# Mecflu - movimento

(Parte 1 de 2)

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:

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):

334 CHAPTER 10 Now we have

cos sin 0f ra r

Va r

Ri ri v a

Substituting gives 2 cos sinf v a a r cos sin 1f v

We want to maximize fa, or alternatively, we maximize 2 fa:

2 cos cos 2 sin sin

2 2c os sin va v a a va v a r aθ θθθθ

2 cos 2 cos

0 when tan fd av a

(Taking a second derivative shows this point to be a maximum.) n implies cos ar v v ar v θθ==+ta and sin ar

Substituting into (1) arva v a

r ar v a r v ai

This may be written as

MOTION IN A NONINERTIAL REFERENCE FRAME 335

This is the maximum acceleration. The point which experiences this acceleration is at A:

tanθ=

The only forces acting are centrifugal and friction, thus 2smgmrµω=, or

2 sgrµω=

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.

⋅, 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

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.

x (m)

10-6. z m r

Consider a small mass m on the surface of the water. From Eq. (10.25)

In the rotating frame, the mass is at rest; thus, eff0=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

mω r θ′

Since F, the sum of the gravitational and centrifugal forces must also be normal to the surface. eff 0=

Thus θ′ = θ.

tan tan rg ωθθ==′

MOTION IN A NONINERTIAL REFERENCE FRAME 337 but tan dzdr θ=

Thus

2 constant 2

The shape is a circular paraboloid.

zr 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:

2 polesequatorggRω−=

For and R = 6370 km, this difference is only 3457.310radsω−=×⋅1−2 mms−⋅. 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.

y z

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

cos , 0, sinz ω λω λ

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:

where v is the initial velocity and is equal to 02gh if the highest point the particle can reach is h:

1 2c os yv t

From (5), the time the particle strikes the ground (z = 0) is

so that

Substituting this value into (9), we have

4 cos v yg v g ωλ ωλ

MOTION IN A NONINERTIAL REFERENCE FRAME 339

y g

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

Now, we treat the z motion of the projectile as if it were undisturbed by the Coriolis force. In this approximation, we have

from which the time T of impact is obtained by setting z = 0:

02s inV T

Substituting this value for T into (2), we find the lateral deflection at impact to be sin cos sinV xT

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 2xyvω− so that

from which the time of flight is obtained by integrating twice, using the initial conditions, and then setting z = 0:

2s in 2c os cos

Now, the acceleration in the y direction is yx zay v

V )gtω ωλ α

() 32 01 cos cos sin cos3 tgtVtVt0yωλωλαα=−+ (4)

Substituting (2) into (4), the range R′ is sin cos 4 sin cos 2 cos cos8 32 2 cos cos 2 cos cos

Vg V V R

gVgV gV cos cos ω αλ ω α λ α λ ωαλωαλωαλ=−+′−−− (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 02coscVosωαλ compared to g in the denominator. But in the third term we must use

2c os sin 2 cos sin 1 cos cos

2 1c os cos

4 sin cos cos ggV g where is the range when Coriolis effects are neglected [see Example 2.7]: 0R′ 2

4 1 cos sin cos sin

Substituting for in terms of from (7), we have, finally, 0V0R′

MOTION IN A NONINERTIAL REFERENCE FRAME 341 10-1.

θ R sin θ d = R θ

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) sinRTωθ=× (1)

sin

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, 57.2710ω−=× 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.

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 x λ

we try to find the direction of when effFfma (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

r a

Hence, we have cosyRωλ×=reω (3) and (1) becomes eff 0 cos 0 sin 0c os 0 zmg m R ω λ ωωλ =− − − e e λ

Fe (4) from which, we have

2 eff 0 sin cos cosz xmg mR mR zω λλ ω λ=− + +eFe (5) e

Therefore, 2

() sin cos () cosfxfz

Fm R Fm g mR ωλ λ

The angular deviation is given by

() sin cos tan

cos() fxfz

Since ε is very small, we can put εε≅. Then, we have sin cos

MOTION IN A NONINERTIAL REFERENCE FRAME 343 10-13.

