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10homerwork08 - Husain Zeena Homework 8 Due 4:00 am Inst H...

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Husain, Zeena – Homework 8 – Due: Oct 21 2003, 4:00 am – Inst: H L Berk 1 This print-out should have 23 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 2) 10 points A wheel starts from rest and rotates with constant angular acceleration to an angular speed of 8 . 39 rad / s in 1 . 95 s. Find the magnitude of the angular acceler- ation of the wheel. Correct answer: 4 . 30256 rad / s 2 . Explanation: Angular acceleration is defined by, α = dt When the angular acceleration is constant, we can replace the differentials with simply differences, α = Δ ω Δ t = ω f - ω i t The angle through which the wheel rotates during this time interval is, θ = 1 2 αt 2 002 (part 2 of 2) 10 points Find the angle in radians through which it rotates in this time. Correct answer: 8 . 18025 rad. Explanation: 003 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . A car traveling on a flat (unbanked, 9 m radius) circular track accelerates uniformly from rest with a tangential acceleration of 1 . 9 m / s 2 . The car makes it 0 . 4 of the way around the circle before skidding off the track. Determine the coefficient of static friction between car and track. Correct answer: 0 . 993633 . Explanation: Given : θ = 0 . 4 rev = 2 . 51327 rad , r = 9 m , a t = 1 . 9 m / s 2 , and g = 9 . 8 m / s 2 . Just before it starts to skid the force equations are X F r = m a r = m v 2 r = m ω 2 r , X F t = m a t , or μ m g = q ( m a t ) 2 + ( m ω 2 r ) 2 . Therefore, μ = s a t g 2 + ω 2 r g 2 = s a t g 2 + 2 α θ r g 2 = s a t g 2 + 2 a t θ g 2 = a t g q 1 + [2 θ ] 2 = 1 . 9 m / s 2 9 . 8 m / s 2 q 1 + [2 (2 . 51327 rad)] 2 = 0 . 993633 . 004 (part 1 of 4) 10 points A disk 3 . 46 cm in radius rotates at a constant rate of 1230 rev / min about its central axis. Determine its angular speed in radians per second. Correct answer: 128 . 805 rad / s. Explanation: Convert the units from revolutions per min- utes to radians per second, ω = ω 0 2 π rad 60 s . The linear speed at r from the center of the disc is, v = r ω
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Husain, Zeena – Homework 8 – Due: Oct 21 2003, 4:00 am – Inst: H L Berk 2 If the radius of the disc is given by R , the radial acceleration on the rim is, a r = R ω 2 The distance a point on the rim moves in time t is, s = R θ = R ω t 005 (part 2 of 4) 10 points Determine the linear speed at a point 1 . 8 cm from its center. Correct answer: 2 . 3185 m / s. Explanation: 006 (part 3 of 4) 10 points Determine the radial acceleration of a point on the rim. Correct answer: 574 . 043 m / s 2 . Explanation: 007 (part 4 of 4) 10 points Determine the total distance a point on the rim moves in 2 . 73 s. Correct answer: 12 . 1667 m. Explanation: 008 (part 1 of 1) 10 points A circular disk with a mass m and radius R is mounted at its center, about which it can rotate freely. The disk has moment of inertia I = (1 / 2) mR 2 . A light cord wrapped around it supports a weight mg . Find the total kinetic energy of the system, when the weight is moving at a speed v .
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