HW 11-3-solutions - madrid (tmm2353) HW 11-3 Antoniewicz...

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Unformatted text preview: madrid (tmm2353) HW 11-3 Antoniewicz (56445) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 7) 3.0 points A disk of radius 7 . 6 cm and mass 0 . 43 kg is pulled along a frictionless surface with a force of 14 . 1 N by a string wrapped around the edge (Fig. 11.43, displayed below). 24 cm of string has unwound off the disk. At the instant shown the angular velocity is 19 rad / s. What is the magnitude of the torque exerted about the center of the disc at this instant? 7 . 6 cm 14 . 1 N 24 cm Correct answer: 1 . 0716 N m. Explanation: Use vector = vectorr vector F . Therefore, | vector | = | vectorr || vector F | = (0 . 076 m) (14 . 1 N). Thus, | vector | = 1 . 0716 N m. 002 (part 2 of 7) 2.0 points What is the direction of the torque exerted about the center of the disc at this instant? (choice of axes: x to the right, y up, z out of the page) 1. y 2.- y 3.- x 4.- z correct 5. z 6. x Explanation: The direction of the torque, given by the right-hand rule, is in the negative z direction. So, vector = < , , 1 . 0716 N m > N m. 003 (part 3 of 7) 2.0 points At the same instant, what is the magnitude of the angular momentum about the center of the disc? Correct answer: 0 . 023595 kg m 2 / s. Explanation: vector L = Ivector = < , , I > , where I = (1 / 2) MR 2 for the solid disc and vector is directed along the negative z-axis. Using the values given from the beginning of the problem, the equation becomes vector L = < , , (1 / 2) MR 2 > = < , , . 023595 kg m 2 / s > | vector L | = 0 . 023595 kg m 2 / s 004 (part 4 of 7) 2.0 points At the same instant, what is the direction angular momentum about the center of the disc? 1. z 2.- y 3. y 4.- z correct 5. x 6.- x Explanation: The direction of the angular momentum is the same as that of the angular velocity, vector . 005 (part 5 of 7) 2.0 points At time t = 0 . 2 s later, what is the direction of the change of the angular momentum about the center of the disk? madrid (tmm2353) HW 11-3 Antoniewicz (56445) 2 1. x 2.- x 3.- z correct 4. z 5.- y 6. y Explanation: The angular momentum increases in the direction of the angular velocity, so the the change in angular momentum is in the nega- tive z direction. 006 (part 6 of 7) 2.0 points At time t = 0 . 2 s later, what is the mag- nitude of the angular momentum about the center of the disk? Correct answer: 0 . 237915 kg m 2 / s. Explanation: Since the torque acting on the disc is con- stant over time, we conclude the change in angular momentum is given by vector L = vector t vector L = < , , RF t > The final angular momentum is then given by vector L f = vector L + vector L = vector L + < , , RF t > = < , , (1 / 2) MR 2 RF t > = < , , . 023595 . 21432 > kg m 2 /s = < , , . 237915 kg m 2 / s > | vector...
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HW 11-3-solutions - madrid (tmm2353) HW 11-3 Antoniewicz...

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