This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
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....
View Full Document