Homework 11 - homework 11 BAUTISTA, ALDO Due: Feb 15 2006,...

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Unformatted text preview: homework 11 BAUTISTA, ALDO Due: Feb 15 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 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 6 . 54 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. 6 . 54 m What is the minimum angular velocity min needed to keep the person from slip- ping downward? Correct answer: 1 . 41349 rad / s (tolerance 1 %). Explanation: Let : R = 6 . 54 m and = 0 . 75 . Basic Concepts: Centripetal force F = m v 2 r . Frictional force f N = f max . Solution: The maximum force due to static friction is f max = 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 N , is allowed to take on the value m g in the pos- itive vertical direction. In other words, the ceiling 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 Newtons second law, N = mv 2 R . Since f max = N = 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 m v 2 min R = m g , or v min = s g R . From this we immediately find the angular speed min v min R = r g R = s 9 . 8 m / s 2 (0 . 75) (6 . 54 m) = 1 . 41349 rad / s . Question 2 Part 2 of 2. 10 points. Suppose the person, whose mass is m , is being held up against the wall with an angular velocity of = 2 min . The magnitude of the frictional force be- tween the person and the wall is 1. F = 3 m g . 2. F = 2 m g . homework 11 BAUTISTA, ALDO Due: Feb 15 2006, 4:00 am 2 3. F = 1 5 m g . 4. F = 1 3 m g . 5. F = 5 m g . 6. F = 1 4 m g . 7. F = 1 2 m g . 8. F = m g . correct 9. F = 4 m g . Explanation: As discussed above, f f max = N . Once the angular velocity has increased past the minimum angular velocity 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 . Question 3 Part 1 of 2. 10 points. An ant of mass m clings to the rim of a flywheel of radius r , as shown. The flywheel rotates clockwise on a horizontal shaft S with constant angular velocity . As the wheel rotates, the ant revolves past the stationary points I , II , III , and IV . The ant can adhere to the wheel with a force much greater than its own weight....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Homework 11 - homework 11 BAUTISTA, ALDO Due: Feb 15 2006,...

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