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CHAPTER 10 Motion in a Noninertial Reference Frame 10-1. The accelerations which we feel at the surface of the Earth are the following: (1) Gravitational : 2980 cm/sec (2) Due to the Earth’s rotation on its own axis: ( ) ( ) ( ) 2 2 8 28 5 2 rad/day 6.4 10 cm 86400 sec/day 6.4 10 7.3 10 3.4 cm/sec r πω − = × × = × × × = 2 (3) Due to the rotation about the sun: ( ) ( ) 2 2 13 25 13 2 2 rad/year 1.5 10 cm 86400 365 sec/day 7.3 10 1.5 10 0.6 cm/sec 365 r πω − = × × × ×= × × = 10-2. The fixed frame is the ground. y a θ x The rotating frame has the origin at the center of the tire and is the frame in which the tire is at rest. From Eqs. (10.24), (10.25): ( ) 2f f r r= + + × + × × + ×a r r v�� �ω ω ω ωa R 333 334 CHAPTER 10 Now we have 0 0 0 cos sin 0 f r r a a r V a r r θ θ= − + = = = = = R i r i v a k k �� �ω ω j Substituting gives 2 0 cos sinf v a a a r θ θ= − + + −a i j j i (2 0 cos sin 1f v a r θ θ = − + + + a i j ) a (1) We want to maximize fa , or alternatively, we maximize 2 fa : 4 22 2 2 2 2 2 2 2 0 0 4 2 2 2 2 2 0 0 2 cos cos 2 sin sin 2 2 cos sin f v av a a a r r v av a a r r aθ θ θ θ θ = + + + + + = + + + a θ 2 2 2 0 0 2 2 cos 2 cos 0 when tan fd av a d r ar v θ θθ θ = − + = = a (Taking a second derivative shows this point to be a maximum.) 2 0 2 2 2 4 0 n implies cos ar v v a r v θ θ= = +ta and 0 2 2 4 0 sin ar a r v θ = + Substituting into (1) 2 2 0 2 2 4 2 2 4 0 0 0 1f arv av a r a r v a r v = − + + + + + j a i This may be written as 2 4 20f a a v r= + +a MOTION IN A NONINERTIAL REFERENCE FRAME 335 θ A This is the maximum acceleration. The point which experiences this acceleration is at A: where 02 ar v tanθ = 10-3. We desire . From Eq. (10.25) we have eff 0=F ( )eff 2f rm m m m= − − × − × × − ×r r v�� �ω ω ω ωF F R r 0 ω The only forces acting are centrifugal and friction, thus 2smg m rµ ω= , or 2 s gr µ ω= 10-4. Given an initial position of (–0.5R,0) the initial velocity (0,0.5ωR) will make the puck motionless in the fixed system. In the rotating system, the puck will appear to travel clockwise in a circle of radius 0.5R. Although a numerical calculation of the trajectory in the rotating system is a great aid in understanding the problem, we will forgo such a solution here. 10-5. The effective acceleration in the merry-go-round is given by Equation 10.27: 2 2x x yω ω= +�� � (1) 2 2y y xω ω= −�� � (2) These coupled differential equations must be solved with the initial conditions ( )0 0 0.5 mx x≡ = − , ( )0 0 0 my y≡ = , and ( ) ( ) 100 0 2 m sx y v −= =� � 0v ⋅ , since we are given in the problem that the initial velocity is at an angle of 45° to the x-axis. We will vary over some range that we know satisfies the condition that the path cross over . We can start by looking at Figures 10-4e and 10-4f, which indicate that we want . Trial and error can find a trajectory that does loop but doesn’t cross its path at all, such as 0v 1⋅ 0. 0 0( , )x y 0.47 m> s− 0v 153 m s−= ⋅ . From here, one may continue to solve for different values of v until the wanted crossing is eyeball-suitable. This may be an entirely satisfactory answer, depending on the inclinations of the instructor. An interpolation over several trajectories would show that an accurate answer to the problem is , which exits the merry-go-round at 3.746 s. The figure shows this solution, which was numerically integrated with 200 steps over the time interval. 