Goldstein_22_15_21_23 - Homework 7: # 4.22, 5.15, 5.21,...

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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 co-latitude, 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 co-latitude, 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 co-latitude, 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 ....
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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.

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Goldstein_22_15_21_23 - Homework 7: # 4.22, 5.15, 5.21,...

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