Spring07_solutions

Spring07_solutions - Qualifying Exam Spring 2007 Solutions...

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Unformatted text preview: Qualifying Exam Spring 2007 Solutions Solution 1 We use energy conservation for this problem, so that the initial potential energy E=mg(L/2) becomes kinetic energy of rotation The rotation of the pencil takes place around its tip, thus the moment of inertia is: . 2 / 2 ω I E = ∫ = = L L M dx x L M I 3 2 . 3 The velocity of the eraser is related to the angular velocity as v= ω L. From here we get gL v 3 = which can be compared with the free fall speed of . 2 gL v free = Solution 2 Solution 3 a.) c.) Solution 4 a.) b.) c.) Solution 5 Solution 6 Solution 7 Solution 8 a.) b.) c.) Solution 9 (a) Electrical resistivity: ρ , the constant of proportionality that relates R = ρ (L/A), where L is the length of a wire, A is the cross-sectional area, and R is the resistance. To measure the resistivity of a conductor, you would measure the length of the wire, the cross-sectional area, and then the resistance through V=IR or R = V/I (V and I are the voltage and current, respectively) (b) Magnetic susceptibility: χ = M/B, where M is the magnetization, and B is the applied magnetic field. The magnetic susceptibility can be measured in several different ways – the Gouy balance is one method, where you can measure the torque in a magnetic field of a material. The magnet remains stationary while the sample moves, giving an apparent weight loss or gain. sample moves, giving an apparent weight loss or gain....
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This note was uploaded on 03/11/2012 for the course PHY 3900 taught by Professor Staff during the Fall '1 term at FSU.

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Spring07_solutions - Qualifying Exam Spring 2007 Solutions...

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