that Newton's Laws of motion are valid in an inertial (ie, non{accelerated) reference frame. Is this classroom an inertial reference frame? Why or why not? Since there is no reference frame that is absolutely at rest, we need to derive the laws that describe motion in a non{inertial reference frames. In particular, we are going to focus on the motion of bodies moving in a rotating reference frame. This will allow us to calculate the motion of bodies near the surface of a rotat- ing planet, as well as the motion of a rotating rigid body (eg, a spinning top). 1 A rotating coordinate system Let the vector r0 point to particle P from the origin of some ‘fixed’ or inertial coordinate system. Let R point to the origin of some noninertial coordinate system; this accelerated origin can be translating as well as rotating. Let r = the position of particle P in this rotating ref' frame. Thus: r0 = R + r Fig. 10{1. Now use Newton's Law, which is valid in the stationary reference frame, to derive the acceleration of point P, ¨r, as seen in the rotating ref' frame. First, compute r_ and then get ¨r. We do this by first computing r_0 = P's velocity in the fixed reference frame: dr0 dR dr = + dt dt dt fixed fixed fixed where the ()fixed is used to indicate that we are calculating these velocities with respect to the fixed origin. 2 Note that the moving origin can be translating as well as rotating. Thus (dR=dt)fixed should be regarded as the translational velocity of the moving origin (measured in the fixed frame). The right term will thus account for any possible rotation of the moving origin. The position of particle P in the moving ref' frame is: r = xix^i i which is shorthand for = xXx^ + yy^ + z^z dr dx^ Thus its velocity is = x_ x^ + x i dt i i i dt fixed i i X X where the first term on the right is simply P's velocity measured in the rotating frame, ie, (dr=dt)rot, and the second term is due to the rotation of the moving coordinate system. 3 Now calculate dx^i=dt, which is due to the rotation of the moving origin about the rotation axis !~ . We will also use the magnitude of the rotation axis to indicate the rotation rate, ie ! = j!~ j. Fig. 1{19 If α is the angle between x^i and !~ , then x^i moves a distance ∆x^i in time ∆t. Note that ∆x^i is perpendicular to both x^i and !~ , so ∆x^i / (!~ × x^i). We also anticipate that ∆x^i / ∆t, so we write ∆x^i = γ∆t~! × x^i, where γ is some coefficient. 4 Now show that γ = 1. But first note that ` j!~ × x^ij = ! sin α since jx^ij = 1; and that sin α = = ` jx^ij ) j!~ × x^ij = !` And since j∆x^ij = distance the end of x^i has traveled in time ∆t, then j∆x^ij = `!∆t = γ∆tj!~ × x^ij = γ∆t!` this implies that γ = 1 dx^ ∆x^ thus i = i = !~ × x^ dt ∆t i which is the velocity of the x^i axes due to their rotation about the !~ axis. This velocity is measured relative to the origin of the fixed reference frame. Plug this result back into our earlier equation: dr dx^ = x_ x^ + x i dt i i i dt fixed i i X X dr = + x !~ × x^ dt i i rot i X dr = + !~ × r dt rot The above result is of course true for any arbitrary vector Q: dQ dQ = + !~ × Q dt dt fixed rot ie, the velocity of Q as measured in the fixed reference frame = the velocity of Q in the rotating frame + !~ × Q. Keep this result handy...we will use it again. 5 Fig. 10{1. Or if we again adopt r0 = R + r, where r0 is particle P's position vector in the fixed ref' frame, R is the position vector of the moving origin, and r = P's position relative to the moving origin, then dr0 dR dr = + + !~ × r dt dt dt fixed fixed rot Or if we define vf = particle P's velocity relative to the fixed origin V = translational velocity of the moving origin (relative to the fixed origin) vr = P's velocity relative to the rotating axes !~ = angular velocity of the rotating axes !~ × r = translational velocity due to the rotating axes, Then vf = V + vr + !~ × r 6 Centrifugal & Coriolis forces Now calculate particle P's acceleration af as measured in the fixed frame: dv dV dv dr a = f = + r + !~_ × r + !~ × f dt dt dt dt fixed fixed fixed fixed F = m where F is the force on particle P which has mass m. dV set R¨ = = the translational acceleration of the moving origin f dt fixed The (dvr=dt)fixed term is obtained using our formula for dQ=dt: dv dv r = r + !