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hw23 - Overton Mays Homework 23 Due 4:00 am Inst Turner...

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Overton, Mays – Homework 23 – Due: Mar 28 2005, 4:00 am – Inst: Turner 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A girl throws a long stick with a length of 0 . 92 m and a mass of 0 . 93 kg into the air in such a way that the center of mass rises vertically. At the moment it leaves her hand, the stick is horizontal and the speed of the end of the stick nearest to her is zero. When the center of mass of the stick reaches its highest point, the stick is horizontal, and it has made 27 complete revolutions. Assume: The stick’s cross sectional area and mass is uniform. The acceleration of gravity is 9 . 8 m / s 2 . 0 . 46 m 0 . 92 m v cm 0 0 . 93 kg v end 0 How long did it take for the center of mass to reach its highest point? Correct answer: 2 . 82187 s. Explanation: From kinematics v 0 = g t (1) From definition of angular velocity v 0 = r ω = 2 ω . (2) From kinematics θ = ω t ω = θ t . (3) Substituting ω from Eq. 3 into Eq. 2 gives v 0 = 2 θ t . (4) Substituting v 0 from Eq. 1 into Eq. 4 gives 2 θ t = g t . (5) Since, θ = n (2 π ) , we have t = s n π ‘ g (6) = s (27) π (0 . 92 m) (9 . 8 m / s 2 ) = 2 . 82187 s . 002 (part 1 of 2) 10 points Assume: When the disk lands on the surface it does not bounce. A disk (with radius 50 cm , mass 3 kg , and moment of inertia m R 2 / 2) rotates at angular speed 9 rad / s around its axis. The disk is carefully lowered onto a horizontal surface and released with zero initial linear velocity along the surface. The kinetic friction force between the sur- face and the disk slows down the rotation of the disk and at the same time gives it a hor- izontal acceleration. Eventually, the disk’s linear motion catches up with its rotation, and the disk begins to roll without slipping on the surface. The acceleration of gravity is 9 . 8 m / s 2 . 50 cm , radius 3 kg 9 rad / s μ = 0 . 04 Determine the time it takes the disk to start rolling without slipping. Correct answer: 3 . 82653 s. Explanation: Let : r = 50 cm = 0 . 5 m , ω 0 = 9 rad / s , m = 3 kg , and μ = 0 . 04 .
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Overton, Mays – Homework 23 – Due: Mar 28 2005, 4:00 am – Inst: Turner 2 From the perspective of the surface, let the speed of the center of the disk be v surface . Using the frictional force f , we can determine the acceleration X F s = m a , and f = μ m g , so a = μ g . Then v surface = a t = μ g t . (1) After pure rolling begins there is no longer any frictional force and consequently no longer any acceleration. From the perspective of the center of the disk, let the tangential velocity of the rim of the disk be v disk , the angular velocity be ω , and the angular acceleration is X τ = I α , so α = τ I = μ m g R m R 2 2 = 2 μ g R (2) = 2 (0 . 04) (9 . 8 m / s 2 ) (0 . 5 m) = 1 . 568 rad / s 2 . The time dependence of ω is ω = ω 0 - α t = ω 0 - 2 μ g R t , so v disk R ω = R ω 0 - 2 μ g R t . (3) When the disk reaches pure rolling, the veloc- ity from the perspective of the surface will be the same as the velocity from the perspective of the center of the disk; that is, there will be no slipping. Setting the velocity v disk from Eq. 3 equal to v surface from Eq. 1 gives
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