09032009

# 09032009 - Straight-Line Motion • Acceleration as a...

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Unformatted text preview: Straight-Line Motion • Acceleration as a function of position (e.g., gravitational forces or forces in springs) By chain rule Substitute and re-arrange Integrating from v o at s o to v at position s ( ) ( ) 17 . 13 s a dt dv = v ds dv dt ds ds dv dt dv = = ( ) ( ) 18 . 13 ds s a dv v = ( ) ( ) ∫ ∫ = s s v v ds s a dv v 19 . 13 Worked Example In terms of distance s from the center of the earth, the magnitude of the acceleration due to gravity is gR E 2 /s 2 , where R E is the earth’s radius. If a spacecraft is a distance so from the center of the earth, what outward velocity vo must be given to reach a specified distance h from the earth’s center? Acceleration due to gravity is toward the center of earth and decreases inversely to square of distance from center of earth a = - gR E 2 /s 2 dv/dt = (dv/ds)(ds/dt) = v(dv/ds) = - gR E 2 /s 2 Therefore, vdv = -(gR E 2 )(ds/s 2 ) Integrate from v o (initial velocity at s o ) to final velocity v = 0 at s = h 0 – v o 2 /2 = gR E 2 (1/h – 1/s o ) v o = [2 gR E 2 (1/s o – 1/h)] 0.5 Escape velocity v esc = [2 gR E 2 /s o ] 0.5 • Acceleration as a function of velocity (aerodynamic and hydrodynamic forces) Rearrange Integrating from v o at t o to velocity v at time t Use chain rule to write Re-arrange Integrate from v o at s o to velocity v at displacement s ( ) v a dt dv = Straight-Line Motion ( ) dt v a dv = ( ) υ a v ds dv dt ds ds dv dt dv = = = ( ) ds v a dv v = Worked Example A boat is moving at 10 m/s when the engine is shut down. Due to hydrodynamic drag, its subsequent acceleration is a = -0.05v 2 , where v is the velocity of the boat in m/s. What is the boat’s velocity 4 seconds after the engine is shut down and what distance has the boat moved in the 4 seconds following the shutdown of the engine? (dv/dt) = a = -0.05v 2 Î- (dv/0.05v 2 ) = dt Integrate from v = 10 m/s at t = 0 (when engine is shut down) to velocity, v, at t = 4s (1/0.05)(1/v – 1/10) = (4 – 0) Î 1/v = 0.3 or v = 3.33 m/s [or, 1/v – 1/10 = 0.05t Î v = 10/(1 + 0.5t) (ds/dt) = v = 10/(1 + 0.5t) Î ds = 10[dt/(1 + 0.5t)] Integrate from t = 0 to t = 4 s for Δ s = distance moved in 4 s Let u = 1 + 0.5t; du = 0.5dt or dt = 2(du) and 10[dt/(1 + 0.5t)] = 20(du/u) Δ s = (20)[ln(1 + 0.5t)] = 21.97 m Curvilinear Motion – Cartesian Coordinates • Let r be the position vector of point P relative to origin of a Cartesian reference frame k j i r z y x + + = ( ) 13.21 k j i r v dt dz dt dy dt dx dt d + + = = k j i r z y x + + = ( ) 13.22 k j i v z y x v v v + + = ( ) 23 . 13 , , dt dz v dt dy v dt dx v z y x = = = k j i dt dv a dt dv dt dv dt dv z y x + + = = ( ) 13.24 k j i a z y x a a a + + = ( ) 25 . 13 , , dt dv a dt dv a dt dv a z z y y x x = = = Projectile Problem An object is thrown through the air with negligible drag. It accelerates due to gravity. In terms of a fixed cartesian coordinate system with its y-axis upward, a x = 0, a y = - g and a z = 0....
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## This note was uploaded on 04/28/2010 for the course CIVIL CIVL1010 taught by Professor Unknown during the Spring '09 term at HKU.

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09032009 - Straight-Line Motion • Acceleration as a...

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