PHY 203: Solutions to Problem Set 9

Transcription

1 PHY 03: Solutions to Problem Set 9 December 1, Problem 10.9 We want to calculate the deflection on a projectile underoin (almost) parabolic motion due to the inertial forces of the rotatin frame of the Earth. In order to solve this problem we are oin to consider a series of approximations. We consider the motion to take place in a much shorter scale than that of the curvature of the Earth. That means that we are zoomin in on a reion where the Earth looks flat. The way to understand how small inertial forces are is to compare them to the force that dominates the motion: ravity. For centrifual forces the relevant parameter is ω R Earth , where ω is the anular frequency of Earth s rotation. For Coriolis forces the important parameter is ωv0 10 1, where V 0, the characteristic speed involved, was taken to be 5000km/h (which is more than enouh for all kinds of missiles and jets). The conclusion is that our results will be expressed with reat accuracy to first order in those parameters. Finally, since our paths are short, we will use the uncorrected parabolic trajectory when we calculate the forces. Then iven these forces we will compute the deviation from the oriinal path. Havin discussed the approximations, we turn to the problem. We pick Cartesian coordinates on the surface of the Earth where we perform the experiment. The z unit vector ˆk will point upwards. To first order, it is not important whether we correct this up direction by the centrifual force or not. So we can think of it as coincidin with the radial vector ˆr pointin normal to the surface of the Earth. We let the y unit vector ĵ point East. Finally, î is directed South, in the x-direction (as in T&M Fi. 10-9). In these coordinates the anular velocity of the Earth and the velocity of the particle on the uncorrected parabolic trajectory are ω = ( ω cos λ, 0, ω sin λ), (1) v r = (0, v 0 cos α, v 0 sin α t). () 1

2 The Coriolis force is iven by and from above we have Therefore F cor = m ω v r, (3) ω cos λ 0 ω sin λ 0 v 0 cos α v 0 sin α t = ( v 0 ω cos α sin λ, ω cos λ(v 0 sin α t), v 0 ω cos α cos λ). a = (v 0 ω cos α sin λ, ω cos λ(v 0 sin α t), + v 0 ω cos α cos λ), (4) where the corrections to the y and z components are not important to first order; they basically just deform the parabola, chanin the time of impact slihtly. The deviation from the oriinal direction of motion towards the South is iven by x = v 0 t ω cos α sin λ, (5) and the time of fliht is v 0 sin α t = v 0 ω cos α cos λ v 0 sin α. (6) Thus the deflection due to the Coriolis force to leadin order is d cor = 4ωv3 0 sin α cos α sin λ. (7) The same procedure can be applied to calculate the deflection due to the centrifual force (in the same direction), which is not neliible unless v 0 is very lare: Problem d cen = ω R Earth v 0 sin α cos λ sin λ. (8) Once aain, the Coriolis force is iven by the expression: F = m ω v, (9) where m is the mass of the particle, ω the anular velocity of the non-inertial frame and v the velocity in that frame. In our case the data is: m = 1300k (10) ω = π 4 h 1 ẑ (11) v = 100 km h ˆθ (North) (1) λ = 65 N, (13)

3 where λ is the latitude. The direction of the force is iven by the riht hand rule; it points East. Its manitude is F = mωv sin λ = 4.78N. (14) 3 Frictionless Puck on Turntable Takin into account Coriolis and centrifual effects, the inertial force in the rotatin frame of the turntable is F eff = m a r = m( ω ( ω r) + ω v r. (15) The coordinate system is defined with respect to the centre of the turntable, with the z-axis pointin out of the pae. Thus ω = (0, 0, ω), (16) r = (x, y, 0), (17) v r = (ẋ, ẏ, 0), (18) and hence Also, ω ( ω r) = 0 0 w yω xω 0 = ( xω, yω, 0). 0 0 ω ẋ ẏ 0 (19) = ( ẏω, ẋω, 0). (0) Therefore we find the acceleration a r = ω ( ω r) (xω + ẏω, yω ẋω, 0). (1) Splittin this up into components, we see that ẍ = xω + ẏω, ÿ = yω ẋω. () Now consider the complex coordinate q x + iy, such that these two real equations can be written as q ω q + iω q = 0. (3) 3

4 We make the ansatz q = Ae iγt γ = ω ± 4ω 4ω and since both roots are the same the solution is = ω, q = Ae iωt + Bte iωt, (4) where A and B are complex. Writin A = C + Di and B = E + F i, the initial conditions show that C = x 0, D = 0, E = v x and F = v y + x 0 ω. Thus the final answer is x = (x 0 + v x t) cos ωt + (v y + x 0 ω)t sin ωt, y = (v y + x 0 ω)t cos ωt (x 0 + v x t) sin ωt. (5) 4 Compton Generator We define our coordinate axes as in the first problem such that The position of a point alon the water-filled tube is ω = ( ω cos λ, 0, ω sin λ). (6) r = (R cos α cos θ, R sin θ, R sin α cos θ), (7) where θ is the anle alon the rin (defined with respect to the x-axis), and α is the anle of rotation. Therefore, since α is chanes with time Thus we find v r = ṙ = ( R cos θ α sin α, 0, R cos θ α cos α). (8) ω cos λ 0 ω sin λ R cos θ α sin α 0 R cos θ α cos α = ĵ( Rω α cos λ cos θ cos α + Rω α sin λ cos θ sin α). Hence the acceleration due to the Coriolis force is a = Rω α cos θ cos(λ + α)ĵ. (9) It points in the y-direction, thouh we really need the tanential acceleration, which we obtain by multiplyin by cos θ. We then averae over the rin to find the mean acceleration (which is what matters for an incompressible fluid) a = 1 π π 0 Rω α cos θ cos(λ + α)dθ = Rω α cos(λ + α). 4

5 Now interatin over half a period of revolution we find the final averae tanential velocity. Thouh it is clear that it does not make a difference, we may assume we rotate the rin at a constant anular velocity α = π τ t. v t = τ 0 Rω α cos(λ + α)dt = Rω sin λ. (30) Therefore, the velocity of the water in the tube is iven by v t = Rω sin λ, where λ is defined from the equator. It is interestin to note that a rotation from the vertical to the vertical position (i.e. from - τ to τ ) produces a cos λ dependence, while a full rotation by π does not produce any motion. 5