Hw26 - Overton Mays Homework 26 Due Apr 4 2005 4:00 am Inst Turner This print-out should have 13 questions Multiple-choice questions may continue

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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 – fnd 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 ±ind 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 µ 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 fnal energy oF the system is E = 1 2 I ω 2 = 1 2 •µ 3 m + M 12 d 2 ‚‰ 6 m v 0 [3 m + M ] d ¾ 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 Fxed symmetry axis. What is the postcollision angular speed of
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This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Hw26 - Overton Mays Homework 26 Due Apr 4 2005 4:00 am Inst Turner This print-out should have 13 questions Multiple-choice questions may continue

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