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Unformatted text preview: homework 05 GADHIA, TEJAS Due: Feb 21 2007, 4:00 am 1 Question 1 part 1 of 1 10 points Given: g = 9 . 8 m / s 2 . To test the performance of its tires, a car trav els along a circular track of radius R = 379 m. The track is perfectly flat (no banking). The car increases its speed at uniform rate a t d  v  dt = 2 . 07 m / s 2 until the tires start skidding. If the tires start skidding when the car reaches the speed v = 30 . 2 m / s, what is the coefficient of static friction between the tires and the road? Correct answer: 0 . 323903 (tolerance 1 %). Explanation: First, let us calculate the cars acceleration just before the tires skidded away. The car travels in a circle rather than along a straight line, hence it has the centripetal accleleration a c = v 2 R = 2 . 40644 m / s 2 . But in addition, the car increases its speed, hence we also have a tangential acceleration a t d  v  dt = 2 . 07 m / s 2 . The net acclerations of the car is the vector sum vector a = vector a c + vector a t , and since the tangential and the centripetal accelerations have perpendicular directions, the magnitude of the net accelerations is  a  = radicalBig a 2 c + a 2 t = 3 . 17425 m / s 2 . The force providing this acceleration is the static friction force vector F s between the tires and the track indeed, it is the only horizontal force available on the flat track. The static friction force may have any horizontal direc tion, and any magnitude up the the maximum of F max s = s N = s Mg. Consequenly, the net horizontal acceleration of the car is limited by M  a  s Mg =  a  s g. When the car tries to exceed this acceleration limit, the tires skid. Therefore, we may find the static friction coefficient according to s =  a  max g = 0 . 323903 . Question 2 part 1 of 3 10 points A curve of radius r is banked at angle so that a car traveling with uniform speed v can round the curve without relying on friction to keep it from slipping to its left or right. The acceleration of gravity is 9 . 8 m / s 2 . m Find the component of the net force parallel to the incline summationdisplay vector F bardbl . 1. F = mg tan 2. F = mv 2 r cos 3. F = mv 2 r sin 4. F = mg cot 5. F = mg cos 6. F = mv 2 r sin homework 05 GADHIA, TEJAS Due: Feb 21 2007, 4:00 am 2 7. F = mv 2 r tan 8. F = mv 2 r tan 9. F = mv 2 r cos correct 10. F = m radicalbigg g 2 + v 4 r 2 Explanation: Basic Concepts: To keep an object mov ing in a circle requires a force directed toward the center of the circle; the magnitude of the force is F c = ma c = mv 2 r Also remember: vector F = summationdisplay i vector F i Solution: Solution in an Inertial Frame: Watching from the Point of View of Some one Standing on the Ground....
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This note was uploaded on 03/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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