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mat-sln-asn-pt5-tst3-sum05

mat-sln-asn-pt5-tst3-sum05 - Solution for Practice Test 5...

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Solution for Practice Test 5 for Test 3 Solution to Practice Test Problem 5.1() Problem: Score for multiple choice Solution Total Points for Problem: 30 Points Solution to Practice Test Problem 5.2() Problem: Let’s think back to some of the data analysis we have done in lab. (a)What is the difference between systematic and random error? You don’t have to get too technical, but you have to make it clear you understand. (b)Why do we like to plot things in a straight line relationship and get our answer from the slope when we can? Solution to Part (a) A systematic error always affects the data the same way, like having a slightly chipped ruler, or having friction in the system. A random error can fluctuate, so someitmes you err a little high, and sometimes a little low, like running a timer, or using any other measuring device. Grading Key: Part (a) 1 Points Solution to Part (b) It is a graphical way to use the brain to calculate the weighted error. Grading Key: Part (b) 1 Points Total Points for Problem: 2 Points Solution to Practice Test Problem 5.3() Problem: A string is wound around a uniform disk of radius R and mass M . The disk is released from rest with the string vertical and its top end tied to a fixed bar. (a)Draw the free-body diagram and find the tension ( T = F c s,d ) in the string. Show all work. (b)Find the magnitude of the acceleration of the center of mass. (c)Find the speed of the center of mass is (4 gh/ 3) 1 / 2 after the disk has descended through distance h . Show all work, do not just state the answer. R h Solution to Part (a)
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We can use Newton II for forces and torques, as well as the condition of rolling motion for acceleration, a = . ma = Σ F = T mg a = T m g = Σ τ = RT The last negative sign comes from the fact that upward is positive, and counter- clockwise rotations are positive. parenleftbigg 1 2 mR 2 parenrightbigg parenleftBig a R parenrightBig = 1 2 maR = RT T = 1 2 ma = 1 2 m parenleftbigg T m g parenrightbigg T = 1 3 mg T F g a Grading Key: Part (a) 4 Points 1 point(s) : Newton II forces. 1 point(s) : Newton II torques 1 point(s) : Condition of rolling motion for acceleration. 1 point(s) : Correct Answer. Solution to Part (b) | a | = 2 T m = 2 3 g Grading Key: Part (b) 2 Points 2 point(s) : Correct answer (1pt IBC). Solution to Part (c) Conserve energy, noting condition of rolling motion for velocity, mgh = 1 2 mv 2 f + 1 2 2 f mgh = 1 2 mv 2 f + 1 2 parenleftbigg 1 2 mR 2 parenrightbigg parenleftBig v f R parenrightBig 2 = 1 2 mv 2 f + 1 4 mv 2 f mgh = 3 4 mv 2 f v 2 f = 4 3 gh v f = radicalbigg 4 gh 3
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Grading Key: Part (c) 2 Points 1 point(s) : Conserve energy correctly.
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