Motion_and_Curvature9-2&3

9 10 engineering and physics examples engineering and

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Unformatted text preview: ctor, s is the moved distance by particle. 9 10 Engineering and Physics Examples Engineering and Physics Examples Example: Velocity and Acceleration Example: Centripetal Acceleration The position of a moving particle is given by r r = r0 cos ωtˆ + r0 sin ωtˆ i j r v = − r0ω sin ωtˆ + r0ω cos ωtˆ i j r r a(t ) = r′′(t ) i j = −ω 2 (r0 cos ωtˆ + r0 sin ωtˆ) 2r = −ω r , r r i.e., is in the opposite i.e., a is in the opposite dierction of r of r (t ) = t 2 i + tj + 2.5tk Graph the curve defined Graph the curve defined by r (t ) and the vectors v (2) and a(2) . the Solution: x = t 2 , y = t , z = 2.5t x = y 2 ⇒ Parabola v (t ) = r′(t ) = 2ti + j + 2.5k a(t ) = r′′(t ) = 2i with a =|| a ||, v =|| v || r (2) = 4i + 2 j + 5k ⇒ P(4,2,5) v (2) = 4 i + j + 2.5k ∴ a = v 2 / r0 11 12 3 1/2/2011 Engineering and Physics Examples Example (Cont. ) Example: Curvilinear Motion in the Plane A projectile is launched with an initial velocity r v 0 = v0 cos θ ˆ + v0 sin θ ˆ i j r ˆ. and an initial height : s0 = s0 j v(t ) = (v0 cos θ)i + (− gt + v0 sin θ) j r r ⎡1 ⎤r r (t ) = ∫ v (t )dt = (v0 cos θ )...
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