14exam3 - Husain, Zeena Exam 3 Due: Nov 12 2003, 9:00 pm...

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Unformatted text preview: Husain, Zeena Exam 3 Due: Nov 12 2003, 9:00 pm Inst: H L Berk 1 This print-out should have 16 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points A wheel rotates about a fixed axis with an initial angular velocity of 79 rad / s. During a 9 s interval the angular velocity increases to 122 rad / s. Assume: The angular acceleration was con- stant during this time interval. How many revolutions does the wheel turn through during this time interval? Correct answer: 143 . 956 . Explanation: First find = f- i t = 4 . 77778 rad / s 2 . The total angle rotated in time t = 9 s is given by = t + 1 2 t 2 = (79 rad / s) (9 s) + 1 2 (4 . 77778 rad / s 2 ) (9 s) 2 = 904 . 5 rad . Finally, the number of revolutions rotated is simply N = 2 = 143 . 956 . 002 (part 1 of 1) 10 points A horizontal disk with a radius of 17 m ro- tates about a vertical axis through its center. The disk starts from rest and has a constant angular acceleration of 1 . 9 rad / s 2 . At what value of t will the radial and tan- gential components of the linear acceleration of a point on the rim of the disk be equal in magnitude? Correct answer: 0 . 725476 s. Explanation: The tangential and radial accelerations are given by a t = r a r = 2 r Equating the two gives us the expression 2 = = (1) Now consider the equation = - t (2) Plug the Eq. (1) into (2) and solve for t , where = t = 0. t = = = 1 = 1 p (1 . 9 rad / s 2 ) = 0 . 725476 s 003 (part 1 of 3) 10 points Three particles of mass m 1 = 3 . 4 kg, m 2 = 2 kg, and m 3 = 3 . 5 kg are connected by rigid rods of negligible mass lying along the y axis and are placed at y 1 = 2 . 7 m, y 2 =- 2 . 5 m, and y 3 =- 4 . 5 m respectively as in the figure. The system rotates about the x axis with an angular speed of 2 . 04 rad / s. y x m m m y y y 1 1 2 2 3 3 Find the moment of inertia about the x axis. Correct answer: 108 . 161 kg m 2 . Explanation: Husain, Zeena Exam 3 Due: Nov 12 2003, 9:00 pm Inst: H L Berk 2 The total rotational inertia of the system about the x axis is, I = X m i r 2 i = 108 . 161 kg m 2 where, r i = | y i | . 004 (part 2 of 3) 10 points Find the total rotational energy of the sys- tem. Correct answer: 225 . 061 J. Explanation: Since = 2 . 04 rad / s, the total rotational energy is, E = 1 2 I 2 = 1 2 (108 . 161 kg m 2 )(2 . 04 rad / s) 2 = 225 . 061 J . 005 (part 3 of 3) 10 points Find the linear speed of the particle of mass m 1 = 3 . 4 kg. Correct answer: 5 . 508 m / s....
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This note was uploaded on 03/23/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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14exam3 - Husain, Zeena Exam 3 Due: Nov 12 2003, 9:00 pm...

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