hw 10 - homework 10 – HETTIGER CHRISTOF – Due Feb 9...

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Unformatted text preview: homework 10 – HETTIGER, CHRISTOF – Due: Feb 9 2008, 4:00 am 1 Question 1, chap 12, sect 2. part 1 of 1 10 points What angular acceleration is necessary to increase the angular speed of a fan blade from 8.6 rad/s to 16.0 rad/s in 5.6 s? Correct answer: 1 . 32143 rad / s 2 (tolerance ± 1 %). Explanation: Basic Concept: α avg = ω 2- ω 1 Δ t Given: ω 1 = 8 . 6 rad / s ω 2 = 16 . 0 rad / s Δ t = 5 . 6 s Solution: α avg = 16 rad / s- 8 . 6 rad / s 5 . 6 s = 1 . 32143 rad / s 2 Question 2, chap 12, sect 2. part 1 of 4 10 points A potter’s wheel of radius 16 cm starts from rest and rotates with constant angular accel- eration until at the end of 35 s it is moving with angular velocity of 16 rad / s. What is the angular acceleration? Correct answer: 0 . 457143 rad / s 2 (tolerance ± 1 %). Explanation: R = 16 cm = 0 . 16 m Since the angular acceleration is constant, α = Δ ω Δ t = ω t = 16 rad / s 35 s = 0 . 457143 rad / s 2 Question 3, chap 12, sect 2. part 2 of 4 10 points What is the linear velocity of a point on the rim at the end of the 35 s? Correct answer: 2 . 56 m / s (tolerance ± 1 %). Explanation: v = ω R = (16 rad / s) (0 . 16 m) = 2 . 56 m / s Question 4, chap 12, sect 2. part 3 of 4 10 points What is the average angular velocity of the wheel during the 35 s? Correct answer: 8 rad / s (tolerance ± 1 %). Explanation: ω = ω + ω 2 = ω 2 = 16 rad / s 2 = 8 rad / s Question 5, chap 12, sect 2. part 4 of 4 10 points Through what angle did the wheel rotate in the 35 s? Correct answer: 280 rad (tolerance ± 1 %). Explanation: θ = ω t = (8 rad / s) (35 s) = 280 rad or, θ = θ + ω t + 1 2 α t 2 = 1 2 α t 2 = 1 2 (0 . 457143 rad / s 2 ) (35 s) 2 = 280 rad Question 6, chap 5, sect 1....
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hw 10 - homework 10 – HETTIGER CHRISTOF – Due Feb 9...

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