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Rotating Coordinate Systems. For Translating Systems: We just got Newton’s 2 nd Law (in the accelerated frame): ma = F where F = F - ma 0 ma 0 A non-inertial or fictitious force Solely from the COORDINATE TRANSFORMATION !

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## Rotating Coordinate Systems

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**Rotating Coordinate Systems**• For Translating Systems: We just got Newton’s 2nd Law (in the accelerated frame): ma = F where F = F - ma0 ma0 A non-inertial or fictitious force Solely from the COORDINATE TRANSFORMATION! Rotating systems are accelerating systems! • For Rotating Systems: Follow a similar procedure as for translating systems & similarly get: Newton’s 2nd Law (in the accelerated frame): ma = F where F = F - mar mar A non-inertial or fictitious force: Solely from the COORDINATE TRANSFORMATION! For rotating systems: mar = much more complicated expressions than ma0!**As in our treatment of rigid bodies, consider again 2 sets**of axes: • A “Fixed” or inertial set (space axes in Goldstein notation): (x1,x2,x3) & A “Rotating” (body axes in Goldstein notation): (x1,x2,x3). CAUTION!The treatment here & the associated figures are from Marion’s book. His unprimed coordinate system = the accelerating system & his primed system is fixed. The rotating system treatment in Goldstein uses the OPPOSITE notation:The primed system is accelerating (rotating) & the unprimed is fixed.**2 axis sets: “Fixed”: (x1,x2,x3),**“Rotating”: (x1,x2,x3). • Consider point P in space. r in fixed system, r in rotating system. R = position of the origin of the rotating system with respect to the fixed axes:See fig: • Clearly: r = R + r**To relate time derivatives in the fixed & rotating systems,**use the results we had for rigid bodies (letting space = s fixed = f & r rotating ): (d/dt)fixed = (d/dt)rotating + ω or, for G = arbitrary vector, (dG/dt)fixed = (dG/dt)rotating + ω G (1) • First, special case: Let G = ω (angular velocity) Relation between time derivatives of ω(between angular accelerations) in fixed & rotating frames is: (dω/dt)fixed = (dω/dt)rotating + ω ω But ω ω = 0 (dω/dt)fixed = (dω/dt)rotating ω • Angular acceleration (sometimes called α ω) is the same in the fixed & rotating frames!**Consider again point P: Position in fixed frame = r.**Position in rotating frame = r. Position of origin of rotating frame in fixed frame = R. r = R + r (1) • Goal:Express velocity of point P in fixed system in terms of ω & velocity in rotating system: 1. Differentiate (1) in the fixed system: (dr/dt)fixed = (dR/dt)fixed + (dr/dt)fixed 2. Use (dr/dt)fixed = (dr/dt)rotating + ω r (dr/dt)fixed = (dR/dt)fixed + (dr/dt)rotating + ω r Moving frame is translating AND rotating!**(dr/dt)fixed = (dR/dt)fixed + (dr/dt)rotating+ ω**r (2) • Define: vf rf (dr/dt)fixed = Velocity relative to the fixed axes. V R (dR/dt)fixed =Velocity (linear) of the moving origin (fixed frame). vr rr (dr/dt)rotating =Velocity (linear) relative to the rotating axes. ω Angular velocity ω r Velocity due to the rotation of the moving axes. (2) becomes: vf = V + vr + ω r**“Fictitious” Centrifugal & Coriolis Forces**• Procedure (the same as for translational acceleration): Use Newton’s 2nd Law & transform from the fixed axis system to rotating the axis system, using the operator: (d/dt)fixed = (d/dt)rotating + ω (1) on the velocity equation we just got! vf = V + vr + ω r (2) • A particle of mass m at point P under the influence of a net force F: Newton’s 2nd Law is validONLY in the fixed, inertial frame (primed coordinates!): F = maf m(dvf/dt)fixed(3)**vf = V + vr + ω r (2)**• Differentiate (2) in fixed frame: (dvf/dt)fixed = [d(V + vr + ω r)/dt]fixed Or: (dvf/dt)fixed = (dV/dt)fixed + (dvr/dt)fixed + [(dω/dt)fixed r]+ ω (dr/dt)fixed (4) • Define:Af (dV/dt)fixed (5) • Recall that (dω/dt)fixed ω(same in both frames). • By discussion we just had: (dvr/dt)fixed (dvr/dt)rotating + ω vr Or: (dvr/dt)fixed = ar+ ω vr (6) ar= Acceleration in rotating frame. • We know: (dr/dt)fixed (dr/dt)rotating + ω r Or: (dr/dt)fixed = vr+ ω r(7) vr= Velocity in rotating frame.