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Unformatted text preview: of a centripetal force, and object in circular motion would resume its original path, tangent to its point on the circle. Calculations 1. • N 1 = number of revolutions/t 1 N 1 = 30/16.049 N 1 =1.87 •N 2 = number of revolutions/t 2 N 2 = 30/13.903 • N = ( N 1 + N 2 )/2 N= ( 1.87 +2.16 )/2 N= 2.015 • ∆ N= abs(( N 1N 2 )/2) ∆ N= abs ((1.872.16)/2) ∆ N=.145 2. • ∆ (M/L)=(M/L)( ∆ L/L) ∆ (M/L)=(150/.5)(.0025/.5) ∆ (M/L)=1.5 • ∆( N 2 )= N2∆N ∆( N 2 )=(2.015)(2(.145)) 3....
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 Spring '08
 Caffey
 Acceleration, Circular Motion

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