hw8 - cerutti(cpc566 HW08 TSOI(58160 This print-out should...

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cerutti (cpc566) – HW08 – TSOI – (58160) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rigid circular wheel spins at constant angu- lar velocity about a stationary axis. Choose the picture below that correctly describes the relative magnitudes and directions of the ve- locity vector of points on the wheel. 1. 2. correct 3. 4. 5. 6. 7. 8. Explanation: Any point on the wheel travels in a circle around the axis, so the velocities of all points are directed parallel to tangents to the wheel. The wheel is rigid, so all points on the wheel travel with the same angular velocity. v = ωr so the points farther from the axis travel faster than the points closer to the axis. keywords: 002 (part 1 of 4) 10.0 points A potter’s wheel of radius 15 cm starts from rest and rotates with constant angular accel- eration until at the end of 30 s it is moving with angular velocity of 19 rad / s. What is the angular acceleration? Correct answer: 0 . 633333 rad / s 2 . Explanation: Let : ω = 19 rad / s , ω 0 = 0 , and t = 30 s . Since the angular acceleration is constant, α = Δ ω Δ t = ω t = 19 rad / s 30 s = 0 . 633333 rad / s 2 . 003 (part 2 of 4) 10.0 points What is the linear velocity of a point on the rim at the end of the 30 s?
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cerutti (cpc566) – HW08 – TSOI – (58160) 2 Correct answer: 2 . 85 m / s. Explanation: Let : r = 15 cm = 0 . 15 m . v = ω r = (19 rad / s) (0 . 15 m) = 2 . 85 m / s . 004 (part 3 of 4) 10.0 points What is the average angular velocity of the wheel during the 30 s? Correct answer: 9 . 5 rad / s. Explanation: ω = ω 0 + ω 2 = ω 2 = 19 rad / s 2 = 9 . 5 rad / s . 005 (part 4 of 4) 10.0 points Through what angle did the wheel rotate in the 30 s? Correct answer: 285 rad. Explanation: θ = ω t = (9 . 5 rad / s) (30 s) = 285 rad . Alternate Solution : θ = θ 0 + ω 0 t + 1 2 α t 2 = 1 2 α t 2 = 1 2 (0 . 633333 rad / s 2 ) (30 s) 2 = 285 rad . 006 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 77 cm apart on the same axle. From the angular displacement 15 . 5 of the two bul- let holes in the disks and the rotational speed 1201 rev / min of the disks, we can determine the speed of the bullet. 15 . 5 v 1201 rev / min 77 cm What is the speed of the bullet? Correct answer: 357 . 975 m / s. Explanation: Let : ω = 1201 rev / min , d = 77 cm , and θ = 15 . 5 . θ = ω t t = θ ω , so the speed of the bullet is v = d t = d ω θ = (77 cm) (1201 rev / min) 15 . 5 × 360 1 rev 1 m 100 cm 1 min 60 s = 357 . 975 m / s . keywords: 007 (part 1 of 2) 10.0 points A grinding wheel, initially at rest, is ro- tated with constant angular acceleration of 5 . 18 rad / s 2 for 10 . 3 s. The wheel is then brought to rest with uniform deceleration in 16 rev. Find the angular deceleration required to bring the wheel to rest.
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cerutti (cpc566) – HW08 – TSOI – (58160) 3 Correct answer: - 14 . 1581 rad / s 2 . Explanation: Let : ω 0 = 0 rad / s , α 1 = 5 . 18 rad / s 2 , t 1 = 10 . 3 s , and θ 2 = 16 rev .
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