# HW 12 - Morby Grant Homework 12 Due noon Inst Drummond This...

• Notes
• 7

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

Morby, Grant – Homework 12 – Due: Feb 20 2006, noon – Inst: Drummond 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points The coefficient of static friction between the person and the wall is 0 . 75 . The radius of the cylinder is 5 . 36 m . The acceleration of gravity is 9 . 8 m / s 2 . An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away. ω 5 . 36 m What is the minimum angular velocity ω min needed to keep the person from slip- ping downward? Correct answer: 1 . 56135 rad / s. Explanation: Let : R = 5 . 36 m and μ = 0 . 75 . Basic Concepts: Centripetal force F = m v 2 r . 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 m g , so that the actual force, which is equal to or less than the maximum μ N , is allowed to take on the value m g 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 m g . The normal force supplies the centripetal ac- celeration v 2 R on the person, so from Newton’s second law, N = mv 2 R . Since f max = μ N = μ m v 2 R m g , the minimum speed required to keep the per- son supported is at the limit of this inequality, which is μ m v 2 min R = m g , or v min = s g R μ . From this we immediately find the angular speed ω min v min R = r g μ R = s 9 . 8 m / s 2 (0 . 75) (5 . 36 m) = 1 . 56135 rad / s . 002 (part 2 of 2) 10 points Suppose the person, whose mass is m , is be- ing held up against the wall with an angular velocity of ω 0 = 2 ω min . The magnitude of the frictional force be- tween the person and the wall is 1. F = 2 m g . 2. F = 1 5 m g .

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

Morby, Grant – Homework 12 – Due: Feb 20 2006, noon – Inst: Drummond 2 3. F = m g . correct 4. F = 1 3 m g . 5. F = 4 m g . 6. F = 1 2 m g . 7. F = 5 m g . 8. F = 3 m g . 9. F = 1 4 m g . Explanation: As discussed above, f f max = μ N . Once the angular velocity has increased past the minimum angular velocity ω min required to keep the person pinned against the wall, there is no “incentive” for the force of friction to increase any more. Therefore, regardless of the maximum frictional force allowed by the angular speed, f stays at its value m g .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern