CBSE Class 11-science Answered

An object is projected from ground with the speed of 20m/s at an angle 30° with horizontal. It's centripetal acceleration 1 second after projection is ?

Asked by arunavamitra50 | 09 Jul, 2018, 06:03: PM

Expert Answer

Centripetal force = v²/R . where v is tangential velocity and R is radius of curvature.

Projectile equation is given by,

begin mathsize 12px style y space equals space left parenthesis tan space alpha right parenthesis space x space minus space fraction numerator g space x squared over denominator 2 u squared cos squared alpha end fraction end style

.................... (1)

where α is angle of projection, u is velocity of projection,

g is acceleration due to gravity (let us assume g = 10 m/s²for convenience of calculation).

x any y are position coordinate by taking the point of projection as origin.

after substituting the given values of α, u and g, we rewrite the eqn(1) as

begin mathsize 12px style y space equals space fraction numerator x over denominator square root of 3 end fraction space minus space x squared over 60 end style

.........................(2)

Radius of curvature R

begin mathsize 12px style equals fraction numerator open square brackets 1 plus open parentheses begin display style fraction numerator d y over denominator d x end fraction end style close parentheses squared close square brackets to the power of begin display style bevelled 3 over 2 end style end exponent over denominator open vertical bar begin display style fraction numerator d squared y over denominator d x squared end fraction end style close vertical bar end fraction end style

....................(3)

from eqn.(2), we get derivatives as,

begin mathsize 12px style fraction numerator d y over denominator d x end fraction space equals space fraction numerator 1 over denominator square root of 3 end fraction space minus space x over 30 space semicolon space space fraction numerator d squared y over denominator d x squared end fraction space equals space minus 1 over 30 end style

...............(4)

after 1 sec, x = u cosα × t = 20×(√3/2)×1 = 10×√3 ;

if we substitute x = 10×√3 in the expression for (dy/dx), we get (dy/dx) = 0; Hence from (3) and (4), radius of curvature = 30 m.

vertical component of velocity after 1 sec, u_y = u×sinα - g×t = (20×0.5) - (10×1) = 0;

Horizontal component of velocity after 1 sec, u_x= u×cosα×t = (20×√3)/2 = 10×√3 m/s;

resultant velocity v = 10×√3 m/s ;

centripetal acceleration = v² / R = 300/30 = 10 m/s²

Answered by Thiyagarajan K | 10 Jul, 2018, 10:19: AM

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