0 10.512 m s−= ⋅0v 336 CHAPTER 10 –0.5 0 0.5 1 0.5 0 0.5 x (m) y (m ) 1 –1 –1 10-6. z m r z = f(r) Consider a small mass m on the surface of the water. From Eq. (10.25) ( )eff 2f rm m m r m= − − × − × × − ×r v�� �ω ω ω ωF F R In the rotating frame, the mass is at rest; thus, eff 0=F . The force F will consist of gravity and the force due to the pressure gradient, which is normal to the surface in equilibrium. Since , we now have 0f r= = =R v�� �ω ( )0 pm m= + − × ×g F rω ω where pF is due to the pressure gradient. Fp mg mω2r θ′ θ Since F , the sum of the gravitational and centrifugal forces must also be normal to the surface. eff 0= Thus θ′ = θ. 2 tan tan r g ωθ θ= =′ MOTION IN A NONINERTIAL REFERENCE FRAME 337 but tan dz dr θ = Thus 2 2 constant 2 The shape is a circular paraboloid. z r g ω= + 10-7. For a spherical Earth, the difference in the gravitational field strength between the poles and the equator is only the centrifugal term: 2poles equatorg g Rω− = For and R = 6370 km, this difference is only 3457.3 10 rad sω −= × ⋅ 1− 2 mm s−⋅ . The disagreement with the true result can be explained by the fact that the Earth is really an oblate spheroid, another consequence of rotation. To qualitatively describe this effect, approximate the real Earth as a somewhat smaller sphere with a massive belt about the equator. It can be shown with more detailed analysis that the belt pulls inward at the poles more than it does at the equator. The next level of analysis for the undaunted is the “quadrupole” correction to the gravitational potential of the Earth, which is beyond the scope of the text. 10-8. x y z λ ω Choose the coordinates x, y, z as in the diagram. Then, the velocity of the particle and the rotation frequency of the Earth are expressed as ( ) ( ) 0,0, cos , 0, sin z ω λ ω λ = = − v � ω (1) so that the acceleration due to the Coriolis force is ( )2 2 0, coszω= − × = −a r ��ω , 0λ (2) 338 CHAPTER 10 This acceleration is directed along the y axis. Hence, as the particle moves along the z axis, it will be accelerated along the y axis: 2 cosy zω λ= −�� � (3) Now, the equation of motion for the particle along the z axis is 0z v gt= −� (4) 20 1 2 z v t gt= − (5) where v is the initial velocity and is equal to 0 2gh if the highest point the particle can reach is h: 0 2v = gh c (6) From (3), we have 2 cosy zω λ= −� + (7) but the initial condition ( )0y z = =� 0 implies c = 0. Substituting (5) into (7) we find ( ) 2 0 2 2 0 1 2 cos 2 cos 2 y v t gt v t ω λ ω λ = − − = − � gt (8) Integrating (8) and using the initial condition y(t = 0) = 0, we find 2 20 1 cos 3 y gtω λ v t = − (9) From (5), the time the particle strikes the ground (z = 0) is 0 1 0 2 v gt = − t so that 0 2v t g = (10) Substituting this value into (9), we have 3 2 0 0 03 2 3 0 2 8 41 cos 3 4 cos 3 v v y g v g g v g ω λ ω λ = − = − (11) If we use (6), (11) becomes MOTION IN A NONINERTIAL REFERENCE FRAME 339 34 cos 3 h y g ω λ= − 8 (12) The negative sign of the displacement shows that the particle is displaced to the west. 10-9. Choosing the same coordinate system as in Example 10.3 (see Fig. 10-9), we see that the lateral deflection of the projectile is in the x direction and that the acceleration is ( ) ( )02 2 sin cosx z ya x v Vω ω λ α= = =��(1) Integrating this expression twice and using the initial conditions, ( )0x 0=� and ( )0x = 0 , we obtain ( ) 20 cos sinx t V tω α= λ (2) Now, we treat the z motion of the projectile as if it were undisturbed by the Coriolis force. In this approximation, we have ( ) 20 1sin 2z t V t gtα= − (3) from which the time T of impact is obtained by setting z = 0: 0 2 sinV T g α= (4) Substituting this value for T into (2), we find the lateral deflection at impact to be ( ) 3 2024 sin cos sinVx T g ω λ α= α (5) 10-10. In the previous problem we assumed the z motion to be unaffected by the Coriolis force. Actually, of course, there is an