# HW14 - homework 14 PAPAGEORGE MATT Due 4:00 am Question 1...

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homework 14 – PAPAGEORGE, MATT – Due: Feb 21 2008, 4:00 am 1 Question 1, chap 6, sect 99. part 1 of 1 10 points A car with mass 924 kg passes over a bump in a road that follows the arc of a circle of radius 42 . 4 m as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . 42 . 4 m v 924 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 . 3843 m / s (tolerance ± 1 %). 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 ) (42 . 4 m) = 20 . 3843 m / s . Question 2, chap 6, sect 99. part 1 of 2 10 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 sin θ = + m v 2 r tan θ 3. T - m g sin θ = - m v 2 r 4. T - m g sin θ = - m v 2 r tan θ 5. T + m g cos θ = + m v 2 r 6. T - m g sin θ = + m v 2 r cos θ 7. T - m g cos θ = + m v 2 r correct 8. T + m g sin θ = + m v 2 r 9. None of these 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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homework 14 – PAPAGEORGE, MATT – Due: Feb 21 2008, 4:00 am 2 direction, so F c = m v 2 r = T - m g cos θ . Question 3, chap 6, sect 99. part 2 of 2 10 points Hint: To solve this part, first find both the radial and the tangential component of the acceleration. The magnitude of a , the total acceleration, is 1. None of these 2. | vectora | = radicalBigg parenleftbigg T m parenrightbigg 2 sin 2 θ + g 2 cos 2 θ . 3. | vectora | = radicalBigg parenleftbigg T m parenrightbigg 2 + g 2 sin 2 θ . 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 + 2 g cos θ T m + g 2 . 6. | vectora | = radicalBigg parenleftbigg T m parenrightbigg 2 cos 2 θ + g 2 sin 2 θ . 7. | vectora | = radicalBigg parenleftbigg T m parenrightbigg 2 + g 2 . 8. | vectora | = radicalBigg parenleftbigg T m parenrightbigg 2 + g 2 cos 2 θ .
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