HW 11-1-solutions

# HW 11-1-solutions - madrid(tmm2353 – HW 11-1 –...

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Unformatted text preview: madrid (tmm2353) – HW 11-1 – Antoniewicz – (56445) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points Given: Use counter-clockwise as the positive angular direction. A particle of mass m is shot with an initial velocity v , making an angle θ , with the hori- zontal as shown in figure. The particle moves in the gravitational field of the Earth. x y v R v v h θ R h O C A Using the origin as the pivot, find the an- gular momentum when the particle is at the origin, 1. vector L = mv sin θ R ˆ k 2. vector L = 0 correct 3. vector L = mv R ˆ k 4. vector L = mv R 2 ˆ k 5. vector L = mv cos θ R ˆ k Explanation: Basic Concepts vector L = vectorr × vectorp vector L = I vectorω We also need the vector cross product | vectora × vector b | = a b sin φ , where φ is the angle between the vectors vectora and vector b . Solution: The angular momentum is vector L = vectorr × vectorp , where vectorr is measured from the origin. At the origin vectorr = 0. Thus, vector L = vectorr × vectorp = 0 . 002 (part 2 of 3) 10.0 points Using the origin as the pivot, find the angular momentum when the particle is at the highest point of the trajectory. 1. vector L = + mv 2 2 g sin 2 θ cos θ ˆ k 2. vector L = − mv 2 2 g sin θ cos 2 θ ˆ k 3. vector L is in the ˆ ı direction. 4. vector L = − mv 3 2 g sin 2 θ cos θ ˆ k correct 5. vector L is in the ˆ direction. 6. vector L = − mv 2 2 g sin 2 θ cos θ ˆ k 7. vector L = + mv 3 2 g sin 2 θ cos θ ˆ k 8. vector L = + mv 2 2 g sin θ cos 2 θ ˆ k 9. vector L = − mv 3 2 g sin θ cos 2 θ ˆ k 10. vector L = + mv 3 2 g sin θ cos 2 θ ˆ k Explanation: At the highest point (denote the position vector to that point with vector r 1 ) we know that vector v 1 = vector v x = v cos θ ˆ ı and vector L = vector r 1 × vector p 1 = vector r 1 × m vector v 1 And, | vector L | = m | vector v 1 || vector r 1 | sin θ 1 . Let us analyze this expression. We see that vector r 1 sin θ 1 is the maximum height of y max = h h = | vector r 1 | sin θ 1 = v 2 sin θ 2 2 g . madrid (tmm2353) – HW 11-1 – Antoniewicz – (56445) 2 Since | vector v 1 | = | vector v x | = v x , then, the magnitude of the angular momentum is | vector L | = mh v x = mh v cos θ = mv 3 2 g sin 2 θ cos θ . The direction is clearly − ˆ k by the right hand rule of the cross product. 003 (part 3 of 3) 10.0 points Using the origin as the pivot, find the angular momentum of the particle just before it hits the ground. 1. vector L = + 2 mv 2 g sin 2 θ cos θ ˆ k 2. vector L = + 2 mv 3 g sin 2 θ cos θ ˆ k 3....
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HW 11-1-solutions - madrid(tmm2353 – HW 11-1 –...

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