upward acceleration given by 2 x yvω− so that 02 cos cosz V gω α λ= −�� (1) from which the time of flight is obtained by integrating twice, using the initial conditions, and then setting z = 0: 0 0 2 sin 2 cos cos V T g V α ω α λ=′ − (2) Now, the acceleration in the y direction is ( ) ( 0 2 2 cos sin y x za y v V )gt ω ω λ α = = = − − �� (3) Integrating twice and using the initial conditions, ( ) 00 cosy V α=� and ( )0 0y = , we have 340 CHAPTER 10 ( ) 3 201 cos cos sin cos3t gt V t V t0y ω λ ω λ α α= − + (4) Substituting (2) into (4), the range R′ is ( ) ( ) 3 3 3 3 2 0 0 0 3 2 00 0 sin cos 4 sin cos 2 cos cos8 3 22 cos cos 2 cos cos V g V V R g Vg V g V cos cos ω α λ ω α λ α λ ω α λω α λ ω α λ= − +′ −− − (5) We now expand each of these three terms, retaining quantities up to order ω but neglecting all quantities proportional to 2ω and higher powers of ω. In the first two terms, this amounts to neglecting 02 cos cV osω α λ compared to g in the denominator. But in the third term we must use 2 2 0 0 0 0 3 20 0 2 2 cos sin 2 2 cos sin 1 cos cos 2 1 cos cos 4 sin cos cos V V V g gV g g V R g α α ωα α αω α λ ω α α λ λ ≅ + − = +′ (6) where is the range when Coriolis effects are neglected [see Example 2.7]: 0R′ 2 0 0 2 cos sin V R g α α=′ (7) The range difference, , now becomes 0R R R∆ = −′ ′ ′ 3 20 2 4 1 cos sin cos sin 3 V R g ω 3λ α α α −′ ∆ = (8) Substituting for in terms of from (7), we have, finally, 0V 0R′ 1 2 3 20 2 1 cos cot tan 3 R R g ω λ α′ −′ α ∆ = (9) MOTION IN A NONINERTIAL REFERENCE FRAME 341 10-11. θ R sin θ d = Rθ This problem is most easily done in the fixed frame, not the rotating frame. Here we take the Earth to be fixed in space but rotating about its axis. The missile is fired from the North Pole at some point on the Earth’s surface, a direction that will always be due south. As the missile travels towards its intended destination, the Earth will rotate underneath it, thus causing it to miss. This distance is: ∆ = (transverse velocity of Earth at current latitude) × (missile’s time of flight) sinR Tω θ= × (1) sin d R d v R ω = (2) Note that the actual distance d traveled by the missile (that distance measured in the fixed frame) is less than the flight distance one would measure from the Earth. The error this causes in ∆ will be small as long as the miss distance is small. Using R = 6370 km, 57.27 10ω −= × rad ⋅ , we obtain for the 4800 km, T = 600 s flight a miss distance of 190 km. For a 19300 km flight the missile misses by only 125 km because there isn’t enough Earth to get around, or rather there is less of the Earth to miss. For a fixed velocity, the miss distance actually peaks somewhere around d = 12900 km. 1s− Doing this problem in the rotating frame is tricky because the missile is constrained to be in a path that lies close to the Earth. Although a perturbative treatment would yield an order of magnitude estimate on the first part, it is entirely wrong on the second part. Correct treatment in the rotating frame would at minimum require numerical methods. 10-12. z Fsr0 x λ ε 342 CHAPTER 10 Using the formula ( )eff 2f rm m m= − × × − ×F a r vω ω ω (1) we try to find the direction of when effF fma (which is the true force) is in the direction of the z axis. Choosing the coordinate system as in the diagram, we can express each of the quantities in (1) as 0 0 ( cos , 0, sin ) (0,0, ) (0,0, ) r f R m mg ω λ ω λ = = − = = − v r a ω (2) Hence, we have cos yRω λ× =r eω (3) and (1) becomes eff 0 cos 0 sin 0 cos 0 x y zmg m R z ω λ ω ω λ = − − − e e e λ 2 2 F e (4) from which, we have 2eff 0 sin cos cosz xmg mR mR zω λ λ ω λ= − + +eF e (5) e Therefore, 2 2 2 0 ( ) sin cos ( ) cos f x