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A-particle-is-moving-in-parabolic-path-x-2-y-with-constant-speed-u-Find-the-acceleration-of-the-particle-when-it-crossess-origin-Also-find-the-radius-of-curvature-at-origin-




Question Number 16499 by Tinkutara last updated on 23/Jun/17
A particle is moving in parabolic path  x^2  = y, with constant speed u. Find the  acceleration of the particle when it  crossess origin. Also find the radius of  curvature at origin.
Aparticleismovinginparabolicpathx2=y,withconstantspeedu.Findtheaccelerationoftheparticlewhenitcrossessorigin.Alsofindtheradiusofcurvatureatorigin.
Answered by ajfour last updated on 23/Jun/17
x^2 =y   ⇒  (dy/dx)=2x    which is zero   at the origin. And (d^2 y/dx^2 ) = 2  radius of curvature,         r = ∣(([1+(dy/dx)^2 ]^(3/2) )/(d^2 y/dx^2 ))∣=(1/2) .               x^2 =y   2x(dx/dt)=(dy/dt) =v_y   (=0 at origin) ..(i)             u^2 =((dx/dt))^2 +((dy/dt))^2       ....(ii)  ⇒   ((dx/dt))^2 =(u^2 /(1+4x^2 ))    ....(iii)     so ((dx/dt))^2  = u^2   at the origin          2((dx/dt))^2 +2x(d^2 x/dt^2 )=(d^2 y/dt^2 )  .....(iv)  means at origin, (d^2 y/dt^2 )=a_y =2u^2                                                     ......(v)  ⇒   ((dx/dt))^2 =(u^2 /(1+4x^2 ))      2((dx/dt))((d^2 x/dt^2 ))=− ((8u^2 x)/((1+4x^2 )^2 ))((dx/dt))  a_x = (d^2 x/dt^2 ) = −((4u^2 x)/((1+4x^2 )^2 ))  , this is zero   at the origin.  acceleration a=(d^2 x/dt^2 )+(d^2 y/dt^2 )   at origin a = 0 +2u^2      [see (v)].
x2=ydydx=2xwhichiszeroattheorigin.Andd2ydx2=2radiusofcurvature,r=[1+(dy/dx)2]3/2d2y/dx2∣=12.x2=y2xdxdt=dydt=vy(=0atorigin)..(i)u2=(dxdt)2+(dydt)2.(ii)(dxdt)2=u21+4x2.(iii)so(dxdt)2=u2attheorigin2(dxdt)2+2xd2xdt2=d2ydt2..(iv)meansatorigin,d2ydt2=ay=2u2(v)(dxdt)2=u21+4x22(dxdt)(d2xdt2)=8u2x(1+4x2)2(dxdt)ax=d2xdt2=4u2x(1+4x2)2,thisiszeroattheorigin.accelerationa=d2xdt2+d2ydt2atorigina=0+2u2[see(v)].
Commented by Tinkutara last updated on 23/Jun/17
Thanks Sir!
ThanksSir!

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