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

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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(part1of7)3.0points 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 τ = vectorr × vector F . Therefore, | vector τ | = | vectorr || vector F | = (0 . 076 m) (14 . 1 N). Thus, | vector τ | = 1 . 0716 N · m. 002(part2of7)2.0points 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 τ = < 0 , 0 , 1 . 0716 N · m > N · m. 003(part3of7)2.0points 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 = Ivectorω = < 0 , 0 , > , where I = (1 / 2) MR 2 for the solid disc and vectorω is directed along the negative z -axis. Using the values given from the beginning of the problem, the equation becomes vector L = < 0 , 0 , (1 / 2) MR 2 ω > = < 0 , 0 , 0 . 023595 kg · m 2 / s > | vector L | = 0 . 023595 kg · m 2 / s 004(part4of7)2.0points 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, vectorω . 005(part5of7)2.0points At time Δ t = 0 . 2 s later, what is the direction of the change of the angular momentum about the center of the disk?
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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(part6of7)2.0points At time Δ 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 Δ vector L = vector τ Δ t Δ vector L = < 0 , 0 , RF Δ t > The final angular momentum is then given by vector L f = vector L + Δ vector L = vector L + < 0 , 0 , RF Δ t > = < 0 , 0 , (1 / 2) MR 2 ω RF Δ t > = < 0 , 0 , 0 . 023595 0 . 21432 > kg · m 2 /s = < 0 , 0 , 0 . 237915 kg · m 2 / s > | vector L f | = 0 . 237915 kg · m 2 / s The change in angular momentum is in the same direction as the initial angular momen- tum, so the magnitude of vector L increases in the negative z direction.
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