3.Motion_2or3D

# 3.Motion_2or3D

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Unformatted text preview: motion Uniform circular motion uniform A particle is in uniform circular motion if it travels around a circle or a circular arc at constant speed. r r v 神舟五号 r a r v Centripetal acceleration & the period: a= v2 r T= r 2πr v yp θ r ⎛ vy ⎞ ⎛ vx ⎞ ⎜ ⎟i ⎜ ⎟j v = ⎜ − p ⎟ ˆ + ⎜ p ⎟ˆ r⎠ ⎝r⎠ ⎝ r r d v ⎛ v dy p ⎞ ˆ ⎛ v dx p ⎞ ˆ ⎟i + ⎜ ⎟j a= = ⎜− dt ⎜ r dt ⎟ ⎜ r dt ⎟ ⎝ ⎠⎝ ⎠ ⎛ v2 ⎞ˆ ⎛ v 2 ⎞ ⎜ ⎟⎜ ⎟j = ⎜− cos θ ⎟ i + ⎜ − sin θ ⎟ ˆ r r ⎝ ⎠⎝ ⎠ r a r v θ = ( − v sin θ ) ˆ + ( v cos θ ) ˆ i j v r a y r v = vxˆ + v yˆ i j a= v2 r tan φ = tan θ , φ = θ The Department of Physics, CUHK The Department of Physics, CUHK 19 20 §8. Relative motion in one dimension Sample problem y y “Top gun ” pilots have long worried about taking a turn too tightly. As a pilot’s body undergoes centripetal acceleration, with the head toward the center of curvature, the blook pressure in brain decreases, leading to loss of brain function. Frame B Frame A There are several warning signs to signal a pilot to ease up. When the centripetal acceleration is 2g or 3g, the pilot feels heavy. At about 4g, the pilot’s vision switches to black and white and narrows to “tunnel vision ” . If that acceleration is sustained or increased, vision ceases and soon after, the pilot is unconscious – a condition know as g-LOC for “g-induced loss of consciousness ”. r v BA x BA x PB x PA = x PB x x + x BA dx PA dx PB dx BA = + dt dt dt What is the centripetal acceleration, in g units, of a pilot flying an F-22 at speed v=2500 km/h through a circular arc having a radius curvature v PA = v PB + v BA r = 5.80km? The Department of Physics, CUHK a PA = a PB The Department of Physics, CUHK 21 §9. Relative motion two dimension Frame B y x xp y Frame A r rBA §10. Review & summary r r r rPA = rPB + rBA r rPB r v BA Displacement: r rrr ˆ r = x ˆ + yˆ + z k ∆ r = r2 − r1 i j r ˆ ∆r = ( x2 − x1 )ˆ + ( y2 − y1 )ˆ + ( z 2 − z1 ) k i j x x Average velocity and instantaneous velocity: r r r d rPA d rPB drBA = + dt dt dt r r r v PA = v PB + v BA 22 r ˆ v = vxˆ + v yˆ + vz k i j dx dy dz vx = , vy = , vz = dt dt dt r r a PA = a PB The Department of Physics, CUHK The Department of Physics, CUHK 23 24 4 Review & summary Review & summary Average acceleration and instantaneous acceleration Uniform circular motion r ˆ a = a xˆ + a yˆ + az k i j dv dv dv ax = x , a y = y , az = z dt dt dt a= Projectile motion x − x 0 = v0 x t x − x 0 = (v 0 cos θ 0 )t v2 r T= 2πr v Relative motion 1 2 y − y 0 = v 0 y t − gt 2 r r r v PA = v PB + v BA = ( v 0 sin θ 0 )t − 1 gt 2 2 r r a PA = a PB v y = v0 sin θ 0 − gt The Department of Physics, CUHK The Department of Physics, CUHK 25 26 5...
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## This document was uploaded on 09/16/2013.

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