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Unformatted text preview: salazar (ggs332) – HW08 – TSOI – (58160) 1 This printout should have 24 questions. Multiplechoice 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. 3. 4. correct 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 17 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 15 rad / s. What is the angular acceleration? Correct answer: 0 . 5 rad / s 2 . Explanation: Let : ω = 15 rad / s , ω = 0 , and t = 30 s . Since the angular acceleration is constant, α = Δ ω Δ t = ω t = 15 rad / s 30 s = . 5 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? salazar (ggs332) – HW08 – TSOI – (58160) 2 Correct answer: 2 . 55 m / s. Explanation: Let : r = 17 cm = 0 . 17 m . v = ω r = (15 rad / s) (0 . 17 m) = 2 . 55 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: 7 . 5 rad / s. Explanation: ω = ω + ω 2 = ω 2 = 15 rad / s 2 = 7 . 5 rad / s . 005 (part 4 of 4) 10.0 points Through what angle did the wheel rotate in the 30 s? Correct answer: 225 rad. Explanation: θ = ω t = (7 . 5 rad / s) (30 s) = 225 rad . Alternate Solution : θ = θ + ω t + 1 2 αt 2 = 1 2 αt 2 = 1 2 (0 . 5 rad / s 2 ) (30 s) 2 = 225 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 62 cm apart on the same axle. From the angular displacement 15 . 8 ◦ of the two bul let holes in the disks and the rotational speed 552 rev / min of the disks, we can determine the speed of the bullet. 15 . 8 ◦ v 552 rev / min 62 cm What is the speed of the bullet? Correct answer: 129 . 965 m / s. Explanation: Let : ω = 552 rev / min , d = 62 cm , and θ = 15 . 8 ◦ . θ = ω t t = θ ω , so the speed of the bullet is v = d t = dω θ = (62 cm) (552 rev / min) 15 . 8 ◦ × 360 ◦ 1 rev 1 m 100 cm 1 min 60 s = 129 . 965 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 3 . 08 rad / s 2 for 5 . 54 s. The wheel is then brought to rest with uniform deceleration in 13 . 2 rev....
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This note was uploaded on 11/03/2010 for the course PHYSICS 58160 taught by Professor Tsoi during the Spring '10 term at University of Texas.
 Spring '10
 TSOI
 Physics

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