h6circularmotion_solution - honea (cth632) H6CircularMotion...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: honea (cth632) H6CircularMotion Avram (1955) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A ball rolls around a circular wall, as shown in the figure below. The wall ends at point X . A B C D E X When the ball gets to X , which path does the ball follow? 1. Path B 2. Path D 3. Path A 4. Path C correct 5. Path E Explanation: As soon as the ball reaches point X the centripetal force is removed, so the ball moves in a straight line (tangent to the circle at oint X). Path C is the correct answer. 002 (part 1 of 3) 10.0 points An air puck of mass 0 . 206 kg is tied to a string and allowed to revolve in a circle of radius 1 . 5 m on a horizontal, frictionless table. The other end of the string passes through a hole in the center of the table and a mass of 0 . 45 kg is tied to it. The suspended mass remains in equilibrium while the puck revolves. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 5 m v . 45 kg . 206 kg What is the tension in the string? Correct answer: 4 . 41 N. Explanation: r v Mg m Since the suspended mass is in equilibrium, the tension is T = M g = (0 . 45 kg) ( 9 . 8 m / s 2 ) = 4 . 41 N . 003 (part 2 of 3) 10.0 points What is the horizontal force acting on the puck? Correct answer: 4 . 41 N. Explanation: The horizontal force acting on the puck is the tension in the string, so F c = T = 4 . 41 N . 004 (part 3 of 3) 10.0 points What is the speed of the puck? Correct answer: 5 . 66671 m / s. Explanation: F c = mv 2 r honea (cth632) H6CircularMotion Avram (1955) 2 v = radicalbigg F c r m = radicalBigg (4 . 41 N) (1 . 5 m) . 206 kg = 5 . 66671 m / s . 005 (part 1 of 2) 10.0 points 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 (see figure). The coefficient of static friction between the person and the wall is and the radius of the cylinder is R . R What is the minimum tangential velocity needed to keep the person from slipping down- ward? 1. v = radicalbig 2 g R 2. v = radicalbig g R 3. v = 2 radicalbig g R 4. v = 1 radicalbig g R 5. v = radicalbig g R 6. v = radicalbig 2 g R 7. v = 2 radicalbig g R 8. v = radicalBigg g R correct 9. v = 1 2 radicalbig g R 10. v = radicalbig 2 g R Explanation: Basic Concepts: Centripetal force: F = mv 2 r Frictional force: f s N = f max s Solution: The maximum frictional force due to friction is f max = N , where N is the inward directed normal force of the wall of the cylinder on the person. To support the person vertically, this maximal friction force f max s must be larger than the force of gravity mg so that the actual force, which is less than N , can take on the value mg in the positive vertical direction. Now, the normal force supplies the centripetal acceleration v 2 R on the person, so from Newtons second law,...
View Full Document

This note was uploaded on 04/08/2011 for the course PHYS 1402 taught by Professor Avram during the Spring '10 term at Austin Community College.

Page1 / 8

h6circularmotion_solution - honea (cth632) H6CircularMotion...

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