# chap04 - 41 Chapter 4 Acceleration 4.1 The position vector...

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Unformatted text preview: 41 Chapter 4 Acceleration 4.1 The position vector of a point is defined by the equation ( 29 3 ˆ ˆ 4 3 10 t t =- + R i j where R is in inches and t is in seconds. Find the acceleration of the point at t = 2 s. ( 29 ( 29 3 ˆ ˆ 4 3 10 t t t =- + R i j ( 29 ( 29 2 ˆ 4 t t =- R i & ( 29 ˆ 2 t t = - R i && ( 29 ( 29 2 ˆ ˆ 2 s 2 2 4 in/s = - = - R i i && Ans. 4.2 Find the acceleration at t = 3 s of a point which moves according to the equation ( 29 ( 29 2 3 3 ˆ ˆ = 6 3 t t t- + R i j . The units are meters and seconds. ( 29 ( 29 ( 29 2 3 3 ˆ ˆ = 6 3 t t t t- + R i j ( 29 ( 29 2 2 ˆ ˆ = 2 2 t t t t- + R i j & ( 29 ( 29 ˆ ˆ = 2 2 t t t- + R i j && ( 29 ( 29 ( 29 2 ˆ ˆ ˆ ˆ 3 s = 2 3 2 3 6 m/s- + = - + R i j i j && Ans. 4.3 The path of a point is described by the equation 2 /10 = ( 4) j t t e π- + R where R is in millimeters and t is in seconds. For t = 20 s, find the unit tangent vector for the path, the normal and tangential components of the point’s absolute acceleration, and the radius of curvature of the path. ( 29 2 /10 = ( 4) j t t t e π- + R ( 29 /10 2 /10 = 2 ( 4) 10 j t j t j t te t e π π π--- + R & ( 29 2 /10 /10 2 /10 = 2 ( 4) 5 5 100 j t j t j t j t j t t e e t e π π π π π π--- --- + ÷ R && Noticing, at t = 20 s, that /10 2 1.0 j t j e e π π-- = = , we find that ( 29 2 20 s = (20 4) 404 mm + = R 42 ( 29 ( 29 2 20 s = 2 20 (20 4) 40.00 126.92 133.07 72.5 mm/s 10 j j π- + =- = ∠ - ° R & ( 29 2 2 2 20 20 20 s = 2 (20 4) 37.873 25.133 mm/s 5 5 100 j j j π π π --- + = -- ÷ R && From the direction of the velocity we find the unit tangent and unit normal vectors ˆ ˆ ˆ ˆ ˆ 1 72.5 cos 72.5 sin 72.5 0.30058 0.95376 = ∠ - ° =- ° +- ° =- τ i j i j Ans. ˆ ˆ ˆ ˆ ˆ ˆ ˆ sin 72.5 cos 72.5 0.95376 0.30058 = =- ° -- ° = -- ρ τ k i j i j × From these, the components of the point’s absolute acceleration are ( 29 ( 29 2 2 ˆ ˆ ˆ ˆ ˆ 0.95376 0.30058 37.873 25.133 mm/s 43.676 mm/s n A = = ---- = ρ R i j i j && g g Ans. ( 29 ( 29 2 2 ˆ ˆ ˆ ˆ ˆ 0.30058 0.95376 37.873 25.133 mm/s 12.586 mm/s t A = =--- = τ R i j i j && g g Ans. Then, from Eq. (4.2) or Eq. (4.14), the radius of curvature is ( 29 2 2 2 133.07 mm/s 405.4 mm 43.676 mm/s n A ρ = - = - = - R & Ans. Where the negative sign indicates that the point is in the negative ˆ ρ direction from the center of curvature of the point’s path. 4.4 The motion of a point is described by the equations 3 4 cos x t t π = and ( 29 3 6 sin 2 y t t π = where x and y are in feet and t is in seconds. Find the acceleration of the point at 1.40 s t = . ( 29 ( 29 3 3 ˆ ˆ 4 cos 6 sin 2 t t t t t π π = + R i j ( 29 ( 29 ( 29 3 3 3 2 3 ˆ ˆ 4 cos 12 sin 2 sin 2 3 cos 2 t t t t t t+ t t π π π π π π =- + R i j & ( 29 ( 29 2 3 2 5 3 2 2 2 ˆ ˆ 48 sin 36 cos 1 2 3 sin 2 cos2 t t t t t t t t+2 t t π π π π π π π π = -- +-...
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## This note was uploaded on 04/03/2012 for the course MECHATRONI 111 taught by Professor Jung during the Spring '11 term at Hanyang University.

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chap04 - 41 Chapter 4 Acceleration 4.1 The position vector...

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