Homework 05

# Homework 05 - Bautista Aldo – Homework 5 – Due Oct 4...

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Unformatted text preview: Bautista, Aldo – Homework 5 – Due: Oct 4 2005, 4:00 am – Inst: Maxim Tsoi 1 This print-out should have 25 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 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 ω . 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. r S I II III IV Ant ω It will be most difficult for the ant to adhere to the wheel as it revolves past which of the four points? 1. It will be equally difficult for the ant to adhere to the wheel at all points. 2. I 3. III correct 4. IV 5. II Explanation: The sum of the ant’s gravity and its ad- hesion force is the centripetal force ~ F c with magnitude mω 2 r . Thus ~ F c = m~g + ~ F ad ~ F ad = ~ F c- m~g . The maximum ~ F ad is when ~ F c and m~g are in opposite directions; i.e. , at III . 002 (part 2 of 2) 10 points What is the magnitude of the minimum adhe- sion force necessary for the ant to stay on the flywheel at point III ? 1. F = mω 2 r + mg correct 2. F = mω 2 r- mg 3. F = mω 2 r 2 + mg 4. F = mω 2 r 2 5. F = mg Explanation: According to the explanation in the previ- ous part, the minimum adhesion force is k ~ G k + k ~ F c k = mω 2 r + mg . 003 (part 1 of 2) 10 points The coefficient of static friction between the person and the wall is 0 . 79 . The radius of the cylinder is 6 . 61 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 6 . 61 m What is the minimum angular velocity ω min needed to keep the person from slip- ping downward? Correct answer: 1 . 36993 rad / s. Explanation: Bautista, Aldo – Homework 5 – Due: Oct 4 2005, 4:00 am – Inst: Maxim Tsoi 2 Let : R = 6 . 61 m and μ = 0 . 79 . Basic Concepts: Centripetal force F = mv 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 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 ....
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Homework 05 - Bautista Aldo – Homework 5 – Due Oct 4...

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