Motion_and_Curvature9-2&3

No this issue is discussed in the next section 93 17

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Unformatted text preview: (t ) ||= v dt With is At that time, the vertical component of rp(t) is Or, Time rate of distance= speed d 2r d 2s = a, can we say 2 = a =|| a || ? d 2t dt which equals to the vertical component of rt(t). No This issue is discussed in the next section (9.3) 17 9.3 Curvature and Components of Acceleration For a given curve C: Unit Tangent Vector: Tangent: Also: ds = v =|| r′(t ) || dt r′( t ) ˆ r′(t ) ∴Unit Tangent: T = ′ || r ( t ) || Thus: r ˆ dr T= ds κ= r r ˆ dT d 2r | T' | =|| 2 ||= r ds ds | r'| ρ = 1/ κ Radius of Curvature: κ depends on the speed. For a given curvature, the speed of a moving car can be recommended to avoid skidding. Example: Curvature of a Circle Find the curvature of a circle with radius a. Solution: r (t ) = a cos ti + a sin tj and r′(t ) = − a sin ti + a cos tj r′(t ) ˆ ˆ T(t ) = = − sin ti + cos tj and T′(t ) = − cos ti − sin tj || r′(t ) || Fast Definition changes Curvature of a smooth curve: κ= 18 ˆ || T′(t ) || 1 = || r′(t ) || a Slow changes 19 20 5 1/2/2011 Acceleration: Tangential and Normal Components Acceleration: Tangential and Normal Unit Vectors ˆ dT dv ˆ dv T =v + dt dt dt ˆ || T ′ ( t ) || ˆ ⇒ || T ′ ( t ) || = κ v o Qκ = || r ′ ( t ) || Suppose a particle moves on a smooth curve C Recall: o Q a ( t ) = r ˆ ˆ Since T and v are parallel & || T ||= 1 and || v ||= v r ˆ ∴ v = vT Also: ˆ ˆ o Q dT / dt = N ˆ ˆˆ ˆ dT = 0 T⋅T =1 ⇒ T⋅ dt dT dT ∴T ⊥ ⇒ is normal t...
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This note was uploaded on 06/30/2011 for the course ENGR 233 taught by Professor S.samuelli during the Winter '10 term at Concordia Canada.

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