To determine The angle between the positon vector and velocity vector and then angle between the position vector and acceleration vector of the given particle. Explanation Write the position vector using the given coordinates. r = 30 cos 2 t i + 40 sin 2 t j + ( 20 t + 3 t 2 ) k (I). Differentiate the equation (I) with respect to time “ t ”. v = d d t ( r ) v = d d t ( 30 cos 2 t i + 40 sin 2 t j + ( 20 t + 3 t 2 ) k ) = [ ( − 30 sin 2 t ) × 2 ] i + [ ( 40 cos 2 t ) × 2 ] j + [ 20 + 6 t ] k = − 60 sin 2 t i + 80 cos 2 t j + ( 20 + 6 t ) k (II). Again differentiate the equation (II) with respect to time “ t ”. a = d v d t = d ( − 60 sin 2 t i + 80 cos 2 t j + ( 20 + 6 t ) k ) d t = − 120 cos 2 t i − 160 sin 2 t j + 6 k (III). Write the relation between the position vector and velocity vector of the particle. v ⋅ r = v r cos θ 1 (IV). Here, e magnitude of the velocity is v , magnitude of the position vector is r , and the angle is θ . Write the relation between the position vector and acceleration vector of the particle. a ⋅ r = a r cos θ 1 (V). Here, e magnitude of the acceleration is a , magnitude of the position vector is r , and the angle is θ . Conclusion: Substitute 2 s for " t " in equation (I). r = 30 cos ( 2 × 2 ) i + 40 sin ( 2 × 2 ) j + ( 20 × 2 + 3 × 2 2 ) k = − 19.609 i − 30.27 j + 52 k (VI). Calculate the magnitude of position vector. r = ( − 19.609 ) 2 + ( − 30.27 ) 2 + ( 52 ) 2 = 63.28 mm Substitute 2 s for " t " in equation (II). v = − 60 sin ( 2 × 2 ) i + 80 cos ( 2 × 2 ) j + ( 20 + 6 × 2 ) k = 45.41 + − 52.29 j + 32 k \ (VII). Calculate the magnitude of velocity vector

8th Edition

ISBN: 9781118885840

Author: James L. Meriam, L. G. Kraige, J. N. Bolton

Publisher: WILEY

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Chapter 2.7, Problem 165P

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The angle between the positon vector and velocity vector and then angle between the position vector and acceleration vector of the given particle.

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Chapter 2.6, Problem 164P

Chapter 2.7, Problem 166P

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