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solution_pdf - tokarski(at23678 Angular momentum...

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tokarski (at23678) – Angular momentum – Morozova – (11104) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A particle is located at the vector position vectorr = (2 . 1 m)ˆ ı + (4 . 1 m)ˆ and the force acting on it is vector F = (5 . 2 N)ˆ ı + (2 . 4 N)ˆ  . What is the magnitude of the torque about the origin? Correct answer: 16 . 28 N m. Explanation: Basic Concept: vector τ = vectorr × vector F Solution: Since neither position of the par- ticle, nor the force acting on the particle have the z -components, the torque acting on the particle has only z -component: vector τ = [ x F y y F x ] ˆ k = [(2 . 1 m) (2 . 4 N) (4 . 1 m) (5 . 2 N)] ˆ k = [ 16 . 28 N m] ˆ k . 002 (part 2 of 2) 10.0 points What is the magnitude of the torque about the point having coordinates [ a, b ] = [(2 m) , (8 m)]? Correct answer: 20 . 52 N m. Explanation: Reasoning similarly as we did in the pre- vious section, but with the difference that relative to the point [(2 m) , (8 m)] the y - component of the particle is now [ y (8 m)], we have vector τ = { [ x a ] F y [ y b ] F x } ˆ k = { [(2 . 1 m) (2 m)] [2 . 4 N] [(4 . 1 m) (8 m)] [5 . 2 N] } ˆ k = { 20 . 52 N m } ˆ k . 003 (part 1 of 2) 10.0 points Given: Two vectors vector A = A x ˆ ı + A y ˆ and vector B = B x ˆ ı + B y ˆ  , where A x = 2, A y = 1, B x = 5, and B y = 5. Find the z component of vector A × vector B . Correct answer: 15. Explanation: It follows from the definition of cross prod- uct that ( vector A × vector B ) x = A y B z A z B y = 0 ( vector A × vector B ) y = A z B x A x B z = 0 ( vector A × vector B ) z = A x B y A y B x = ( 3) (3) (4)(2) = 15 . 004 (part 2 of 2) 10.0 points Find the angle between vector A and vector B. Correct answer: 108 . 435 . Explanation: vector A · vector B = A x B x + A y B y = ( 2) (5) + (1) (5) = 5 , A = radicalBig A 2 x + A 2 y = radicalBig ( 2) 2 + (1) 2 = 2 . 23607 , and B = radicalBig B 2 x + B 2 y = radicalBig (5) 2 + (5) 2 = 7 . 07107 . Therefore, using vector A · vector B = | A | | B | cos θ ,
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tokarski (at23678) – Angular momentum – Morozova – (11104) 2 and solving for θ , gives θ = cos 1 bracketleftBigg vector A · vector B A B bracketrightBigg = cos 1 bracketleftbigg ( 5) (2 . 23607) (7 . 07107) bracketrightbigg = 1 . 89255 rad = 108 . 435 . 005 10.0 points A ball having mass 3 . 4 kg is fastened at the end of a flagpole that is connected to the side of a tall building at point P . The length of the flagpole is 4 . 6 m, and it makes an angle 1 . 0472 rad (60 ) with the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . If the ball becomes loose and starts to fall, determine the magnitude of its angular mo- mentum after 38 s about P . Neglect air resis- tance. Correct answer: 2912 . 17. Explanation: Basic Concepts: vector L = vectorr × ( mvectorv ) . The vector from P to the falling ball is vectorr = l cos θ ˆ ı + l sin θ ˆ 1 2 g t 2 ˆ Its velocity is vectorv = gt ˆ . So, vector L = vectorr × ( mvectorv ) = m ( l cos( θ ) ˆ ı + l sin( θ ) ˆ 1 2 g t 2 ˆ × ( g t ˆ ) = m l g t cos( θ ) ˆ k = (3 . 4 kg) (4 . 6 m) (9 . 8 m / s 2 ) (38 s) × cos(1 . 0472 rad) ˆ k = 2912 . 17 kg m 2 / s 2 ˆ k .
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