f z F mR F mg mR ω λ λ ω λ = = − + (6) The angular deviation is given by 2 2 2 0 ( ) sin cos tan cos( ) f x f z F R g RF ω λ λε ω λ= = − (7) Since ε is very small, we can put ε ε≅ . Then, we have 2 2 2 0 sin cos cos R g R ω λ λε ω λ= − (8) It is easily shown that ε is a maximum for 45λ °� . Using , , , the maximum deviation is 86.4 10 cmR = × 5 17.3 10 secω − −= × 2980 cm/secg = 1.7 0.002 rad 980 ε ≅ ≅ (9) MOTION IN A NONINERTIAL REFERENCE FRAME 343 10-13. ω λ ε z′ z x x′ Earth The small parameters which govern the approximations that need to be made to find the southerly deflection of a falling particle are: height of fall radius of Earth h R δ ≡ = (1) and 2 0 centrifugal force purely gravitational force R g ωα ≡ = (2) The purely gravitational component is defined the same as in Problem 10-12. Note that although both δ and α are small, the product 2 0h gδα ω= is still of order 2ω and therefore expected to contribute to the final answer. Since the plumb line, which defines our vertical direction, is not in the same direction as the outward radial from the Earth, we will use two coordinate systems to facilitate our analysis. The unprimed coordinates for the Northern Hemisphere-centric will have its x-axis towards the south, its y-axis towards the east, and its z-axis in the direction of the plumb line. The primed coordinates will share both its origin and its y′-axis with its unprimed counterpart, with the z′- and x′-axes rotated to make the z′-axis an outward radial (see figure). The rotation can be described mathematically by the transformation cos sinx x zε ε= +′ ′ (3) (4) y y= ′ sin cosz x zε ε= − +′ ′ (5) where 2 sin cos R g ωε λ≡ λ (6) as found from Problem 10-12. a) The acceleration due to the Coriolis force is given by 2X ≡ − × ′a vω (7) Since the angle between ω and the z′-axis is π – λ, (7) is most appropriately calculated in the primed coordinates: 344 CHAPTER 10 2 sinx yω λ=′ ′�� � (8) ( )2 cos siny z xω λ= − +′ ′ ′�� � � λ (9) 2 cosz yω λ=′ ′�� � (10) In the unprimed coordinates, the interesting component is ( )2 sin cos cos sinx yω λ ε λ= +�� � ε (11) At our level approximation this becomes 2 sinx yω λ�� �� (12) Using the results for and y� z� , which is correct to order ω (also found from Example 10.3), 2 22 sin cosx gtω λ�� � λ (13) Integrating twice and using the zeroth order result for the time-of-fall, 2h=t , we obtain for the deflection g 2 22 sin cos 3X h d g ω λ= λ (14) b) The centrifugal force gives us an acceleration of ( )c ≡ − × × ′a rω ω (15) The component equations are then ( )2 sin sin cosx x R zω λ λ= + +′ ′ ′ λ �� (16) (17) 2y ω=′�� y′ ( )2 0cos sin cosz x R zω λ λ λ= + +′ ′ ′��g− (18) where we have included the pure gravitational component of force as well. Now transform to the unprimed coordinates and approximate ( )2 0sin cos sinx R z gω λ λ+ −�� � ε (19) We can use Problem 10-12 to obtain sin ε to our level of approximation 2 0 sin sin cos R g ωε ε λ� � λ (20) The prompts a cancellation in equation (19), which becomes simply 2 sin cosx zω λ� λ (21) Using the zeroth order result for the height, 2 2z h gt= − , and for the time-of-fall estimates the deflection due to the centrifugal force 2 25 sin cos 6c h d g ω λ� λ (22) MOTION IN A NONINERTIAL REFERENCE FRAME 345 c) Variation in gravity causes the acceleration 03g GM g r ≡ − +a r k (23) where ( )x y R z= + + +′ ′ ′jr i is the vector pointing to the particle from the center of the spherical Earth. Near the surface k ( )22 2 2 2 2r x y z R R Rz= + + + +′ ′ ′ � ′ (24) so that (23) becomes, with the help of the binomial theorem, (0 2g g x y zR− + − )′ ′ ′a i j� k (25) Transform and get the x component ( )0 cos 2 singx x z R ε ε− +′ ′�� � (26) ( ) (0 cos sin cos 2 sin cos sing x z x z R )ε ε ε ε ε = − − + + ε (27) (0 3 sing x z R )ε− +� (28) Using (20), 23 sin cosx zω λ�� � λ (29) where we have neglected the x term. This is just thrice the part (b) result, R 2 25 sin cos 2g h d g ω λ� λ (30) Thus the total deflection, correct to order 2ω , is 2 24 sin cos h d g ω λ� λ (31) (The solution to this and the next problem follow a personal communication of Paul Stevenson, Rice University.) 