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Unformatted text preview: pampalone (tp8327) – HW#7 – swaminathan – (10500) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A curve of radius r is banked at angle θ 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. m μ ≈ θ Find the component of the net force on the car parallel to the incline summationdisplay vector F bardbl . 1. F = m v 2 r cos θ 2. F = m v 2 r sin 2 θ cos θ 3. F = m v 2 r cot θ 4. F = m v 2 r tan θ 5. F = m v 2 r tan θ 6. F = m v 2 r sin θ 7. F = m v 2 r sin θ 8. F = m v 2 r sin θ cos θ 9. F = m v 2 r cos θ correct 10. F = m v 2 r sin 2 θ cos 2 θ 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 = m a c = m v 2 r Also remember: vector F = summationdisplay i vector F i Solution: In an Inertial Frame: The car is performing circular motion with a constant speed, thus its acceleration is just the cen tripetal acceleration, a c = v 2 r . The net force on the car is F net = m a c = m v 2 r . Watching from the “Point of View of Some one Standing on the Ground”. m g F net F bardbl N r y pampalone (tp8327) – HW#7 – swaminathan – (10500) 2 The component of this force parallel to the incline is summationdisplay vector F bardbl = m g sin θ = F net cos θ = m v 2 r cos θ . 002 (part 2 of 2) 10.0 points If r = 56 m and v = 145 km / hr, what is θ ? The acceleration due to gravity is 9 . 8 m / s 2 . Correct answer: 71 . 3101 ◦ . Explanation: F bardbl is the component of the weight of the car parallel to the incline, so m g sin θ = F bardbl = m v 2 r cos θ tan θ = v 2 g r = (145 km / hr) 2 (9 . 8 m / s 2 )(56 m ) × parenleftbigg 1000 m km parenrightbigg 2 parenleftbigg hr 3600 s parenrightbigg 2 = 2 . 95609 θ = arctan(2 . 95609) = 71 . 3101 ◦ . 003 10.0 points A 64 kg child sits in a swing supported by two chains, each 1 . 4 m long. The acceleration of gravity is 9 . 8 m / s 2 . If the tension in each chain at the lowest point is 408 N, find the child’s speed at the lowest point. Note: Neglect the mass of the seat. Correct answer: 2 . 03224 m / s. Explanation: Let : m = 64 kg , ℓ = 1 . 4 m , t = 408 N , and g = 9 . 8 m / s 2 . At the lowest point, Newton’s second law states summationdisplay F y = 2 T m g m v 2 ℓ = 0 . Solving for v , v = radicalbigg ℓ m (2 T m g ) = braceleftbigg 1 . 4 m 64 kg [2 (408 N) (64 kg) ( 9 . 8 m / s 2 )bracketrightbig bracerightbigg 1 / 2 = 2 . 03224 m / s . 004 (part 1 of 3) 10.0 points A race car starts from rest on a circular track of radius 398 m. The car’s speed increases at the constant rate of 0 . 368 m / s 2 ....
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This note was uploaded on 04/28/2011 for the course PHYS 105 taught by Professor Ken during the Spring '08 term at NJIT.
 Spring '08
 KEN

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