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 EarthhR δ≡= (1) and centrifugal force purely gravitational forceRg 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 20hgδαω= 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

where

sin cos as found from Problem 10-12. a) The acceleration due to the Coriolis force is given by

Since the angle between ω and the z′-axis is π – λ, (7) is most appropriately calculated in the primed coordinates:

In the unprimed coordinates, the interesting component is ()2sincoscossinxyωλελ=+ ε (1)

Using the results for and y z , which is correct to order ω (also found from Example 10.3),

Integrating twice and using the zeroth order result for the time-of-fall, 2h=t, we obtain for the deflection g

sin cos3X h d g b) The centrifugal force gives us an acceleration of

where we have included the pure gravitational component of force as well. Now transform to the unprimed coordinates and approximate

We can use Problem 10-12 to obtain sin ε to our level of approximation 2 sin sin cos

Using the zeroth order result for the height, 22zhgt=−, and for the time-of-fall estimates the deflection due to the centrifugal force

sin cos6c h d g

MOTION IN A NONINERTIAL REFERENCE FRAME 345 c) Variation in gravity causes the acceleration

Transform and get the x component

()0 cos 2 sing x zR

() (0 cos sin cos 2 sin cos sing

where we have neglected the x term. This is just thrice the part (b) result, R

sin cos2g h d g

Thus the total deflection, correct to order 2ω, is

24s in cosh d g ωλ 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

0g g gR

As before, we approximate r for near the surface and (1) becomes

(0g g xy zR

In the unprimed coordinates,

0 x xg R

To estimate the order of this term, as we probably should have done in part (c) of Problem 10-13, we can take 2~xhgω, so that

which is reduced by a factor hR 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

10-15. The Lagrangian in the fixed frame is

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 where v and are measured in the rotating frame. The Lagrangian becomes r

The Hamiltonian is then

H is a constant of the motion since 0Lt∂∂=, but H ≠ E since the coordinate transformation equations depend on time (see Section 7.9). We can identify

as the centrifugal potential energy because we may find, with the use of some vector identities,

MOTION IN A NONINERTIAL REFERENCE FRAME 347

which is the centrifugal force. Computing the derivatives of (3) required in Lagrange’s equations

r r d m

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

( )2s in sin cos coscy x z yvv v vω λλ λ =− + + j λ kai (1) where we have chosen the usual coordinates as shown in Figure 10-9 of the text.

a) Our acceleration is

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 7 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) v g

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.

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 ββWater surfaceTangent to Earth surface

The length BC is (using cosine theorem) where AC is the centrifugal force 2cosωα=ACmR with α = 47° and Earth’s radius , 6400kmR≅

The angle β that the water surface is deviated from the direction tangential to the Earth’s surface is

sin sin αβαβ−=⇒ ==×BCACAC BC

So the distance the lake falls at its center is sinβ=hr 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.

x y The projectile’s velocity is v where β = 37° 0

cos sin 0

G xyvv v gt

The Earth’s angular velocity is cos sin 0

So the Coriolis acceleration is

The velocity generated by Coriolis force is

2 cos sin sin cos cosω βα β α ω− −∫t ccadtvtgtα==v

And the distance of deviation due to the Coriolis force is

0 cossin 3 ω αωα β== − − −∫tcc gtdt v tzv

The flight time of the projectile is 02sin2 β=vt. If we put this into , we find the deviation distance due to Coriolis force to be cz

10-19. The Coriolis force acting on the car is

2 sinω ωα=× ⇒ =G GccFm v F mv where α = 65°, m = 1300 kg, v = 100 km/hr.

So 4.76N.cF=G

pole GMg R

The magnitude of the gravitational field vector at the equator is

eq GMg R where ω is the angular velocity of the Earth about itself. If one use the book’s formula, we have

10-21. The Coriolis acceleration acting on flowing water is

2 sinω ωα=× ⇒ =G Gccav 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

(Parte 1 de 2)