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Unformatted text preview: Homework 3 Classical Mechanics (Phys701) Due: September 26, 2011 Problem 4-21 A particle is thrown up vertically with initial speed v , reaches a maximum height and falls back to ground. Show that the Corilois deection when it again reaches the ground is opposite in direction, and four times greater in magnitude, than the Coriolis deection when it is dropped t rest from the same maximum height. Solution: Consider a particle thrown up from the ground. Its vertical velocity is v z = v- gt . Total travel time is T = 2 v /g . Coriolis force creates a small deviation from the original trajectory and can be taken into account as a small perturbation which does not alter the vertical motion. Then the Coriolis acceleration creates a horizontal velocity v y ( t ) = t 2 v z ( s ) ds = 2 ( v t- gt 2 2 ) where x-axis is chosen along the projection of the Earths angular velocity on the horizontal plane. The horizontal displacement is then given by y 1 = T v y ( t ) dt = 2 ( v T 2 2- gT 3 6 ) = 4 v 3 3 g 2 Next, consider the particle falling from the height. Its vertical velocity is v z =- gt and the time of ight is T/ 2. We get v y =- 2 t gs ds =- 2 gt 2 2 y 2 =- T/ 2 gt 2 dt =- g 3 ( T 2 ) 3 =- v 3 3 g 2 =- y 1 4 Problem 4-22 A projectile is fired horizontally along Earths surface. Show that to a first approximation the angular deviation from the direction of the fire resulting from the Coriolis effect varies linearly with time at a rate cos where is the angular velocity of Earths rotation and is the co-latitude, the direction of deviation being to the right in the northern hemisphere. Solution: Let us point x-axis along the fire direction. The z-axis is up, and y-axis points to the left with respect to the fire direction. The Coriolis acceleration is- 2[ v ] = (0 , 2 v sin , 0) where is the angle between v and...
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