10-14. The solution to part (c) of the Problem 10-13 is modified when the particle is dropped down a mineshaft. The force due to the variation of gravity is now 0 0g g g R ≡ − +a r k (1) As before, we approximate r for near the surface and (1) becomes (0g g x y zR− + + )′ ′ ′a i j� k (2) In the unprimed coordinates, 346 CHAPTER 10 0 x x g R −�� � (3) To estimate the order of this term, as we probably should have done in part (c) of Problem 10-13, we can take 2 2~x h gω , so that 2~ h x h R ω ×�� (4) which is reduced by a factor h R from the accelerations obtained previously. We therefore have no southerly deflection in this order due to the variation of gravity. The Coriolis and centrifugal forces still deflect the particle, however, so that the total deflection in this approximation is 2 23 sin cos 2 h d g ω λ� λ (5) 10-15. The Lagrangian in the fixed frame is ( )212 f fL mv U r= − (1) where fv and fr are the velocity and the position, respectively, in the fixed frame. Assuming we have common origins, we have the following relation f r r= + ×v v rω (2) where v and are measured in the rotating frame. The Lagrangian becomes r rr ( ) ( ) (22 2 2 r r r r m U= + ⋅ × + × − v r rω ω )rrL v (3) The canonical momentum is (r r r L m m )r∂≡ = + ×∂p vv ω r (4) The Hamiltonian is then ( ) ( 221 1 2 2r r r r r mv U r m≡ ⋅ − = − − ×v p rω )H L (5) H is a constant of the motion since 0L t∂ ∂ = , but H ≠ E since the coordinate transformation equations depend on time (see Section 7.9). We can identify ( 21 2c U m= − × rω )r (6) as the centrifugal potential energy because we may find, with the use of some vector identities, ( )22 2 2c r m U rω r −∇ = ∇ − ⋅ rω (7) ( )2 r rm ω = − ⋅ r rω ω (8) MOTION IN A NONINERTIAL REFERENCE FRAME 347 ( )rm= − × ⋅ rω ω (9) which is the centrifugal force. Computing the derivatives of (3) required in Lagrange’s equations r r d m m dt ∂ r= + ×∂ L a v ω v (10) ( ) (r r c r m ∂ = ∇ × ⋅ −∇ + ∂ L v r r ω )U U (11) ( ) ( )r rm m U= − × − × × −∇v rω ω ω (12) The equation of motion we obtain is then ( ) ( )2r rm m= −∇ − × × − ×a rω ω ω rv a m U (13) If we identify F and F , then we do indeed reproduce the equations of motion given in Equation 10.25, without the second and third terms. eff rm= U= −∇ 10-16. The details of the forces involved, save the Coriolis force, and numerical integrations in the solution of this problem are best explained in the solution to Problem 9-63. The only thing we do here is add an acceleration caused by the Coriolis force, and re-work every part of the problem over again. This is conceptually simple but in practice makes the computation three times more difficult, since we now also must include the transverse coordinates in our integrations. The acceleration we add is ( )2 sin sin cos cosc y x z yv v v vω λ λ λ= − + + j λ ka i (1) where we have chosen the usual coordinates as shown in Figure 10-9 of the text. a) Our acceleration is Cg= − +a k a (2) As a check, we find that the height reached is � 1800 km, in good agreement with the result of Problem 9-63(a). The deflection at this height is found to be � 77 km, to the west. b) This is mildly tricky. The correct treatment says that the equation of motion