hw1sol - HW 1 Physics 1B Spring 2010 C M Cooper Teaching...

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HW 1: Physics 1B Spring 2010 C. M. Cooper, Teaching Assitant W. Gekelman, Professor April 7, 2010 Abstract This hw covers all of chapter 13 in the text except section 13.6. Topics include Simple Harmonic Motion and Damped-Driven Harmonic motion. Two good problems to memorize are problems 71 and 82 Note that problems 55, 71, and 93 of these solutions involve the physical pendulum (section 13.6) which was not covered in section but consists of a simple one line formula f = ω 2 π = 1 2 π r mgd I (1) A good way to remember this formula is to consider that I used to drink a lot of Miller Genuine Draft playing kings as a freshman and although I’d like to think I’m past that point in my life (I am over MGD), in actuality its the opposite (MGD is over I). Also, remember whenever we see a moment of inertia, we have to remember 3 things: 1) Use the table on page 299 for regular bodies, 2) the parallel axis theorum, and 3) how to integrate and find the moment of inertia of a solid body. The parallel axis theorem is used when you are rotating something with a standard moment of inertia a distance away from you (on the end of a stick or string) you need to add an additional mL 2 where L is the distance to the center of mass and in many cases needs to be calculated. We do not use integration to find the moment of inertia of anything. .. YET! 1 Question 13.15: Weighing an Astronaut As mentioned in the intro, basic one and two step problems like this only require you to have the right equation as the first line then substitute in. These kinds of problems are easy to ace. We start with the equation ω 2 = k M (2) In the second case, M = M C + M A , ω = ω C + A = 2 π/T C + A , but k and M C are both unknown. In order to solve a problem, you must have at least as many unknowns as equations! We have two unknowns so we need a second equation. From the first case, have M = M C , ω = ω C = 2 π/T C , and only k is unknown, we solve for k in the first case to get k = M C ω 2 C (3) 1
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We rearrange the second case and sub in the first case to get M A = k ω 2 C + A - M C = M C ω 2 C ω 2 C + A - M C = M C ( T 2 C + A T 2 C - 1) = 120 .kg (4) 2 Question 13.33: Two masses on a spring 2.1 (a) Why does the glue fail at the lowest part?
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hw1sol - HW 1 Physics 1B Spring 2010 C M Cooper Teaching...

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