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hw1_sol - Dynamics 19 Problem 2.7 Let r D t cfw_ C .2 C 3t...

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Dynamics 19 Problem 2.7 Let E r D Ot O { C .2 C 3t C 2t 2 / O | ± ft describe the motion of a point P relative to the Cartesian frame of reference shown. Recalling that for any two vectors E p and E q we have that E p ± E q D j E p jjE q j cos ˇ , where ˇ is the angle formed by E p and E q , and recalling that the velocity vector is always tangent to the trajectory, determine the function ² .x/ describing the angle between the acceleration vector and the tangent to the path of P . Solution The velocity vector is the time derivative of the position vector: E v D O1 O { C .3 C 4t/ O | ± ft = s : (1) The acceleration vector is the time derivative of the velocity vector: E a D 4 O | ft = s 2 : (2) ² is the angle between the acceleration vector and the tangent to the path of P: E a ± E v D ˇ ˇ E a ˇ ˇ ˇ ˇ E v ˇ ˇ cos ² : (3) From Eqs. (1) and (2) we have that ˇ ˇ E v ˇ ˇ D ± q 1 2 C .3 C 2 ² ft = s D ³ p 10 C 24t C 16 t 2 ´ ft = s and ˇ ˇ E a ˇ ˇ D 4 ft = s 2 : (4) Substituting the expressions for ˇ ˇ E v ˇ ˇ and ˇ ˇ E a ˇ ˇ from Eq. (4) in Eq. (3) and solving for ² we obtain cos ² D E a ± E v ˇ ˇ E a ˇ ˇ ˇ ˇ E v ˇ ˇ ) ² D cos ± 1 ± 4.3 C 4 p 10 C 24t C 16 t 2 ² : (5) Recall that r x D x D t and we have our ±nal answer: ² .x/ D cos ± 1 µ 3 C 4x p 10 C 24x C 16x 2 :
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38 Solutions Manual Problem 2.26 Two Coast Guard patrol boats P 1 and P 2 are stationary while monitoring the motion of a surface vessel A . The velocity of A with respect to P 1 is expressed by E v A D . ± 23 O { 1 ± 6 O | 1 / ft = s ; whereas the acceleration of A , expressed relative to P 2 is given by E a A D . ± 2 O { 2 ± 4 O | 2 / ft = s 2 : Determine the velocity and the acceleration of A expressed with respect to the land-based component system . O {; O |/ . Solution DeFne ± D 72 ı . The unit vectors of P 1 can be expressed in terms of the . O O component system as O { 1 D sin ± O { C cos ± O |; D 0:9511 O { C 0:3090 O |: O | 1 D ± cos ± O { C sin ± O D ± 0:3090 O { C 0:9511 O The velocity of the surface vessel A in the land-based frame is E v A D ± ± 23. sin ± O { C cos ± O ± 6. ± cos ± O { C sin ± O ² ft = s ; E v A D . ± 20:0 O { ± 12:8 O ft = s : DeFne ˇ D 45 ı . The unit vectors of P 2 can be expressed in terms of the . O O component system as O { 2 D ± cos ˇ O { ± sin ˇ O D ± 1 p 2 O { ± 1 p 2 O D ± 0:7071 O { ± 0:7071 O O | 2 D sin ˇ O { ± cos ˇ O D 1 p 2 O { ± 1 p 2 O D 0:7071 O { ± 0:7071 O As done for the velocity vector, the acceleration vector in the land-based frame is found to be E a A D ± ± 2 . ± cos ˇ O { ± sin ˇ O ± 4. sin ˇ O { ± cos ˇ O ² ft = s 2 ; E a A D ³ ± p 2 O { C 3 p 2 O | ´ ft = s 2 D . ± 1:41 O { C 4:24 O ft = s 2 : (1)
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68 Solutions Manual Problems 2.60 through 2.62 As we will see in Chapter 3, the acceleration of a particle of mass m suspended by a linear spring with spring constant k and unstretched length L 0 (when the spring length is equal to L 0 the spring exerts no force on the particle) is given by R x D g
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hw1_sol - Dynamics 19 Problem 2.7 Let r D t cfw_ C .2 C 3t...

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