8 - test 02 – JACOBS, AMBER – Due: Mar 20 2008, 6:00 pm...

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Unformatted text preview: test 02 – JACOBS, AMBER – Due: Mar 20 2008, 6:00 pm 1 Question 1, chap 6, sect 5. part 1 of 1 10 points As viewed by a bystander, a rider in a “barrel of fun” at a carnival finds herself stuck with her back to the wall. ω Which diagram correctly shows the forces acting on her? 1. correct 2. 3. 4. 5. None of the other choices 6. Explanation: The normal force of the wall on the rider provides the centripetal acceleration neces- sary to keep her going around in a circle. The downward force of gravity is equal and oppo- site to the upward frictional force on her. Note: Since this problem states that it is viewed by a bystander, we assume that the free-body diagrams are in an inertial frame. Question 2, chap 9, sect 4. part 1 of 1 30 points What is the momentum of a 0.133 kg base- ball thrown with a velocity of 35 m / s toward home plate? 1. 4 . 356 kg · m / s 2. 4 . 512 kg · m / s 3. 4 . 655 kg · m / s correct 4. 4 . 8 kg · m / s 5. 4 . 95 kg · m / s 6. 5 . 106 kg · m / s 7. 5 . 27 kg · m / s 8. 5 . 434 kg · m / s 9. 5 . 616 kg · m / s 10. 5 . 809 kg · m / s Explanation: Let : m = 0 . 133 kg and v = 35 m / s . vectorp = mvectorv p = mv = (0 . 133 kg) (35 m / s) = 4 . 655 kg · m / s toward home plate. Question 3, chap 6, sect 5. part 1 of 1 10 points A rock on a string is whirled fast enough (in a clockwise direction) to move in a vertical circle as shown. The rock has only enough velocity at the top of the loop to keep a very small (negligible) tension in the string. test 02 – JACOBS, AMBER – Due: Mar 20 2008, 6:00 pm 2 rock R 2 2 5 ◦ ω center The angle 225 ◦ is measured from the vertical axis when the rock is at its lowest position ( i.e. , measured from the negative y axis in a counter- clockwise direction). What is the direction of the rock’s acceler- ation when θ = 225 ◦ ? 1. 2. 3. 4. 5. 6. correct 7. 8. 9. 10. None of these Explanation: Given : θ = 225 ◦ . Hint: The mass m of the rock is not re- quired and the length R of the string is not required. The free body diagram is T mg F net The tension along the string must point toward the center of the circle and the gravi- tational force is down, therefore we have summationdisplay F r : T- mg cos θ = ma r summationdisplay F θ : mg sin θ = ma θ . Thus, the free body diagram gives us the centripetal acceleration a r and the tangential acceleration a θ a r = T m- g cos θ = v 2 R and (1) a θ = g sin θ . (2) The tension in the string T is T = m v 2 R + mg cos θ > . v 2 R should always be larger than g cos θ since the rock has just enough velocity to make a full vertical circle. test 02 – JACOBS, AMBER – Due: Mar 20 2008, 6:00 pm 3 The vector sum of the radial acceleration a r and tangential acceleration a θ is shown below....
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This note was uploaded on 04/28/2008 for the course PHY 317k taught by Professor Kopp during the Spring '07 term at University of Texas.

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8 - test 02 – JACOBS, AMBER – Due: Mar 20 2008, 6:00 pm...

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