hw6 - a p = ¨ x ˆ i + ¨ y ˆ j = 6 ˙ x ˆ j = 18 ˆ j...

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TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 6 (1.101, 1.107) 1.101 a B = ( ¨ r - r ˙ θ 2 ) ˆ e r + ( r ¨ θ + r ˙ θ ) ˆ e θ We are given r = 3 0 , ˙ r = 4 00 / s and ¨ r = 5 00 / s 2 . For the total acceleration to be equal to zero, each of the individual components (the radial and the tangential components) have to be individually equal to zero. ( 5 - 3 ˙ θ 2 ) = 0 and ( 3 ¨ θ + 2(4) ˙ θ ) = 0 ˙ θ = ± r 5 3 and ¨ θ = - 8 3 ± ± r 5 3 ! Sp, ( ˙ θ, ¨ θ ) = either ( 1 . 29 rad / sec , - 3 . 44 rad / sec 2 ) or ( 1 . 29 rad / sec , + 3 . 44 rad / sec 2 ) . 1.107 In terms of the rectangular components, the velocity vector can be written as v p = ˙ x ˆ i + ˙ y ˆ j = ˙ x ˆ i + 2 x ˙ x ˆ j , since y = x 2 = 3 ˆ i + 6 x ˆ j , since ˙ x = 3 At (1,1) , v p = 3 ˆ i + 6 ˆ j Similarly, the acceleration
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Unformatted text preview: a p = ¨ x ˆ i + ¨ y ˆ j = 6 ˙ x ˆ j = 18 ˆ j Now, lets obtain the radial and transverse components v r = v · ˆ e r = 3 ² 1 √ 2 ³ + 6 ² 1 √ 2 ³ = 9 ² 1 √ 2 ³ v θ = v · ˆ e θ = 3 ²-1 √ 2 ³ + 6 ² 1 √ 2 ³ = 3 ² 1 √ 2 ³ 1 a r = a · ˆ e r = 18 ± 1 √ 2 ² a θ = a · ˆ e θ = 18 ± 1 √ 2 ² v = ± 9 √ 2 ² ˆ e r + ± 3 √ 2 ² ˆ e θ f t / s a = ± 18 √ 2 ² ˆ e r + ± 18 √ 2 ² ˆ e θ f t / s 2 2...
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This note was uploaded on 05/24/2010 for the course TM 212 taught by Professor Richard during the Spring '10 term at University of Illinois, Urbana Champaign.

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hw6 - a p = ¨ x ˆ i + ¨ y ˆ j = 6 ˙ x ˆ j = 18 ˆ j...

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