# Radius of Curvature of a Projectile Motion

Let at any time t, the velocity vector be inclined at an angle θ with horizontal at a point, say, P as shown in Figure . Since gravitational acceleration is always acting vertically downwards, the components of perpendicular to the velocity vector can be treated as a radial acceleration towards, the centre of a circular path of radius (let r) as shown in Figure;  r is known as radius of curvature of the parabola at P.

$\displaystyle a_r = \frac{V^2}{r}$

$\displaystyle r = \frac{V^2}{a_r}$

$\displaystyle r = \frac{V^2}{g cos\theta}$

where V and θ can be found in terms of t as

$\displaystyle V = \sqrt{V_0^2 + (gt)^2 – 2 V_0 gt sin \theta_0 }$

$\displaystyle tan\theta = \frac{V_0 sin\theta_0 – gt}{V_0 cos\theta_0}$

V and θ can be found in terms of y as

$\displaystyle V = \sqrt{V_0^2 – 2 g y}$

$\displaystyle tan\theta = \frac{\sqrt{V_0^2 sin^2\theta_0 – 2 g y}}{V_0 cos\theta_0}$

Solved Example: A boy is standing inside a train moving with a constant velocity of magnitude 10 m/sec. He throws a ball vertically up with a speed 5 m/sec relative to the train. Find the radius of curvature of the path of the ball just at the time of projection. Solution: The ball continues to move horizontally with (Vx)0 = 10 m/sec. It begins to move up with (Vy)0 = 5 m/sec. Therefore θ0 is given as,

$\displaystyle tan\theta_0 = \frac{(V_y)_0}{(V_x)_0}$

$\displaystyle tan\theta_0 = \frac{5}{10} = \frac{1}{2}$

Now the required radius of curvature is given as

$\displaystyle r = \frac{V^2}{g cos\theta_0}$

Putting , $\displaystyle V = \sqrt{10^2 + 5^2} = 5 \sqrt{5}$ m/s

g = 10 m/s2

we obtain , r  ≈ 14 m

Solved Example : Referring to the previous illustration, if the boy releases the ball from rest, what will be the radius of curvature of the path at the instant of its release?

$\displaystyle r = \frac{v^2}{g cos\theta}$
$\displaystyle r = \frac{v^2}{g cos0}$
$\displaystyle r = \frac{10^2}{10 \times cos0}$