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hw26 - Overton Mays Homework 26 Due Apr 4 2005 4:00 am Inst...

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Overton, Mays – Homework 26 – Due: Apr 4 2005, 4:00 am – Inst: Turner 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A projectile of mass m = 1 . 11 kg moves to the right with speed v 0 = 15 m / s. The projectile strikes and sticks to the end of a stationary rod of mass M = 6 . 08 kg and length d = 2 . 89 m that is pivoted about a frictionless axle through its center. d O O v 0 ω m M Find the angular speed of the system right after the collision. Correct answer: 3 . 67348 rad / s. Explanation: The initial angular momentum of the pro- jectile about the pivot O is L i = d 2 m v 0 . With the projectile stuck to the end of the rod, the rotational inertia of the projectile and the rod combined about O is, I = m d 2 2 + 1 12 M d 2 = m 4 + M 12 d 2 (1) = (1 . 11 kg) 4 + (6 . 08 kg) 12 (2 . 89 m) 2 = 6 . 54944 kg m 2 . Using conservation of angular momentum, we have L i = I f ω (2) = ~r × ~p = d 2 m v 0 (3) Therefore, combining Eqs. (1), (2), and (3), we have ω = d 2 m v 0 m d 2 4 + M d 2 12 · 12 12 = 6 m v 0 (3 m + M ) d = 6 (1 . 11 kg) (15 m / s) [3 (1 . 11 kg) + (6 . 08 kg)] (2 . 89 m) = 3 . 67348 rad / s . 002 (part 2 of 2) 10 points How much kinetic energy is lost in the col- lision relative to the initial kinetic energy? (Determine the ratio of the kinetic energy lost to the initial kinetic energy) Correct answer: 0 . 646121 . Explanation: The initial energy of the system is E 0 = 1 2 m v 2 0 = 1 2 (1 . 11 kg) (15 m / s) 2 = 124 . 875 J . The final energy of the system is E = 1 2 I ω 2 = 1 2 3 m + M 12 d 2 ‚ ‰ 6 m v 0 [3 m + M ] d 2 = 1 2 m v 2 0 3 m 3 m + M = 1 2 (1 . 11 kg) (15 m / s) 2 × 3 (1 . 11 kg) 3 (1 . 11 kg) + (6 . 08 kg) = 44 . 1906 J . Therefore, the fractional loss of the energy is f = E 0 - E E 0 = 1 2 m v 2 0 1 - 3 m 3 m + M 1 2 m v 2 0
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Overton, Mays – Homework 26 – Due: Apr 4 2005, 4:00 am – Inst: Turner 2 = M 3 m + M = (6 . 08 kg) 3 (1 . 11 kg) + (6 . 08 kg) = 0 . 646121 . 003 (part 1 of 2) 10 points A small blob of putty of mass m falls from the ceiling and lands on the outer rim of a turntable of radius R and moment of inertia I 0 that is rotating freely with angular speed ω i about its vertical fixed symmetry axis. What is the postcollision angular speed of the turntable plus putty? 1. ω f = ω i 1 + m R I 0 2. ω f = ω i 2 + m R 2 I 0 3. ω f = ω i 1 + m R 3 I 0 4. ω f = ω i 1 + m R 2 I 0 correct Explanation: The final rotational inertia of the turntable- plus-blob is I f = I 0 + I blob = I 0 + m R 2 Using conservation of angular momentum, I 0 ω i = I f ω f ω f = I 0 ω i I f ω f = I 0 ω i I 0 + m R 2 = ω i 1 + m R 2 I 0 004 (part 2 of 2) 10 points
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