Lecture 13

# Lecture 13 - Lecture 13 pls-13.1 Uniform Circular Motion...

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Lecture 13 pls-13.1 Uniform Circular Motion Applications 1. Summary Clicker questions 1 . r v a c 2 = r v m F c 2 = 2. Flat Curves Clicker question 2 . Example . A car goes around a corner whose radius of curvature is 5 m. The coefficient of static friction between the tires and the road is 0.9. What is the maximum speed without slippage? (i) F.B. diagram (looking at the car from behind, if going around a left- hand corner) (ii) x,y components, resolve forces (iii) N’s second x : c MAX S ma f = y : mg F N = (iv) Algebra… N F mg MAX S f x y The centripetal force

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Lecture 13 pls-13.2 Auxiliary eqns.: N S MAX S F f µ = , r v a c 2 = . This second equation we can rearrange to solve for r a v c = 2 . Using the other equations, the following substitutions should be obvious: gr r m mg r m F r m f r a v S S N S MAX S c = = = = = 2 Æ m/s 6.6 = = gr v S . Can multiply by m/s 0.447 mi/hr 1 to convert answer to m/hr 15 . 3.
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Lecture 13 - Lecture 13 pls-13.1 Uniform Circular Motion...

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