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Unformatted text preview: homework 14 – YOO, HEE – Due: Feb 21 2008, 4:00 am 1 Question 1, chap 6, sect 99. part 1 of 1 10 points A car with mass 828 kg passes over a bump in a road that follows the arc of a circle of radius 45 . 8 m as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . 4 5 . 8 m v 828 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: 21 . 1858 m / s (tolerance ± 1 %). Explanation: At the highest point, we have mg N = mv 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 ) (45 . 8 m) = 21 . 1858 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 mg cos θ = + mv 2 r correct 2. T mg sin θ = + mv 2 r 3. T mg sin θ = + mv 2 r cos θ 4. None of these 5. T + mg cos θ = + mv 2 r 6. T + mg sin θ = + mv 2 r 7. T mg sin θ = + mv 2 r tan θ 8. T mg sin θ = mv 2 r 9. T mg sin θ = mv 2 r tan θ Explanation: O θ θ mg T The centripetal force is F c = mv 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 homework 14 – YOO, HEE – Due: Feb 21 2008, 4:00 am 2 direction, so F c = mv 2 r = T mg 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.  vectora  = radicalBigg parenleftbigg T m parenrightbigg 2 cos 2 θ + g 2 sin 2 θ . 2.  vectora  = radicalBigg parenleftbigg T m parenrightbigg 2 2 g cos θ T m + g 2 . cor rect 3.  vectora  = radicalBigg parenleftbigg T m parenrightbigg 2 + g 2 cos 2 θ . 4. None of these 5.  vectora  = radicalBigg parenleftbigg T m parenrightbigg 2 + g 2 . 6.  vectora  = radicalBigg parenleftbigg T m parenrightbigg 2 sin 2 θ + g 2 cos 2 θ . 7.  vectora  = radicalBigg parenleftbigg T m parenrightbigg 2 + 2 g cos θ T m + g 2 ....
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 Spring '08
 Turner
 Acceleration, Force, Mass, Work, Correct Answer, 184 N force

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