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?

Sol: Radius of curvature ,

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

θ  = angle between v & horizontal = 0 when the ball is released, it starts moving horizontally with a speed v = speed of the train = 10 m/sec.

$ \displaystyle r = \frac{v^2}{g cos0} $

$ \displaystyle r = \frac{10^2}{10 \times cos0} $

r = 10 m

Also Read :

→ Position vector , Instantaneous Velocity
→ Equation of Trajectory , Time of Flight , Max. height & Horizontal Range
→ Angle of Projection for given Ratio of Range & Max. Height
→ Speed & Angle of Projection so that projectile Passes through Two given Points
→ Horizontal Projection from a given height
→ Projectile on an inclined plane

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