Final Exam Solutions

Final Exam Solutions - Physics 40B Final Exam March 17,...

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Physics 40B Final Exam Name_________ Solutions________ March 17, 2008 Signature______________________ ID # ______XXX-XX-__________ Instructions: Show all work on test papers. You may use the back of the page if necessary. You may tear off the formula pages if you want. This is a Closed Notes- Closed Book exam. You may use any calculator. Score (200 points max) 1. 2. 3. 4. 5. 6. _______________
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Problem 1 (30 points): A rectangular dam 30 m wide ( L ) supports a body of water to a depth of 25 m ( H ). Part A (25 points): Find the total horizontal force on the dam due to the water. Hint: first consider the force on a narrow horizontal strip of height dh . Consider the fluid force dF on strip with area dA = Ldh . dF = p o + " gh ( ) dA = p o + gh ( ) Ldh The total force is F = dF = p o + gh ( ) Ldh 0 H # h = 0 h = H # $ F = p o LH + 1 2 gLH 2 (Here, we used hdh = 1 2 # h 2 evaluted at H and 0) $ F = (1.01 x 10 5 Pa )(30 m)(25 m ) + 1 2 1000 kg / m 3 ( ) (9.8 m / s 2 )(30 m )(25 m ) 2 F = 7.57 x 10 7 N + 9.19 x 10 7 N = 1.68 x 10 8 N However, it should be noted (see Part B) that there is an equal force on the right side of the dam due to atmospheric pressure so the net force really is F = 9.19 x 10 7 N No points will be taken off for either answer. Part B (5 points): What is the net effect of the atmospheric air pressure on this total force? Give a qualitative answer. The net effect of atmospheric pressure is to cancel out because it exerts the same horizontal force on both sides of the dam. L H h dh
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Problem 2 (40 points): A uniform rod of mass M and length L is free to swing about a horizontal axis that is a distance x from the center of the rod. The moment of inertia of the rod about an axis of rotation a distance of x from the center of the rod is Part A (20 points): Find the expression for the period, T , of oscillation for small angular displacements of the rod in term of M, L, g and x. Using the formula for the period of a physical pendulum T = 2 " I mgd = 2 1 12 ML 2 + Mx 2 ( ) Mgx = 2 1 12 L 2 + x 2 ( ) gx Part B (20 points): It can be shown that for x 0 and x the period, T , becomes infinite. Find the value of x in terms of M and L for which the period is a minimum value. To find the minimum value of the period, T , set dT dx = 0 and solve for x . dT dx = d 2 1 12 L 2 + x 2 gx # $ % % & ( ( dx = d 2 1 12 L 2 + x 2 gx # $ % & ( 1 2 # $ % % & ( ( dx = 2 1 2 1 12 L 2 + x 2 ( ) gx # $ % % & ( ( ) 1 2 2 x gx ) 1 12 L 2 + x 2 ( ) gx 2 # $ % % & ( ( # $ % % & ( ( = 0 Simplifying: g gx 1 12 L 2 + x 2 ( ) # $ % % & ( ( 1 2 2 ) L 2 12 x 2 ) 1 # $ % & ( = 0 This equation is satisfied if the second term in brackets is zero. * 2 ) L 2 12 x 2 ) 1 # $ % & ( = 0 + x 2 = 1 12 L 2 + x = 1 12 L = 0.289 L I = 1 12 ML 2 + Mx 2 L L/2 x axis center
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Problem 3 (30 points): An asteroid crashing into the ocean generates a tsunami. When the tsunami wave strikes an island it does a lot of damage. After the wave passed, an operating tape recorder was found in a tree. The tape recorder apparently recorded the blasts from the warning siren. Also on the tape was a faint echo of the siren in between the blasts of the stationary warning siren. This was interpreted as a reflection from the moving wave. The frequency of the warning siren was 4000 Hz and the frequency of
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This note was uploaded on 06/18/2008 for the course PHYS 40B taught by Professor Johnellison during the Spring '08 term at UC Riverside.

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Final Exam Solutions - Physics 40B Final Exam March 17,...

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