Test 4 - Version 093 – TEST04 – TSOI – (58160) 1 This...

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Unformatted text preview: Version 093 – TEST04 – TSOI – (58160) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The figure below shows a complex wave pat- tern on a string moving towards a rigid hook at the wall on the right. After some time, the wave is reflected from the wall. v Select the wave pattern for the reflected wave. 1. v 2. v correct 3. 4. v 5. v Explanation: Consider the wave pattern image reflected about the rigid hook on the wall. v v After the time it takes for the wave to be reflected from the wall, this image is the wave pattern traveling to the left along the string. Note: Reflection about a point (hook) is the same as reflection about the y-axis (wall) fol- lowed by reflection about the x-axis (string). The leading part of the wave must remain in front and the wave is flipped over. 002 10.0 points Hint: Consider a uniform rod with a mass m and length L pivoted on a frictionless horizontal bearing at a point O parenleftbigg 2 3 L from the lower end parenrightbigg , as shown in the figure. 2 3 L L O θ What is the period of this pendulum? Use the small angle approximation. The moment of inertia of a uniform rod about its center of mass is 1 12 M L 2 . 1. T = 2 π radicalBigg 14 15 L g 2. T = 2 π radicalBigg 14 9 L g 3. T = 2 π radicalBigg 7 12 L g 4. T = 2 π radicalBigg 2 3 L g correct Version 093 – TEST04 – TSOI – (58160) 2 5. T = 2 π radicalBigg 62 105 L g 6. T = 2 π radicalBigg 38 63 L g 7. T = 2 π radicalBigg 19 24 L g 8. T = 2 π radicalBigg 43 72 L g 9. T = 2 π radicalBigg 26 21 L g 10. T = 2 π radicalBigg 26 45 L g Explanation: Using the parallel axis theorem, the mo- mentum of inertia is I ≡ M d 2 . In this case there are two masses with I O = I cm + M D 2 , where D = parenleftbigg 2 3 − 1 2 parenrightbigg L = 1 6 L , so (1) I O = M bracketleftBigg 1 12 + parenleftbigg 1 6 parenrightbigg 2 bracketrightBigg L 2 = 1 9 M L 2 . (2) When the rod is at an angle θ with respect to the vertical direction, the lever arm due to the weight about point O is D sin θ ≈ D θ using the small angle approximation. In turn the torque due to the weight about point O is given by the equation of motion τ = Iα, so I O d 2 θ dt 2 = − mg D θ . D is given in Eq. 1 and I O is given in Eq. 2, and for simple-harmonic-motion, we have d 2 θ dt 2 = − ω 2 θ = − M g D I O θ = − 1 6 M g L 1 9 M L 2 θ = − 3 2 g L θ . (3) The expression for simple harmonic motion and Eqs. 1 and 2, yields ω 2 = 2 M g D I O , so T ≡ 2 π ω = 2 π radicalBigg I O 2 M g D = 2 π radicaltp radicalvertex radicalvertex radicalvertex radicalvertex radicalvertex radicalbt parenleftbigg 1 9 parenrightbigg M L 2 parenleftbigg 1 6 parenrightbigg M g L = 2 π radicalBigg 2 3 L g . (4) keywords: 003 10.0 points A liquid of density 1123 kg / m 3 flows with speed 1 . 24 m / s into a pipe of diameter 0 . 18 m ....
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This note was uploaded on 11/21/2010 for the course PHY 58160 taught by Professor Tsio during the Fall '10 term at University of Texas.

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Test 4 - Version 093 – TEST04 – TSOI – (58160) 1 This...

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