Question

Find the radius of curvature of a projectile at the highest point, fired with a velocity v, making an angle $\theta$ with the horizontal.
$\text{A}\text{. }\dfrac{{{v}^{2}}{{\sin }^{2}}\theta }{g}$
$\text{B}\text{. }\dfrac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}$
$\text{C}\text{. }\dfrac{{{v}^{2}}{{\cos }^{2}}\theta }{2g}$
$\text{D}\text{. }\dfrac{{{v}^{2}}{{\cos }^{2}}\theta }{g}$

Answer 1

Hint: At the highest point, the velocity and its acceleration are perpendicular. Hence, we can say that the projectile is in a circular motion, like it is in the case of a body in uniform circular motion. Thus, the acceleration of the projectile at point is $a=\dfrac{{{v}^{2}}}{R}$, where R is the radius of curvature.
Formula used:
$a=\dfrac{{{v}^{2}}}{R}$

Complete step-by-step answer:
Radius of curvature of a path at a point is a circle to which the curve of the path touches the circle tangentially. It tells us how much the curve is at this point. Less the radius of curvature, more pointed is the curve at the given point. If the radius of curvature is infinite then it means that the curve is a straight line.
Let us calculate the radius of curvature (R) for a projectile at its height point from the ground. It is already given that the projectile is fired with a velocity v, making an angle $\theta$ with the horizontal.
At the highest point, the vertical component of velocity of the projectile is zero and it has velocity only along the horizontal i.e. $v\cos \theta$. At this point, the projectile has an acceleration g, which is acceleration due to gravity.
Here, we can say that the projectile is in a circular motion, since the velocity and its acceleration are perpendicular like it is the case of a body in uniform circular motion. And the radius of that circle will be the radius of curvature at the highest point.
Now, let that radius be R.
The relation between acceleration a (centripetal acceleration) and velocity v of the body in uniform circular motion is $a=\dfrac{{{v}^{2}}}{R}$.
In this case, a=g and $v=v\cos \theta$. Substitute these values in equation (i).
$\Rightarrow g=\dfrac{{{\left( v\cos \theta \right)}^{2}}}{R}$.
Therefore, $R=\dfrac{{{v}^{2}}{{\cos }^{2}}\theta }{g}$.
Hence, the correct option is D.

Note: The method we use here is not inversely. We were able to find the radius of curvature at the highest point of the projectile only because its velocity and acceleration were perpendicular and we could relate this to the uniform circular motion.
In addition, the radius of curvature of the projectile at any other point will not be the same.