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

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Dynamics 19 Problem 2.7 Let E r D OEt 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 j jE 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 .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 OE1 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) 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 : (3) From Eqs. (1) and (2) we have that ˇ ˇ E v ˇ ˇ D q 1 2 C .3 C 4t/ 2 ft = s D 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 we obtain cos D E a E v ˇ ˇ E a ˇ ˇ ˇ ˇ E v ˇ ˇ ) D cos 1 4.3 C 4t/ 4 p 10 C 24t C 16 t 2 : (5) Recall that r x D x D t and we have our final answer: .x/ D cos 1 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 Define 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 |/ ft = s ; E v A D . 20:0 O { 12:8 O |/ ft = s : Define 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 |/ ft = s 2 ; E a A D 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
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