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CBSE Class 11-science Answered

A particle is projected with a velocity u at an angle ø with horizontal.find radius of curvature of the parabola traced out by the particle at the point where velocity makes an angle ø/2 with the horizontal 
Asked by ojasmagarwal | 07 Aug, 2019, 07:15: PM
answered-by-expert Expert Answer
Vertical displacement y of projectile is given by,  y = u sinφ t - (1/2)g t2  ..................(1)
Horizontal displacement x is given by,   x = u cosφ t  ....................(2)
 
By substituting t = x/ ( u cosφ )  from eqn.(2),  eqn.(1) can be written as, 
 
y = (u sinφ) [ x/ ( u cosφ )] - (1/2)g [ x/ ( u cosφ )]2 = ( tanφ )x -  [ g/(2u2 cos2φ) ]x2  .....................(3)
 
By differentiating eqn.(1), we get,  dy/dx = tanφ - [ g/(u2 cos2φ) ]x  ............................... (4)
 
By differentiating eqn.(2), we get,  d2y/dx2 = - g/(u2 cos2φ)  ........................ (5)
 
Radius of curvature = [ 1 + (dy/dx)2 ]3/2 / ( d2y/dx2 ) ......................(6)
 
We need to find the radius of curvature at a point where velocity makes angle φ/2 with horizontal.
 
Hence  (dy/dt) / (dx/dt) = dy/dx = tan(φ/2)   ........................ (7)
 
Using eqn.(5) and eqn.(7), Radius of curvature = - [ 1 + tan2(φ/2) ]3/2 (u2 cos2φ) / g  = (u2/g) ( cos2φ ) / [ cos3(φ/2) ]
Answered by Thiyagarajan K | 07 Aug, 2019, 10:21: PM
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