Question

A particle is projected with velocity u at an angle $\theta$ with the horizontal plane. Find the radius of curvature of the parabola traced out by the particle at the point where velocity makes an angle$\dfrac{\theta }{2}$ with the horizontal plane.

Answer 1

The horizontal component of velocity remains constant during projectile motion, and radius of curvature can be calculated using the formula $R = \dfrac{{{v^2}}}{{{a_n}}}$, where v denotes speed of the particle and an denotes component of acceleration perpendicular to the velocity.

Complete step by step solution:
Assuming that the particle is moving with speed v when it makes angle $\dfrac{\theta }{2}$ with the horizontal. We can say that the horizontal speed of the particle will be $v\operatorname{Cos} \dfrac{\theta }{2}$.

Since the particle was thrown with speed u at angle $\theta$, the initial horizontal velocity of the particle is $u\operatorname{Cos} \theta$
Since the horizontal velocity remains constant
$v\operatorname{Cos} \dfrac{\theta }{2} = u\operatorname{Cos} \theta$
From the above equation, we can calculate v
$v = \dfrac{{u\operatorname{Cos} \theta }}{{\operatorname{Cos} \dfrac{\theta }{2}}}$

Since the only force acting on the particle is weight. It has acceleration g in the downward direction, the component of g perpendicular to velocity v will be $g\operatorname{Cos} \dfrac{\theta }{2}$ , hence ${a_n} = g\operatorname{Cos} \dfrac{\theta }{2}$
Using the formula for radius of curvature
$R = \dfrac{{{v^2}}}{{{a_n}}}$,

And substituting the values of v and an, we get :
R = \dfrac{{{{\left\\{ {\dfrac{{u\operatorname{Cos} \theta }}{{\operatorname{Cos} \dfrac{\theta }{2}}}} \right\\}}^2}}}{{g\operatorname{Cos} \dfrac{\theta }{2}}}
$R = \dfrac{{{u^2}{{\operatorname{Cos} }^2}\theta }}{{g{{\operatorname{Cos} }^3}\dfrac{\theta }{2}}}$

Note: Before attempting to solve the problem, the student needs to be able to understand concepts of radius of curvature and the concepts of projectile motion. Generally students focus on formulas of projectile motion without diving into the basics. While calculating the radius of curvature, one must always remember that the acceleration in the denominator is the component of acceleration perpendicular to net velocity.