**We had, F = maf m(dvf/dt)fixed**• Combine (4)-(7) on previous page: (dvf/dt)fixed =Af + ar+ ω vr + ω r + ω [vr+ ω r] • Put into Newton’s 2nd Law (above) & rearrange: F = maf = mAf + mar + m(ω r) + m[ω (ω r)] + 2m(ω vr) (I) • Observer in rotating frame. Measures mar.Insistson writing this in Newtonian form, even rotating though frame is not inertial! So: mar Feff (II) (I) & (II) together We must have:mar Feff F - mAf - m(ω r) - m[ω (ω r)] - 2m(ωvr)**Applying Newton’s 2nd Law to the rotating frame yields:**Feff mar F - mAf - m(ω r) - m[ω (ω r)] - 2m(ω vr) • PhysicalInterpretations: - mAf : Fromtranslational acceleration of the rotating frame. - m(ω r): From the angular acceleration of rotating frame. - m[ω (ω r)]: Centrifugal “Force”.See figure! - 2m(ω vr): Coriolis “Force”.Comes from motion of particle in rotating system (= 0 if vr = 0) More discussion of last two follows**- m[ω (ω r)]: Centrifugal “Force”**If ω r: This has magnitude mω2r. Outwardly directed from center of rotation!**“Fictitious” Forces**• Physical discussion of “Centrifugal Force”&“Coriolis Force”. • These terms have entered the right side of the product mar(mass x acceleration in rotating frame). They came about because we wanted to write a Newton’s 2nd Law-like eqtn in the rotating frame: Feff mar, when, in fact Newton’s 2nd Law, F = maf, is valid only in the fixed (inertial) frame. The transformation from the fixed to the rotating frame gave: Feff F - (non-inertial terms)**Feff F - (non-inertial terms)**• Example: A body rotating about a fixed (attractive) force center: The only real force (defined by Newton!) is force of attraction to the center: Causesthe centripetal acceleration (in the inertial frame!). • However, an observer moving WITH the body (in the rotating frame) notices that body doesn’t fall towards the force center. To that observer, the body is stationary (in equilibrium). Total “force” = 0 in the rotating frame: The observer postulates an additional “force”, the Centrifugal “Force”.It comes solely from the attempt to extend Newton’s 2nd Law to the non-inertial system! Only possible with a correction “force”. • Similar for the Coriolis “Force”: This correction “force” arises when one attempts to describe the motion of the body relative to rotating system using Newton’s 2nd Law.**Bottom Line:In the sense just discussed, the Centrifugal**Force & the Coriolis Force are “artificial” or “fictitious” forces. • However, as long as we understand what they really are(partially a philosophical view) they are very useful concepts. • They can be used with the Newtonian & also the Lagrangian & Hamiltonian methods to treat complicated problems involving rotating bodies, relative motion where one body is translating & the other is rotating, etc.**Example from Marion**• A person does measurements with a hockey puck on a large merry-go-round with frictionless horizontal surface. The merry-go-round has constant angular velocityω& rotates counterclockwise as seen from above. Find the effective force on hockey puck after it is given a push. Plot the path for various initial directions & velocities of the puck, as observed by a person on the merry-go-round who pushes the puck (how does he stay on??). See figure:

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