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solution_pdf4 - garcia(jg46766 Problem Set 9 lacosse(S210-7...

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garcia (jg46766) – Problem Set 9 – lacosse – (S210-7) 1 This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. Angular Motion Problem Set (35 Ques- tions). 001 10.0 points A box with its center oF mass o±-center as indicated by the dot is placed on an inclined plane. In which orientation does the box tip over? 1. b 2. 3. None oF the orientations will cause the box to tip over. 4. 5. correct Explanation: In order to tip over, the box must pivot about its bottom leFt corner. The only orien- tation where gravity creates a torque su²cient to tip the block is 002 10.0 points An automobile engine develops a torque oF 650 N · m and is rotating at a speed oF 2000 rev / min. What horsepower does the engine generate? 1 hp = 746 W. Correct answer: 182 . 488 hp. Explanation: Let : τ = 650 N · m and ω = 2000 rev / min . P = τ ω = (650 N · m) (2000 rev / min) × p 2 π 1 rev Pp 1 min 60 s 1 hp 746 W P = 182 . 488 hp . 003 10.0 points A turntable that is initially at rest is set in motion with a constant angular acceleration α . What is the magnitude oF the angular ve- locity oF the turntable aFter it has made one complete revolution? 1. b b = 2 π α 2. b b = 2 α 3. b b = 4 π α 4. b b = 2 π α 5. b b = 2 π α correct 6. b b = 2 α
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garcia (jg46766) – Problem Set 9 – lacosse – (S210-7) 2 Explanation: Let : ω 0 = 0 rad / s and θ = 1 rev = 2 π rad . Similar to uniform linear accelerated mo- tion, we have ω 2 f = ω 2 0 + 2 αθ = 2 α (2 π ) = 4 π α b f b = 2 π α 004 (part 1 of 3) 10.0 points Four particles with masses 7 kg, 5 kg, 7 kg, and 5 kg are connected by rigid rods of neg- ligible mass as shown. The origin is centered on the mass in the lower left corner. The rectangle is 6 m wide and 5 m long. x y 2 rad / s 5 m 6 m 7 kg 5 kg 7 kg 5 kg If the system rotates in the xy plane about the z axis (origin, O) with an angular speed of 2 rad / s, calculate the moment of inertia of the system about the z axis. Correct answer: 732 kg · m 2 . Explanation: Let : m 1 = 7 kg , top left m 2 = 5 kg , bottom left m 3 = 7 kg , bottom right m 4 = 5 kg , top right = 6 m , and h = 5 m . The moment of inertia is I z = s j m j r 2 j = m 1 h 2 + m 2 · 0 + m 3 2 + m 4 ( h 2 + 2 ) = (7 kg) (5 m) 2 + (7 kg) (6 m) 2 + (5 kg) [(5 m) 2 + (6 m) 2 ] = 732 kg · m 2 . 005 (part 2 of 3) 10.0 points Find the moment of inertia of the four-particle system about an axis that is perpendicular to the plane of the con±guration and passing through the center of mass of the system. Correct answer: 366 kg · m 2 . Explanation: x CM = m 1 · 0 + m 2 · 0 + m 3 + m 4 m 1 + m 2 + m 3 + m 4 = (7 kg) + (5 kg) 7 kg + 5 kg + 7 kg + 5 kg = 2 , and y = m 1 h + m 2 · 0 + m 3 · 0 + m 4 h m 1 + m 2 + m 3 + m 4 = (7 kg) h + (5 kg) h 7 kg + 5 kg + 7 kg + 5 kg = h 2 . For each point r 2 = x 2 + y 2 = p 2 P 2 + p h 2 P 2 , so the moment of inertia is I z CM = s j m j r 2 j = r 2 s j m j = 2 + h 2 4 ( m 1 + m 2 + m 3 + m 4 ) = (6 m) 2 + (5 m) 2 4 × (7 kg + 5 kg + 7 kg + 5 kg) = 366 kg · m 2 .
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garcia (jg46766) – Problem Set 9 – lacosse – (S210-7) 3 006 (part 3 of 3) 10.0 points For the four-particle system ±nd the mo- ment of inertia about the y -axis, which passes through the two masses on the left-hand side of the system.
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