Old Midterm 3 - Solutions

# Old Midterm 3 - Solutions - oldmidterm 03 – FRENNEA KYLE...

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Unformatted text preview: oldmidterm 03 – FRENNEA, KYLE – Due: Apr 3 2007, 3:00 pm 1 Question 1 part 1 of 1 10 points The tub of a washer goes into its spin- dry cycle, starting from rest and reaching an angular speed of 2 . 8 rev / s in 9 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 7 s . Through how many revolutions does the tub turn? Assume Constant angular acceleration while it is starting and stopping. Correct answer: 22 . 4 rev (tolerance ± 1 %). 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 (2 . 8 rev / s) (9 s) = 12 . 6 rev . While slowing down, θ 2 = ω t = ω + 0 2 t 2 = 1 2 ω t 2 = 1 2 (2 . 8 rev / s) (7 s) = 9 . 8 rev . So θ = θ 1 + θ 2 = (12 . 6 rev) + (9 . 8 rev) = 22 . 4 rev . Question 2 part 1 of 1 10 points Assume: When the disk lands on the sur- face it does not bounce. The disk has mass m and outer radius R with a radial mass distribution (which may not be uniform) so that its moment of inertia is 1 3 mR 2 . The disk is rotating at angular speed ω around its axis when it touches the surface, as shown in the figure below. The disk is carefully lowered onto a horizontal surface and released at time t with zero initial linear velocity along the surface. The coefficient of friction between the disk and the surface is μ. 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 (at time t rolling ) without slipping on the surface. The acceleration of gravity is g . I = 1 3 mR 2 R, radius m ω μ Once the disk rolls without slipping, what is its angular speed? 1. ω rot = 1 3 ω 2. ω rot = 2 5 ω 3. ω rot = 5 11 ω 4. ω rot = 4 9 ω 5. ω rot = 1 4 ω correct 6. ω rot = 4 13 ω 7. ω rot = 3 7 ω oldmidterm 03 – FRENNEA, KYLE – Due: Apr 3 2007, 3:00 pm 2 8. ω rot = 7 16 ω 9. ω rot = 7 15 ω 10. ω rot = 4 11 ω 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 f = μmg , and summationdisplay F surface = ma, or ma = μm g , so a = μg , and α = μg R . (1) Since ω surface = αt, we have = μg R t. (2) After pure rolling begins at t rolling there is no longer any frictional force and consequently no acceleration. From the perspective of the center of the disk, let the tangential velocity of the rim of the disk be v disk and the angular velocity be ω ; the angular acceleration is summationdisplay τ = I α, so α = τ I = μmg R 1 3 mR 2 = 3 μg R . (3) The time dependence of ω is ω = ω − αt = ω − 3 μg R t. (4) 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...
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Old Midterm 3 - Solutions - oldmidterm 03 – FRENNEA KYLE...

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