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Kaplunovsky - practice 12

# Kaplunovsky - practice 12 - practice 12 Due noon Therefore...

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practice 12 – – Due: Apr 17 2007, noon 1 Question 1, chap 3, sect 4. part 1 of 2 7 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 = 2, and B y = 1. Find the z component of vector A × vector B . Correct answer: 4 (tolerance ± 1 %). 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) = 4 . Question 2, chap 3, sect 4. part 2 of 2 7 points Find the angle between vector A and vector B. Correct answer: 126 . 87 (tolerance ± 1 %). Explanation: vector A · vector B = A x B x + A y B y = ( 2) (2) + (1) (1) = 3 , 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 (2) 2 + (1) 2 = 2 . 23607 . Therefore, using vector A · vector B = | A | | B | cos θ , and solving for θ , gives θ = cos 1 bracketleftBigg vector A · vector B A B bracketrightBigg = cos 1 bracketleftbigg ( 3) (2 . 23607) (2 . 23607) bracketrightbigg = 2 . 2143 rad = 126 . 87 . Question 3, chap 13, sect 4. part 1 of 1 7 points A particle is located at the vector position vectorr = (2 . 3 m)ˆ ı + (4 . 5 m)ˆ and the force acting on it is vector F = (3 N)ˆ ı + (1 . 7 N)ˆ  . What is the magnitude of the torque about the origin? Correct answer: 9 . 59 N m (tolerance ± 1 %). Explanation: Basic Concept: vector τ = vectorr × vector F Solution: Since neither the position of the particle, nor the force acting on the particle have a z -component, the torque acting on the particle has only a z -component vector τ = [ x F y y F x ] ˆ k = [(2 . 3 m) (1 . 7 N) (4 . 5 m) (3 N)] ˆ k = [ 9 . 59 N m] ˆ k . Question 4, chap 13, sect 3. part 1 of 1 10 points A space station shaped like a giant wheel has a radius of 111 m and a moment of inertia of 4 . 9 × 10 8 kg · m 2 (when it is unmanned). A

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practice 12 – – Due: Apr 17 2007, noon 2 crew of 150 live on the rim, and the station is rotating so that the crew experience an apparent acceleration of 1 g . When 100 people move to the center of the station, the angular speed changes. The acceleration of gravity is 9 . 8 m / s 2 . What apparent acceleration is experienced by those remaining at the rim? Assume: The average mass of for each in- habitant is 67 . 6 kg. Correct answer: 13 . 1112 m / s 2 (tolerance ± 1 %). Explanation: Given : r = 111 m , I = 4 . 9 × 10 8 kg · m 2 , and m = 150 × 67 . 6 kg . For an apparent acceleration of g, summationdisplay F c = m a c = m g v 2 r = g r ω 2 0 = g ω 0 = radicalbigg g r Initially, L 0 = I ω 0 + ( 150 m r 2 ) ω 0 and after 100 people move to the center, L f = I ω + 50 m r 2 ω Angular momentum is conserved, so L f = L i ( I + 150 m r 2 ) ω 0 = ( I + 50 m r 2 ) ω ω = I + 150 m r 2 I + 50 m r 2 radicalbigg g r and the apparent gravity is g n = r ω 2 = parenleftbigg I + 150 m r 2 I + 50 m r 2 parenrightbigg 2 g Since I + 150 m r 2 = 4 . 9 × 10 8 kg · m 2 + 150(67 . 6 kg)(111 m) 2 = 6 . 14935 × 10 8 m / s 2 and I + 50 m r 2 = 4 . 9 × 10 8 kg · m 2 + 50(67 . 6 kg)(111 m) 2 = 5 . 31645 × 10 8 m / s 2 , then g n = parenleftbigg 6 . 14935 × 10 8 m / s 2 5 . 31645 × 10 8 m / s 2 parenrightbigg 2 ( 9 . 8 m / s 2 ) = 13 . 1112 m / s 2 .
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Kaplunovsky - practice 12 - practice 12 Due noon Therefore...

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