# hw13 - alexander(jra2623 homework 13 Turner(92510 This...

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alexander (jra2623) – homework 13 – Turner – (92510) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A car with mass 895 kg passes over a bump in a road that follows the arc of a circle of radius 41 . 6 m as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . 41 . 6 m v 895 kg What is the maximum speed the car can have as it passes the highest point of the bump before losing contact with the road? Correct answer: 20 . 2 m / s. Explanation: At the highest point, we have m g - N = m v 2 r , where N is the normal force. To get the maximum speed, we need N = 0 . Therefore, v max = g r = radicalBig (9 . 8 m / s 2 ) (41 . 6 m) = 20 . 2 m / s . 002 (part 1 of 2) 10.0 points A small sphere of mass m is connected to the end of a cord of length r and rotates in a vertical circle about a fixed point O. The tension force exerted by the cord on the sphere is denoted by T . r O θ What is the correct equation for the forces in the radial direction when the cord makes an angle θ with the vertical? 1. T + m g sin θ = + m v 2 r 2. T + m g cos θ = + m v 2 r 3. T - m g sin θ = - m v 2 r 4. None of these 5. T - m g cos θ = + m v 2 r correct 6. T - m g sin θ = + m v 2 r 7. T - m g sin θ = + m v 2 r cos θ 8. T - m g sin θ = + m v 2 r tan θ 9. T - m g sin θ = - m v 2 r tan θ Explanation: O θ θ mg T The centripetal force is F c = m v 2 r . This centripetal force is provided by the ten- sion force and the radial component of the weight. In this case, they are in opposite

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alexander (jra2623) – homework 13 – Turner – (92510) 2 direction, so F c = m v 2 r = T - m g cos θ . 003 (part 2 of 2) 10.0 points What is the magnitude of the total accel- eration? You may want to first find both the radial and the tangential component of the acceleration. 1. | vectora | = radicalBigg parenleftbigg T m parenrightbigg 2 + g 2 2. | vectora | = radicalBigg parenleftbigg T m parenrightbigg 2 cos 2 θ + g 2 sin 2 θ 3. | vectora | = radicalBigg parenleftbigg T m parenrightbigg 2 - g 2 4. | vectora | = radicalBigg parenleftbigg T m parenrightbigg 2 - 2 g cos θ T m + g 2 cor- rect 5. | vectora | = radicalBigg parenleftbigg T m parenrightbigg 2 + g 2 cos 2 θ 6. None of these 7. | vectora | = radicalBigg
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