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# quiz3 - Muraj Hamza Quiz 3 Due Apr 5 2006 10:00 pm Inst...

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Muraj, Hamza – Quiz 3 – Due: Apr 5 2006, 10:00 pm – Inst: Florin 1 This print-out should have 20 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 The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an an- gular speed of 3 . 4 rev / s in 5 s . At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 11 . 5 s . Through how many revolutions does the tub turn? Assume Constant angular acceleration while it is starting and stopping. Correct answer: 28 . 05 rev. Explanation: We will break the motion into two stages: (1) an acceleration period and (2) a deceleration period. While speeding up, θ 1 = ω t = 0 + ω 2 t 1 = 1 2 ω t 1 = 1 2 (3 . 4 rev / s) (5 s) = 8 . 5 rev . While slowing down, θ 2 = ω t = ω + 0 2 t 2 = 1 2 ω t 2 = 1 2 (3 . 4 rev / s) (11 . 5 s) = 19 . 55 rev . So θ = θ 1 + θ 2 = (8 . 5 rev) + (19 . 55 rev) = 28 . 05 rev . 002 (part 1 of 1) 10 points Assume: When the disk lands on the surface it does not bounce. The disk has mass m and outer radius R with a non-uniform radial mass distribution so that its moment of inertia I = 5 6 m R 2 . A disk is given a hard kick along a hori- zontal surface. The kicking force acts along a horizontal line through the disk’s center, so the disk acquires a linear velocity v 0 but no angular velocity. The kinetic friction force between the sur- face and the disk slows down its linear motion while at the same time making the disk spin on its axis at an accelerating rate. Eventually, the disk’s rotation catches up with its linear motion, and the disk begins to roll without slipping on the surface. The acceleration of gravity is g . I = 5 6 m R 2 R , radius m v 0 μ How long Δ t does it take for the ball to roll without slipping? 1. Δ t = 2 5 v 0 μ g 2. Δ t = 4 11 v 0 μ g 3. Δ t = 7 16 v 0 μ g 4. Δ t = 8 17 v 0 μ g 5. Δ t = 1 3 v 0 μ g 6. Δ t = 3 7 v 0 μ g 7. Δ t = 5 11 v 0 μ g correct 8. Δ t = 1 2 v 0 μ g

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Muraj, Hamza – Quiz 3 – Due: Apr 5 2006, 10:00 pm – Inst: Florin 2 9. Δ t = 3 8 v 0 μ g 10. Δ t = 1 4 v 0 μ g Explanation: 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 = v 0 - a t = v 0 - 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 ; the angular acceleration is X τ = I α , so α = τ I = μ m g R 5 6 m R 2 = 6 5 μ g R . (2) The time dependence of ω is ω = α t = 6 5 μ g R t , so v disk R ω = 6 5 μ g 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 v disk = v surface 6 5 μ g t = v 0 - μ g t , or 11 5 μ g t = v 0 , so t = 5 11
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quiz3 - Muraj Hamza Quiz 3 Due Apr 5 2006 10:00 pm Inst...

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