PHYS_2014_Lecture_30

# PHYS_2014_Lecture_30 - Fall 2010 Oklahoma State University...

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Unformatted text preview: Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 30, Slide 1 Lecture 30: Review for Final Exam Final Exam on Tuesday, 14 December 2010, 12:00 - 1:50 pm The class will be divided into three groups based on their recitation TA. Each group will take the exam in a different room as shown in the table below. Based on your recitation section, please go directly to your designated room. Recitation Section TA Exam Room 1, 2, 3, 19, 20, 21, 22, 23, 24 Ayon Patra PS-141 4, 5, 6, 7, 8, 9, 16, 17, 18, Ben Grossman PS-103 10, 11, 12, 13, 14, 15, 25, 26, 27, 28, 29, 30 Razvan Stoian PS-110 Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 30, Slide 2 Regarding the Final Exam: • Worth (at least) 30% of your final grade. • 6-8 homework type problems. • Otherwise same format as previous exams. • Generally speaking if we covered it in a homework problem, sample example, actual exam and/or in a lecture example, then it may be on the final exam. Review for Final Exam Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 30, Slide 3 20 s = 4 c m A wheel ( m = 0.7 kg, r = 0.75 cm) is attached to a spring with a spring constant of k = 120 N/m. The other end of the spring is attached to a post at the base of a 20 ° ramp as shown above. Assume that the wheel is a solid disk and that there is no slipping when it rolls up and down the ramp. If the wheel is pulled up the ramp such that the spring is stretched a distance Δ s = 4 cm, what is the maximum speed of the wheel as it rolls down the ramp? Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 30, Slide 4 Solution: We solve this problem using Conservation of Energy. Why? Because this approach is the only one that we’ve used that works for springs (since a is not constant). 2 2 2 1 1 1 2 2 2 gi si Tf Rf U U K K mgh k s mv I ω + = + + Δ = + 20 s = 4 c m h The height the wheel will have descended when it reaches its maximum velocity will be: Also: and sin h s θ = Δ 2 1 2 I mr = v r ω = Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 30, Slide 5 Substitute all this into the conservation of energy equation and multiply both sides by 2. 2 2 2 1 2 sin 2 mg s k s mv m r θ Δ + Δ = + 2 2 v r ( ) ( ) ( ) 2 2 2 2 3 2 2 2 sin 3 N m 2 2 0.7kg 9.80 0.04m sin 20 120 0.04m m s m 0.60 s 3 0.7kg mv mg s k s v m v θ = Δ + Δ = ⋅ ⋅ ⋅ ⋅ + = = ⋅ r 2 2 2 1 1 1 2 2 2 mgh k s mv I ω + Δ = + sin h s θ = Δ 2 1 2 I mr = v r ω = 20 s = 4 c m h Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 30, Slide 6 A steel ball having a mass 31.4 g is moving at 2 m/s to the right. A lead ball having a mass of 41.9 g is moving at 3 m/s to the left. The two balls elastically collide head-on. Find the velocity of the each ball following the collision....
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PHYS_2014_Lecture_30 - Fall 2010 Oklahoma State University...

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