Midterm 02 - Version 039/AACBD midterm 02 Turner (58220) 1...

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Unformatted text preview: Version 039/AACBD midterm 02 Turner (58220) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A curve of radius 52 . 9 m is banked so that a car traveling with uniform speed 48 km / hr can round the curve without relying on fric- tion to keep it from slipping to its left or right. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 2 M g What is ? 1. 21.4122 2. 25.633 3. 25.0761 4. 33.1666 5. 21.8825 6. 27.4758 7. 18.928 8. 18.1615 9. 22.3949 10. 28.9469 Correct answer: 18 . 928 . Explanation: Let : m = 2200 kg , v = 48 km / hr , r = 52 . 9 m , and . Basic Concepts: Consider the free body diagram for the car. The forces acting on the car are the normal force, the force due to gravity, and possibly friction. N N N cos mg N sin x y To keep an object moving in a circle re- quires a force directed toward the center of the circle; the magnitude of the force is F c = ma c = m v 2 r . Also remember, vector F = summationdisplay i vector F i . Using the free-body diagram, we have summationdisplay i F x N sin N cos = m v 2 r (1) summationdisplay i F y N cos + N sin = mg (2) ( mg ) bardbl = mg sin (3) ma bardbl = m v 2 r cos (4) and , if = 0 , we have tan = v 2 g r (5) Solution: Solution in an Inertial Frame: Watching from the Point of View of Some- one Standing on the Ground. The car is performing circular motion with a constant speed, so its acceleration is just the centripetal acceleration, a c = v 2 r . The net force on the car is F net = ma c = m v 2 r The component of this force parallel to the incline is summationdisplay vector F bardbl = mg sin = F net cos = m v 2 r cos . In this reference frame, the car is at rest, which means that the net force on the car Version 039/AACBD midterm 02 Turner (58220) 2 (taking in consideration the centrifugal force) is zero. Thus the component of the net real force parallel to the incline is equal to the component of the centrifugal force along that direction. Now, the magnitude of the cen- trifugal force is equal to F c = m v 2 r , so F bardbl = F net cos = F c cos = m v 2 r cos F bardbl is the component of the weight of the car parallel to the incline. Thus mg sin = F bardbl = m v 2 r cos tan = v 2 g r = (48 km / hr) 2 (9 . 8 m / s 2 ) (52 . 9 m) parenleftbigg 1000 m km parenrightbigg 2 parenleftbigg hr 3600 s parenrightbigg 2 = 0 . 342922 = arctan(0 . 342922) = (0 . 330356 rad) bracketleftbigg 180 deg rad bracketrightbigg = 18 . 928 . 002 10.0 points A(n) 16 kg object, initially at rest in free space, explodes into three segments. The masses of two of these segments are both 5 kg and their velocities are 3 . 1 m / s. The an- gle between the direction of motion of these segments is 96 ....
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Midterm 02 - Version 039/AACBD midterm 02 Turner (58220) 1...

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