# 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?

Jun 3, 2018

#### Explanation:

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

$y = x \tan \theta - \frac{g {x}^{2}}{2 {u}^{2} {\cos}^{2} \theta}$

At any point $P$ on the trajectory,

The horizontal component of the velocity is

${v}_{x} = u \cos \theta$

$\implies$, $u \cos \theta = x t$

$\implies$, $t = \frac{x}{u \cos \theta}$

and the vertical component is

${v}_{y} = u \sin \theta - g t$

$\implies$, ${v}_{y} = u \sin \theta - \frac{g x}{u \cos \theta}$

The resultant velocity is

${v}_{r} = \sqrt{{v}_{x}^{2} + {v}_{y}^{2}}$

$= \sqrt{{u}^{2} {\cos}^{2} \theta + {u}^{2} {\sin}^{2} \theta + \frac{{g}^{2} {x}^{2}}{{u}^{2} {\cos}^{2} \theta} - \frac{2 u \sin \theta g x}{u \cos \theta}} |$

$= \sqrt{{u}^{2} + \frac{{g}^{2} {x}^{2} {\sec}^{2} \theta}{{u}^{2}} - 2 g x \tan \theta} |$

The angle $\alpha$ is defined as follows

$\cos \alpha = {v}_{x} / {v}_{r} = \frac{u \cos \theta}{v} _ r$

The centripetal acceleration is

${a}_{n} = {v}_{r}^{2} / \rho$ where $\rho$ is the radius of curvature

${a}_{n} = g \cos \alpha$

$\rho = {v}_{r}^{2} / \left(g \cos \alpha\right) = {v}_{r}^{2} / \left(\frac{u \cos \theta}{{v}_{r}}\right) = {v}_{r}^{3} / \left(g u \cos \theta\right)$

You can also get the same result by applying the formula

$\rho = {\left(1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}\right)}^{\frac{3}{2}} / \left(\frac{{d}^{2} y}{\mathrm{dx}} ^ 2\right)$

Jun 3, 2018

$= {u}^{2} / \left(g \cos \theta\right)$

#### Explanation:

For the motion:

• $\boldsymbol{a} = \left(\begin{matrix}0 \\ - g\end{matrix}\right)$

• $\boldsymbol{v} = \left(\begin{matrix}u \cos \theta \setminus \\ u \sin \theta \setminus - g t\end{matrix}\right)$

$\tan \alpha = \frac{u \sin \theta \setminus - g t}{u \cos \theta \setminus}$

$\implies \boldsymbol{v} = u \cos \theta \left(\begin{matrix}1 \\ \tan \alpha\end{matrix}\right)$

And curvature:

• $R = | \boldsymbol{v} {|}^{3} / | \boldsymbol{v} \times \boldsymbol{a} |$

$= | \left(u \cos \alpha \sqrt{{\tan}^{2} \alpha + 1}\right) {|}^{3} / | u \cos \theta \det \left[\begin{matrix}1 & \tan \alpha \\ 0 & - g\end{matrix}\right] |$

$= | u {|}^{3} / | u g \cos \theta |$

• $u , g , \theta > 0$#

$= {u}^{2} / \left(g \cos \theta\right)$