An object is projected from ground with speed 20 m/s at angle 30º with horizontal. Its centripetal acceleration
one second after the projection is [take g = 10 m/s2]
Answers
Answer:10 m/s^2
Explanation: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/s2 for 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, uy = u×sinα - g×t = (20×0.5) - (10×1) = 0;
Horizontal component of velocity after 1 sec, ux = 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
Answer: Centripetal force = v2/R . where v is tangential velocity and R is radius of curvature.
g is acceleration due to gravity (let us assume g = 10 m/s2 for convenience of calculation).
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, uy = u×sinα - g×t = (20×0.5) - (10×1) = 0;
Horizontal component of velocity after 1 sec, ux = 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
Explanation: