# hw8 - patel(kp9583 HW8 Mackie(20211 This print-out should...

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patel (kp9583) – HW8 – Mackie – (20211) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A(n) 144 g ball is dropped From a height oF 53 . 7 cm above a spring oF negligible mass. The ball compresses the spring to a maximum displacement oF 4 . 49201 cm. The acceleration oF gravity is 9 . 8 m / s 2 . x h Calculate the spring Force constant k . Correct answer: 813 . 955 N / m. Explanation: Let : m = 144 g , h = 53 . 7 cm , and x = 4 . 49201 cm . Using conservation oF energy, we have 1 2 k x 2 = m g ( h + x ) From which k = 2 m g ( h + x ) x 2 = 2 (0 . 144 kg)(9 . 8 m / s 2 )(0 . 537 m + 0 . 0449201 m) (0 . 0449201 m) 2 = 813 . 955 N / m . 002 10.0 points A wheel rotating with a constant angular ac- celeration turns through 13 revolutions during a 6 s time interval. Its angular velocity at the end oF this interval is 15 rad / s. What is the angular acceleration oF the wheel? Note that the initial angular veloc- ity is not zero. Correct answer: 0 . 462144 rad / s 2 . Explanation: Let : N = 13 , t = 6 s , and ω = 15 rad / s . ±rom kinematics α t = ω f ω 0 ω 0 = ω f α t and Δ θ = N 2 ( π ) , so Δ θ = ω 0 t + 1 2 α t 2 2 N π = ( ω f αt ) t + 1 2 α t 2 = ω f t 1 2 α t 2 α = 2 ω f t 2 N π t 2 = 2 (15 rad / s) (6 s) 2 (13) π (6 s) 2 = 0 . 462144 rad / s 2 . keywords: 003 (part 1 oF 3) 10.0 points As a result oF Friction, the angular speed oF a wheel c hanges with time according to d θ d t = ω 0 e σ t , where ω 0 and σ are constants. The angular speed changes From an initial angular speed oF 4 . 58 rad / s to 3 . 08 rad / s in 3 . 92 s . Determine the magnitude oF the angular acceleration aFter 2 . 55 s.

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patel (kp9583) – HW8 – Mackie – (20211) 2 Correct answer: 0 . 358117 rad / s 2 . Explanation: Let : ω 0 = 4 . 58 rad / s , t 0 = 0 , ω 2 = 3 . 08 rad / s , t 2 = 3 . 92 s , and t 3 = 2 . 55 s . The equation of motion is ω = ω 0 e σ t , so ω 2 ω 0 = e σ t 2 ln p ω 2 ω 0 P = σ t 2 σ = ln p ω 2 ω 0 P t 2 = ln p 3 . 08 rad / s 4 . 58 rad / s P 3 . 92 s = 0 . 101217 s 1 . Thus the angular acceleration at t 3 is α ( t 3 ) = d ω dt = ω 0 ( σ ) e σ t 3 = (4 . 58 rad / s) (0 . 101217 s 1 ) × e (0 . 101217 s - 1 ) (2 . 55 s) = 0 . 358117 rad / s 2 b ( t 3 ) b = 0 . 358117 rad / s 2 . 004
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## This note was uploaded on 12/10/2009 for the course PHYSICS 2021 taught by Professor Mihalisin during the Fall '09 term at Temple.

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hw8 - patel(kp9583 HW8 Mackie(20211 This print-out should...

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