This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Homework 7: # 4.22, 5.15, 5.21, 5.23, Foucault pendulum Michael Good Oct 9, 2004 4.22 A projectile is fired horizontally along Earth’s surface. Show that to a first approximation the angular deviation from the direction of fire resulting from the Coriolis effect varies linearly with time at a rate ω cos θ where ω is the angular frequency of Earth’s rotation and θ is the colatitude, the direction of deviation being to the right in the northern hemisphere. Answer: I’ll call the angular deviation ψ . We are to find ψ = ω cos θt We know ω is directed north along the axis of rotation, that is, sticking out of the north pole of the earth. We know θ is the colatitude, that is, the angle from the poles to the point located on the surface of the Earth. The latitude, λ is the angle from the equator to the point located on the surface of the Earth. λ = π/ 2 θ . Place ourselves in the coordinate system of whoever may be fir ing the projectile on the surface of the Earth. Call y the horizontal direction pointing north (not toward the north pole or into the ground, but horizontally north), call x the horizontal direction pointed east, and call z the vertical di rection pointed toward the sky. With our coordinate system in hand, lets see where ω is. Parallel transport it to the surface and note that it is between y and z . If we are at the north pole, it is completely aligned with z , if we are at the equator, ω is aligned with y . Note that the angle between z and ω is the colatitude, θ .( θ is zero at the north pole, when ω and z are aligned). If we look at the components of ω , we can take a hint from Goldstein’s Figure 4.13, that deflection of the horizontal trajectory in the northern hemisphere will depend on only the z component of ω , labeled ω z . Only ω z is used for our approximation. It is clear that there is 1 no component of ω in the x direction. If we took into account the component in the y direction we would have an effect causing the particle to move into the vertical direction, because the Coriolis effect is F c = 2 m ( ω × v ) and ω y × v would add a contribution in the z direction because our projectile is fired only along x and y , that is, horizontally. So following Goldstein’s figure, we shall only be concerned with ω z . Our acceleration due to the Coriolis force is a c = 2( ω × v ) = 2( v × ω ) The component of ω in the z direction is ω z = ω cos θ . Thus the magnitude of the acceleration is a c = 2 vω cos θ The distance affected by this acceleration can be found through the equation of motion, d = 1 2 a c t 2 = vω cos θt 2 And using a small angle of deviation, for ψ we can draw a triangle and note that the distance traveled by the projectile is just x = vt ....
View
Full
Document
This note was uploaded on 03/17/2012 for the course PHYS 202 taught by Professor Atkin during the Spring '12 term at Amity University.
 Spring '12
 atkin
 Work

Click to edit the document details