hw11 - Overton, Mays Homework 11 Due: Feb 19 2005, 4:00 am...

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

View Full Document Right Arrow Icon
Overton, Mays – Homework 11 – Due: Feb 19 2005, 4:00 am – Inst: Turner 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points A mass slides with negligible ±riction on a circular track o± 1 m radius oriented vertically. Its speed at the position shown in the fgure is 3.13 m/s. The acceleration o± gravity is 9.8 m/s 2 . v At the position shown in the fgure, which o± the labeled arrows best represents the di- rection o± the acceleration o± the mass? 1. 2. correct 3. 4. 5. 6. 7. 8. 9. The mass is traveling at a constant veloc- ity, there±ore it has no acceleration. Explanation: The magnitude o± the centripetal is a r = v 2 r = (3 . 13 m / s) 2 1 m 9 . 8 m / s 2 . The centripetal acceleration is inward - ˆ r and gravity is down - ˆ k . a a r g 002 (part 1 o± 2) 10 points The coe²cient o± static ±riction between the person and the wall is 0 . 73 . The radius o± the cylinder is 5 . 59 m . The acceleration o± gravity is 9 . 8 m / s 2 . An amusement park ride consists o± a large vertical cylinder that spins about its axis ±ast enough that any person inside is held up against the wall when the ³oor drops away. ω 5 . 59m What is the minimum angular velocity ω min needed to keep the person ±rom slip- ping downward? Correct answer: 1 . 54969 rad / s. Explanation: Let : R = 5 . 59 m and μ = 0 . 73 . Basic Concepts: Centripetal ±orce F = mv 2 r .
Background image of page 1

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

View Full DocumentRight Arrow Icon
Overton, Mays – Homework 11 – Due: Feb 19 2005, 4:00 am – Inst: Turner 2 Frictional force f μ N = f max . Solution: The maximum force due to static friction is f max = μ N , where N is the inward directed normal force exerted by the wall of the cylinder on the person. To support the person vertically, the maximal friction force must be larger than the force of gravity mg , so that the actual force, which is equal to or less than the maximum μ N , is allowed to take on the value mg in the pos- itive vertical direction. In other words, the “ceiling” μ N on the frictional force has to be raised high enough to allow for the value mg . The normal force supplies the centripetal ac-
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 7

hw11 - Overton, Mays Homework 11 Due: Feb 19 2005, 4:00 am...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online