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Kinematics

# Kinematics - 11 Chapter 1 Kinematics 1.1 Velocity and...

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11 Chapter 1. Kinematics 1.1 Velocity and Acceleration (V & A) A’ s Vector displacement, velocity, acceleration: different path may lead to e same displacement A the same displacement Consider a particle moving from A at time t to A’ at time t’=t+ Δ t . Vector velocity is defined by the rate of change of vector displacement, as follows Path y 0 lim , lim t rd r vd v vr av r d t td t Δ→ Δ Δ = == = = Δ G GG G G G G ±± ± ± ' rr r = + Δ G r Δ G G 00 tt ΔΔ 2 s dv d s x A r 2 If rectilinear (straight line) , ds va dt dt dt =

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12 1.1 V & A ( centripetal acceleration ) Consider a small time interval dt in which a particle turns an angle d α ( ) ' When 0, 1 tt tn dd e d e ee αα →= × G G G G Unit tangent vector nit normal vector t e G G ' v G Since local radius , ds d s ds ρ = Unit normal vector n e ' G ' t e G ,a n d tnn ds e d e e ρρ == = G GG d O n e e G G v G () t t ds s de de v av e e va e v t dt dt dt ⎛⎞ + = + ⎜⎟ G G G G n t e N N Chain rule / 2 entripetal acceleration n v e ds dt dv e e v = ⎝⎠ +⇐ G ±²²²²³²²²²´ G t de G Centripetal acceleration ae dt =+⇐ Almost 90 o d 1 t e = G
13 1.1 V & A ( accelerometer ) The basic task of kinematics is integrating acceleration d 2 x/dt 2 for velocity dx/dt and tegrating velocity dx/dt for displacement integrating velocity dx/dt for displacement x . The acceleration d 2 x/dt 2 is more easily measured. See accelerometer at the right showing the measurement in y direction. Modern method : piezoelectric ansducer – voltage is transducer a voltage is generated when a compression or shear force is applied on the ystal without any power supply crystal without any power supply.

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14 1.1 V & A (basic) Remedial Normally, acceleration ( a ) is measured and the signal shows a as a function of math box time, a (t)= dv/dt ( ) ( ) 22 2 1 vt t d v atd t v v t =→ = + 1 1 const 1 nn xdx x n + =+ + () ( ) () 11 1 2 1 1 21 1 '' t st t t t t ds v t dt s s v a t dt dt = + + ∫∫ 1 but ln const dx x x P 0 For constant acceleration shown on the right t u g d t v g t = −= = N 0 0 0 0 t s t s vdt gt s t gt = = = 2 1 2 Time to hit ground : hg τ = 2/ 2 vg g h ττ →= == Warning: never use these formulas for situations where constant! a
15 1.1 V & A example 1, or ex. 1) (p , ) Example : A rocket has expended all its fuel when it reaches position A , here it has a velocity t an angle where it has a velocity u at an angle θ with respect to the horizontal. It then begins unpowered flight and ttains n added height nd attains an added height h and maximum horizontal distance s at position B after traveling a horizontal istance om Determine the distance s from A . Determine the expressions for h and s , the time t of flight from A to B , and the equation of the path. For the interval concerned, assume a flat earth with a constant gravitational acceleration g and neglect any atmospheric resistance.

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16 1.1 V & A (solution for ex. 1, or soln 1) Solution () 0 n 0 0, cos , cos ,, s i n y t xx x vt yyy av u x v d t u t ag d v g d t v u g t θ == = = =− = = ∫∫ sin 2 0 1 sin 2 u t y y v dt ut gt ( ) () () Position B reached when sin 0 sin / . Insert this time into , : y vu g t ug x
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Kinematics - 11 Chapter 1 Kinematics 1.1 Velocity and...

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