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please explain how final results are derived

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Asked by Prashant DIGHE 18th April 2020, 9:31 PM
Answered by Expert
Answer:
begin mathsize 14px style F space minus space F subscript r space equals space m space a
F space minus space fraction numerator I space alpha over denominator r end fraction space equals space m space a
sin c e space alpha space equals space a divided by r space space comma space w e space g e t space f r o m space a b o v e space e q u a t i o n space comma space
F space minus fraction numerator I space a over denominator r squared end fraction space equals space m space a space space space o r space space space space F space equals space m a space plus space fraction numerator I space a over denominator r squared end fraction space equals space open parentheses m space plus space I over r squared close parentheses space a
H e n c e space f r o m space a b o v e space e q n. comma space w e space g e t comma space space a space equals space fraction numerator F over denominator left parenthesis space I over r squared space plus space m right parenthesis end fraction space space.............. space left parenthesis 1 right parenthesis end style 
 
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begin mathsize 14px style F subscript r space equals space fraction numerator I space alpha over denominator r end fraction space equals space I over r squared a space space end style
If we substitute accelerationa from eqn.(1), then we get
 
begin mathsize 14px style F subscript r space equals space fraction numerator F space I over denominator open parentheses m r squared plus I close parentheses end fraction end style
Answered by Expert 19th April 2020, 8:46 AM
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