hw26 - This print-out should have 12 questions...

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Unformatted text preview: This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A disk has mass 9 kg and outer radius 80 cm with a radial mass distribution (which may not be uniform) so that its moment of inertia is 9 10 mR 2 . The disk is rotating at angular speed 2 rad / s around its axis when it touches the surface, as shown. The disk is carefully lowered onto a horizontal surface and released at time t with zero initial linear velocity along the surface. Assume that when the disk lands on the surface it does not bounce. The coefficient of friction between the disk and the surface is 0 . 01 . 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 disks linear motion catches up with its rotation, and the disk begins to roll (at time t rolling ) without slipping on the surface. 9 kg 2 rad / s = 0 . 01 Once the disk rolls without slipping, what is its angular speed? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 947368 rad / s. Explanation: Let : R = 80 cm = 0 . 8 m , = 2 rad / s , m = 9 kg , = 0 . 01 , and g = 9 . 8 m / s 2 . m From the perspective of the surface, let the speed of the center of the disk be v surface . summationdisplay F surface = f ma = m g a = g , so = a R = g R and surface = t = g R t. After pure rolling begins at t rolling there is no longer any frictional force and conse- quently no acceleration. From the perspective of the center of the disk, let the tangential ve- locity of the rim of the disk be v disk and the angular velocity be , so summationdisplay = I = I = mg R 9 10 mR 2 = 10 9 g R and the time dependence of is disk = - t = - 10 9 g R t. When the disk reaches pure rolling, the velocity 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, so disk = surface - 10 9 g R t = g R t 19 9 g t = R t = 9 19 R g homework 26 Turner (58185) 1 hinojosa (jlh3938) homework 26 Turner (58185) 2 rot = t = g R parenleftbigg 9 19 R g parenrightbigg = 9 19 = 9 19 (2 rad / s) = . 947368 rad / s . 002 10.0 points Two pulley wheels, of respective radii R 1 = . 2 m and R 2 = 0 . 93 m are mounted rigidly on a common axle and clamped together. The combined moment of inertia of the two wheels is I + 3 . 4 kg m 2 . Mass m 1 = 38 kg is attached to a cord wrapped around the first wheel, and another mass m 2 = 17 kg is attached to another cord wrapped around the second wheel: 1 2 m R 1 m R 2 Find the angular acceleration of the system....
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This note was uploaded on 05/02/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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hw26 - This print-out should have 12 questions...

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