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oldhw23sol

# oldhw23sol - rote(cmr2495 oldhomework 23 Turner(58120 This...

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rote (cmr2495) – oldhomework 23 – Turner – (58120) 1 This print-out should have 14 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 4 kg and outer radius 30 cm with a radial mass distribution (which may not be uniform) so that its moment of iner- tia is 9 10 m R 2 . It disk is given a hard kick (impulse) along a horizontal surface at time t 0 . The kicking force acts along a horizontal line through the disk’s center, so the disk ac- quires a linear velocity 0 . 9 m / s but no initial angular velocity. The coefficient of friction between the disk and the surface is 0 . 09 . 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 at time t rolling without slipping on the surface. 30 cm 4 kg 0 . 9 m / s μ = 0 . 09 Once the disk rolls without slipping, what is its linear speed? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 473684 m / s. Explanation: Let : R = 30 cm = 0 . 3 m , v 0 = 0 . 9 m / s , m = 4 kg , and μ = 0 . 09 . From the perspective of the surface, let the speed of the center of the disk be v surface . summationdisplay F surface = f m a = μ m g a = μ g , so v surface = v 0 a t = v 0 μ g 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 ω , so summationdisplay τ = I α α = τ I = f R I = μ m g R 9 10 m R 2 = 10 9 μ g R , the time dependence of ω is ω = α t = 10 9 μ g R t , and v disk = R ω = 10 9 μ g t .

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