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Class notes-- 5.3.11 Using PE, KE and W equations Whiteboard Quiz tomorrow: on Equations Test next week TWINS DAY LABS DUE THURSDAY Work and Power Lab If not there use ________38_______ steps Each Step = 19 cm high Calories of Big Dog *(with bun) = 590 Cal.

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As a kid growing up in Indiana, I always strove to find the perfect fit in bats for me, one that had a sweet spot as big as Indianapolis. I never knew the exact location of that spot other than it was somewhere near the bat barrel's thickness, but I knew it when I hit the pitched ball there. In experiments conducted by Dr. Daniel A. Russell of Kettering University, he examined the sweet spot in terms of two separate criteria: * the location where the measured performance of the bat is maximized, and * the location where the hand sensation, or sting, is minimized. He had numerous opinions on where this location was on a bat. A list follows: * the location producing least vibrational sensation, or sting, in the batter's hands * the location which produces maximum batted ball speed * the location where maximum energy is transferred to the ball * the location where coefficient of restitution is maximum * the center of percussion * the node of the fundamental vibrational mode * the region between nodes of the first two vibrational modes * the region between center of percussion and node of first vibrational mode Dr. Russell explained that most bats have "sweet spots" in different locations, thereby leading to his conclusion that the sweet spot is more a region than a spot, generally located approximately 5- 7 inches from the end of the barrel where the batted-ball speed is the highest and the sensation in the hands is minimized. How to Find the Sweet Spot on the Bat his conclusion that the sweet spot is more a region than a spot

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About the Sweet Spot on the Bat

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Mechanical Energy Height look at PE Speed look at KE Work change in PE or KE Mass in kg & Force in N

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5.3.11 Reference Reading –Vocabulary HW--1 sheet –Mechanical Energy WPE Work--

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Question 3 on Mechanical Energy Find the Work W =F g h W= 630 N(5 m) W = 3150 J

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Question 3 on Mechanical Energy W = 3150 J Increase in PE? (PE=0 at bottom) F g =m a g -630 N =m(-9.8) m=64.3 kg

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Question 3 on Mechanical Energy m=64.3 kg PE =mgh =(64.3kg )(9.8)(5m) PE =3150.7 J = 3.15 kJ Or W = ΔPE = 3150 J PE came from the WORK done by the person

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Question 4. SPEED What is KE at the original speed? KE = ½ mv 2 KE = ½ ( 750 kg)(13.9 m/s) 2 KE = ½ (750 kg)(193.21m 2 /s 2 ) KE = 72,453.75 J = 72.45 kJ

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Question 4. SPEED KE for Double speed ? V = 27.8 m/s KE = ½ mv 2 KE = ½ ( 750 kg)(27.8 m/s) 2 KE = ½ (750 kg)(772.84m 2 /s 2 ) KE = 289,815 J = 289 kJ

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Question 4. SPEED Amount of work to double speed? –KE = 72,454 J –KE = 289,815 J –Work = ΔKE W= KE f – KE i W = 289,815 J - 72,454J W = 217,000 J

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