with air resistance is (cf. equation (2) of Problem 9-63 solution) 2 C t v g v = − + + a k v a (3) The deflection is calculated to be � 8.9 km. c) Adding the vaiation due to gravity gives us a deflection of � 10 km. d) Adding the variation of air density gives us a deflection of � 160 km. 348 CHAPTER 10 Of general note is that the deflection in all cases was essentially westward. The usual small deflection to the north did not contribute significantly to the total transverse deflection at this precision. All of the heights obtained agreed well with the answers from Problem 9-63. Inclusion of the centrifugal force also does not change the deflections to a significant degree at our precision. 10-17. Due to the centrifugal force, the water surface of the lake is not exactly perpendicular to the Earth’s radius (see figure). mg C B β β Water surface Tangent to Earth surfaceαA The length BC is (using cosine theorem) 2 2( ) 2 cosα= + −AC mg ACmgBC where AC is the centrifugal force 2 cosω α=AC m R with α = 47° and Earth’s radius , 6400 kmR ≅ The angle β that the water surface is deviated from the direction tangential to the Earth’s surface is 5 sinsin 4.3 10 sin sin αβα β −= ⇒ = = ×BC AC AC BC So the distance the lake falls at its center is sin β=h r where r = 162 km is the lake’s radius. So finally we find h = 7 m. MOTION IN A NONINERTIAL REFERENCE FRAME 349 10-18. Let us choose the coordinate system Oxyz as shown in the figure. O νx νy ν β α x y The projectile’s velocity is v v where β = 37° 0 0 cos sin 0 0 β β = = − G x y v v v gt The Earth’s angular velocity is cos sin 0 ω α ω ω α − = − G where α = 50° So the Coriolis acceleration is ( )( )0 02 2 cos sin 2 sin cosω ω β α β ω α= × = − + −GG Gc zv v gt ea v The velocity generated by Coriolis force is ( ) 20 0 2 cos sin sin cos cosω β α β α ω− −∫tc ca dt v t gt α= =v And the distance of deviation due to the Coriolis force is ( ) 320 0 cossin 3 ω αω α β= = − − −∫tc c gtdt v tz v The flight time of the projectile is 0 2 sin 2 β= vt . If we put this into , we find the deviation distance due to Coriolis force to be cz ~ 260 mcz 350 CHAPTER 10 10-19. The Coriolis force acting on the car is 2 2 sinω ω α= × ⇒ =G GGGc cF m v F mv where α = 65°, m = 1300 kg, v = 100 km/hr. So 4.76 N.cF = G 10-20. Given the Earth’s mass, M , the magnitude of the gravitational field vector at the poles is 245.976 10 kg=× 22 9.866 m/s= =pole pole GMg R The magnitude of the gravitational field vector at the equator is 2 2e2 R 9.768 m/sω= − =eq q eq GMg R where ω is the angular velocity of the Earth about itself. If one use the book’s formula, we have at the poles 2( 90 ) 9.832 m/sλ = ° =g and g at the equator 2( 0 ) 9.780 m/sλ = ° = 10-21. The Coriolis acceleration acting on flowing water is 2 2 sinω ω α= × ⇒ =GG G Gc ca v a v Due to this force, the water is higher on the west bank. As in problem 10-17, the angle β that the water surface is deviated from the direction tangential to Earth’s surface is 5 2 2 2 2 2 2 2 sin 2.5 10 4 sin ω α ω αsinβ −= = = ×+ + ∝ c c a v g a g v The difference in heights of the two banks is 3sin 1.2 10 mβ −∆ = = ×Ah where A m is the river’s width. 47= 10-22. The Coriolis acceleration is 2ca v ω= ×G G G . This acceleration caG pushes lead bullets eastward with the magnitude 2 coscos 2ω α ωGc gt α= =a v , where α = 42°. The velocity generated by the Coriolis force is MOTION IN A NONINERTIAL REFERENCE FRAME 351 2( ) cosω α= =∫cv t a dt gt and the deviation distance is 3 ( ) cos 3 ω α∆ = =∫c c gtx v t dt The falling time of the bullet is 2= ht . So finally g 3 38 cos 2.26 10 m 3 ω α −∆ = = ×c hx g 352 CHAPTER 10
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