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CBSE Class 11-science Answered

a bead of mass M is constrained along elliptical wireframe in the horizontal plane the equation of the article frame is given by x is equal to 2 x cos theta and Y is equal to a sin theta the work done by a force f is equal to K Poorna square are cap is moving The Beat from point a to point B is alpha ke upon to a the value of Alpha is
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Asked by padmayera | 31 Aug, 2020, 04:54: PM
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Workdone W is given as
 
begin mathsize 14px style W space equals space integral subscript theta equals pi over 2 end subscript superscript theta equals 0 end superscript F with rightwards arrow on top times stack d r with rightwards arrow on top end style .................. (1)
let P (x,y) be a point on the path of bead and point O be origin . Line OP makes angle θ with x-axis . 
This angle θ is given as limits of integration in eqn.(1)
 
Where begin mathsize 14px style F with rightwards arrow on top space times space stack d r with rightwards arrow on top space equals space k over r squared r with hat on top space times stack d r space with rightwards arrow on top space end style...................(2)
let us consider the following substitutions
 
r2 = x2 + y2   ;    begin mathsize 14px style r with hat on top space equals space cos theta space i with hat on top space plus space sin theta space j with hat on top space end style ;  begin mathsize 14px style stack d r with rightwards arrow on top space equals space d x space i with hat on top space plus space d y space j with hat on top space end style
using above substitutions , eqn.(2) is rewritten after simplification as
 
begin mathsize 14px style F with rightwards arrow on top space times space stack d r with rightwards arrow on top space equals space k over a open parentheses fraction numerator sin theta space cos theta space d theta over denominator 4 space minus space 3 space sin squared theta end fraction close parentheses space end style................(3)
 
begin mathsize 14px style W space equals space integral subscript pi divided by 2 end subscript superscript 0 space k over a open parentheses fraction numerator sin theta space cos theta space d theta over denominator 4 space minus space 3 space sin squared theta end fraction close parentheses space end style....................... (4)
Let u = 3 sin2 θ  ; du = 6 sinθ cosθ dθ ; Using these substitution eqn.(4) becomes
 
begin mathsize 14px style W space equals space fraction numerator k over denominator 6 a end fraction integral subscript 3 superscript 0 fraction numerator d u over denominator 4 minus u end fraction space equals space minus space fraction numerator k over denominator 6 a end fraction ln left parenthesis 4 right parenthesis end style  ...................(5)
if W = (α k ) / ( 2 a ) , then by comparing with eqn.(5) , we get α = - ln4 / 3 = -0.462
Answered by Thiyagarajan K | 14 Oct, 2020, 08:05: PM
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