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Homework 25 - homework 25 BAUTISTA ALDO Due 4:00 am Version...

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homework 25 – BAUTISTA, ALDO – Due: Mar 31 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 3. 10 points. A cylindrical 9 . 67 kg pulley with a radius of 0 . 643 m is used to lower a 6 . 36 kg bucket into a well. The bucket starts from rest and falls for 4 . 04 s. The acceleration of gravity is 9 . 81 m / s 2 . a) What is the linear acceleration of the falling bucket? Correct answer: 5 . 57317 m / s 2 (tolerance ± 1 %). Explanation: Basic Concepts: Newton’s second law for the bucket: F net = ma = mg - F T τ = = 1 2 MR 2 α τ = F T R since F T is perpendicular to R . a = αR Given: R = 0 . 643 m M = 9 . 67 kg m = 6 . 36 kg g = 9 . 81 m / s 2 Solution: F T = τ R = ( 1 2 MR 2 ) ( a R ) R = 1 2 Ma Thus from Newton’s second law, ma = mg - 1 2 Ma 2 ma + Ma = 2 mg a = 2 mg (2 m + M ) = 2(6 . 36 kg)(9 . 81 m / s 2 ) 2(6 . 36 kg) + 9 . 67 kg = 5 . 57317 m / s 2 Question 2 Part 2 of 3. 10 points. b) How far does it drop? Correct answer: 45 . 4815 m (tolerance ± 1 %). Explanation: Basic Concept: Δ y = v i Δ t + 1 2 a t ) 2 = 1 2 a t ) 2 since v i = 0 m / s. Given: Δ t = 4 . 04 s Solution: Δ y = 1 2 (5 . 57317 m / s 2 )(4 . 04 s) 2 = 45 . 4815 m Question 3 Part 3 of 3. 10 points. c) What is the angular acceleration of the cylindrical pulley? Correct answer: 8 . 66744 rad / s 2 (tolerance ± 1 %). Explanation: Basic Concept: a = αR Solution: α = a R = 5 . 57317 m / s 2 0 . 643 m = 8 . 66744 rad / s 2 Question 4 Part 1 of 2. 10 points. A 1 . 7 kg particle moves in a circle of ra- dius 1 . 7 m. The magnitude of its angular
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homework 25 – BAUTISTA, ALDO – Due: Mar 31 2006, 4:00 am 2 momentum relative to the center of the circle depends on time according to L = (6 N · m) t . Find the magnitude of the torque acting on the particle. Correct answer: 6 N · m (tolerance ± 1 %).
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