~ × v dt dt r fixed rot = ar + !~ × vr where ar is P's acceleration as measured in the rotating reference frame. The last term is: dr dr !~ × = !~ × + !~ × r dt dt fixed rot = !~ × vr + !~ × (!~ × r) Thus F a = R¨ + a + !~_ × r + 2!~ × v + !~ × (!~ × r) = f f r r m 7 Now suppose you are an observer that co{moves and co{rotates with this moving/rotating reference frame. What is P's acceleration that you would measure in this moving ref' frame? You would thus see particle P move as if it were driven by the effective force ¨ _ Feff = mar = F − mRf − m!~ × r − 2m~! × vr − m~! × (!~ × r) Keep in mind that Feff is a fictitious force! It is merely the invention of the observer who insists writing a Newton{esque law of motion Feff = mar. The only real force in this problem is the inertial force F; the other \noninertial" force terms are a consequence of the observer employing a noninertial reference frame. ¨ _ The first two noninertial force terms, mRf and m!~ × r, are due to: translational acceleration of the moving origin & changes in the orientation of the rotation axis !~ . However these terms are zero if the moving frame is merely rotating about a fixed axis with a constant angular velocity !. This will be typical of the problems studied in this class. 8 The remaining two terms are more important: −m~! × (!~ × r) = centrifugal force −2m~! × vr = Coriolis force Which direction is the centrifugal force pointing? Fig. 10{3 The centrifugal force is perpendicular to & points away from the !~ axis. 9 Example: A hockey puck is initially stationary on a frictionless merry{go{round that rotates counter{clockwise with an angular velocity !. An observer on the merry{go{round releases the puck from an off–axis spot. In which directions does the puck initially move in due to centrifugal & Coriolis forces? ¨ _ First: what is F =?. Rf ? !~ × r? 10 Another observer sitting in a fixed reference frames also watches the puck|what does he see? Motion at the Surface of the Earth Earth spins about its rotation axis with an angular frequency 2π ! = = 7:3 × 10−5 radians/sec 1 day Place the fixed (inertial) coordinates (x^0; y^0; ^z0) at the center of the Earth, and the moving coordinates (x^; y^; ^z) at the surface such that x^ points South, and y^ points East, and ^z points Up: Fig. 10{9 11 Examine how the centrifugal force alters our perception of gravity F = mg. Also assume that the Earth's rotation pole !~ = constant. Then ¨ _ ar = g0 − Rf − !~ × r − 2!~ × vr − !~ × (!~ × r) GM⊕ ^ where g0 = − 2 Rf is the downward pull due to gravity Rf ¨ Is Rf = 0? dQ dQ recall that = + !~ × Q dt dt fixed rot dR_ dR_ thus R¨ = f = f + !~ × R_ f dt dt f !fixed !rot dR_ but R = R ^z so f = 0 f f dt !rot ¨ _ and Rf = !~ × Rf dR dR similarly R_ = f = f + !~ × R f dt dt f fixed rot ¨ ) Rf = !~ × (!~ × Rf ) Plug this result into ar: ar = g0 − 2!~ × vr − !~ × [!~ × (Rf + r)] ' g0 − 2!~ × vr − !~ × (!~ × Rf ) for motion near the Earth's surface where jrj jRf j. Consequently, a plumb line (eg, a mass dangling from a string) does not point directly towards the center of the Earth, due to the centrifugal acceleration −!~ × (!~ × Rf ). And if you cut the string, the Coriolis acceleration −2!~ × vr will cause an additional deflection as the mass falls. 12 Example 10.3 Estimate the deflection caused by the Coriolis force as a particle falls a distance h due to gravity. The particle's acceleration in the rotating reference frame is ar ' g0 − 2!~ × vr − !~ × (!~ × Rf ) ≡ g − 2!~ × vr where g ≡ g0 − !~ × (!~ × Rf ) is the effective gravity that includes the constant centrifugal term. Note that the centrifugal force causes g to deviate slightly from the normal direction, ^z. However we will ignore this slight deviation and write g =' g^z. From the diagram, !~ = −! sin(π=2 − λ)x^ + ! cos(π=2 − λ)^z = −! cos λx^ + ! sin λ^z and vr = vxx^ + vyy^ + vz^z ' vz^z ' −gt^z since we anticipate the horizontal deflections and velocities will be small compared to the vertical motions. 13 Thus x^ y^ ^z !~ × v = −! cos λ 0 ! sin λ = −!gt cos λy^ r 0 0 −gt a y^ ^z So r ' +2 !gt cos λ − g = x ¨x^ + y¨y^ + z¨^z ) x¨ = 0 y¨ = 2!gt cos λ z¨ = −g How do I obtain the particle's vertical motion z(t)? 1 2 z_ = −gt so z(t) = h − 2gt .