# HW23 - Pickering Tracey Homework 23 Due noon Inst Drummond...

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Pickering, Tracey – Homework 23 – Due: Mar 29 2006, noon – Inst: Drummond 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 3) 10 points A cylinder oF mass 15 kg rolls without slipping on a horizontal surFace . At the instant its center oF mass has a speed oF 9 . 5 m / s determine the translational kinetic energy oF its center oF mass. Correct answer: 676 . 875 J. Explanation: The translational kinetic energy is given by T t = 1 2 m v 2 = 676 . 875 J The rotational energy is given by T r = 1 2 I ω 2 = 338 . 438 J Since the moment oF inertia I For a cylinder is I = 1 2 m R 2 and using v = ω R , we get T r = 1 4 m v 2 = 338 . 438 J 002 (part 2 oF 3) 10 points At the instant its center oF mass has a speed oF 9 . 5 m / s determine the rotational energy about its center oF mass Correct answer: 338 . 438 J. Explanation: 003 (part 3 oF 3) 10 points At the instant its center oF mass has a speed oF 9 . 5 m / s determine its total energy. Correct answer: 1015 . 31 J. Explanation: 004 (part 1 oF 3) 10 points A 4 . 65 kg hollow cylinder with inner radius 0 . 26 m and outer radius 0 . 43 m is pulled by a horizontal string with a Force oF 38 . 9 N, as shown in the fgure. 38 . 9 N 4 . 65 kg ω 0 . 26 m 0 . 43 m The Frictional Force exerted on the wheel by the ±oor 1. is to the right. correct 2. is to the leFt. 3. is zero. Explanation: IF there were no Friction the bottom oF the wheel would spin to the leFt; the Frictional Force will oppose this spinning motion. ThereFore, the Frictional Force is directed to the right. 005 (part 2 oF 3) 10 points What must be the magnitude oF the Force oF Friction iF the cylinder is to roll without slipping? Correct answer: 7 . 33243 . Explanation: Let : R in = 0 . 26 m , R out = 0 . 43 m , F = 38 . 9 N , m = 4 . 65 kg , and f = Force oF Friction . Suppose the cylinder has a length oF ; then the density oF the cylinder is ρ = m π ( R 2 out - R 2 in ) . The moment oF inertia about the center is I cm = Z ρ r 2 dV = m π ( R 2 out - R 2 in ) × Z R out R in r 2 r dr Z 2 π 0 Z 0 dz

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Pickering, Tracey – Homework 23 – Due: Mar 29 2006, noon – Inst: Drummond 2 = m π ( R 2 out - R 2 in ) R 4 out - R 4 in 4 2 π ‘ = m 2 ( R 2 out + R 2 in ) . (1) By the parallel axis theorem, the moment of inertia about the point of contact with the ground is I = I cm + m R 2 out = 1 2 m (3 R 2 out + R 2 in ) (2) = 1 2 (4 . 65 kg) h 3 (0 . 43 m) 2 + (0 . 26 m) 2 i = 1 . 44685 kg · m 2 . Using a free body diagram (see part 1 for the direction of f ), F ω 0 . 26 m 0 . 43 m f a we have F + f = m a f = m a - F . (3) From the torque equation (using the contact point of the cylinder and the surface as the axis of rotation), we have τ = I α F (2 R out ) = I α (4) = I a R out , so a = 2 R 2 out I F m a = 2 m R 2 out I F , using Eq . 2 = 4 R 2 out 3 R 2 out + R 2 in F . (5) Substituting m a from Eq. 5 into Eq. 3, we get
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HW23 - Pickering Tracey Homework 23 Due noon Inst Drummond...

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