A projectile is thrown with the speed 'u' at an angle '⍬' to the horizontal. What is the radius of curvature of it's trajectory when the velocity vector of the projectile makes an angle '⍺' with the horizontal?

2 Answers
Narad T.
Jun 3, 2018

Please see the explanation below

Explanation:

The trajectory of the projectile in the #x-y# plane is

#y=xtantheta-(gx^2)/(2u^2cos^2theta)#

At any point #P# on the trajectory,

The horizontal component of the velocity is

#v_x=ucostheta#

#=>#, #ucostheta=xt#

#=>#, #t=x/(ucostheta)#

and the vertical component is

#v_y=usintheta- g t#

#=>#, #v_y=usintheta-(gx)/(ucostheta)#

The resultant velocity is

#v_r=sqrt(v_x^2+v_y^2)#

#=sqrt(u^2cos^2theta+u^2sin^2theta+(g^2x^2)/(u^2cos^2theta)-(2usinthetagx)/(ucostheta))|#

#=sqrt(u^2+(g^2x^2sec^2theta)/(u^2)-2gxtantheta)|#

The angle #alpha# is defined as follows

#cosalpha=v_x/v_r=(ucostheta)/v_r#

The centripetal acceleration is

#a_n=v_r^2/rho# where #rho# is the radius of curvature

#a_n=gcosalpha#

#rho=v_r^2/(gcosalpha)=v_r^2/((ucostheta)/(v_r))=v_r^3/(gucostheta)#

You can also get the same result by applying the formula

#rho=(1+(dy/dx)^2)^(3/2)/((d^2y)/dx^2)#

Ultrilliam
Jun 3, 2018

#= u^2/ (g cos theta)#

Explanation:

For the motion:

  • #bb a = ((0 ),( - g ))#

  • #bb v = ((u cos theta \ ),(u sin theta \ - g t ))#

# tan alpha = (u sin theta \ - g t )/(u cos theta \ )#

#implies bb v = u cos theta ((1 ),(tan alpha ))#

And curvature:

  • # R= |bbv|^3/|bb v times bb a|#

#= |(u cos alpha sqrt(tan^2 alpha + 1) )|^3/|u cos theta det [(1, tan alpha),(0, -g)]|#

#= |u |^3/|u g cos theta |#

  • #u, g, theta gt 0##

#= u^2/ (g cos theta)#

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