PHY303k_-_Spring_2010_-_HW_13

PHY303k_-_Spring_2010_-_HW_13 - saldana(avs387 – Homework...

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Unformatted text preview: saldana (avs387) – Homework 13 – florin – (58140) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The speed of a moving bullet can be deter- mined by allowing the bullet to pass through two rotating paper disks mounted a distance 98 cm apart on the same axle. From the angular displacement 19 . 5 ◦ of the two bul- let holes in the disks and the rotational speed 1223 rev / min of the disks, we can determine the speed of the bullet. 19 . 5 ◦ v 1223 rev / min 98 cm What is the speed of the bullet? Correct answer: 368 . 782 m / s. Explanation: Let : ω = 1223 rev / min , d = 98 cm , and θ = 19 . 5 ◦ . θ = ω t t = θ ω , so the speed of the bullet is v = d t = d ω θ = (98 cm) (1223 rev / min) 19 . 5 ◦ × 360 ◦ 1 rev 1 m 100 cm 1 min 60 s = 368 . 782 m / s . keywords: 002 10.0 points A bug is on the rim of a 78 rev / min, 12 in . diameter record. The record moves from rest to its final angular speed in 2 . 93 s. Find the bug’s centripetal acceleration 1 . 5 s after the bug starts from rest. (1 in = 2.54 cm). Correct answer: 2 . 66489 m / s 2 . Explanation: Let : Δ w = 78 rev / min , Δ t = 2 . 93 s , r = 6 in , and t = 1 . 5 s . α = Δ ω Δ t , so ω = α t = Δ ω Δ t t ω = 78 rev / min 2 . 93 s (1 . 5 s) · 1 min 60 s = 4 . 18165 rad / s , and a r = v 2 t r = r ω 2 = (6 in)(4 . 18165 rad / s) 2 × 1 cm 2 . 54 in × 1 m 100 cm = 2 . 66489 m / s 2 . 003 10.0 points A small wheel of radius 1 . 4 cm drives a large wheel of radius 13 . 9 cm by having their cir- cumferences pressed together....
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PHY303k_-_Spring_2010_-_HW_13 - saldana(avs387 – Homework...

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