{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 11-3-solutions

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

This preview shows pages 1–3. Sign up to view the full content.

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?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}