Projectile Motion Comparison - 3

Let the speed of projection be $u$ . Consider the second particle to be initially at the origin and it is projected along the plane:

$y = \frac{x}{\sqrt{3}}$

Gravity is considered to be acting along the $-Y$ direction.

Initial velocity vector:

$\vec{v}_o = v_{ox} \ \hat{i} +v_{oy} \ \hat{j} +v_{oz} \ \hat{k}$

$\vec{v}_o = u\left(\sin{45^{\circ}}\cos{30^{\circ}} \ \hat{i} + \sin{45^{\circ}}\sin{30^{\circ}} \ \hat{j}+\sin{45^{\circ}}\ \hat{k} \right)$

This situation is a bit tedious to analyse using Newton's laws with the pure rolling constraint and 3D motion. Instead, here, I apply the Lagrangian approach. Consider the kinetic energy of a pure rolling solid sphere at any instant to be:

$\mathcal{T} = \frac{7m}{10}\left(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\right)$

Since $y = \frac{x}{\sqrt{3}}$ :

$\mathcal{T} = \frac{7m}{10}\left(4\dot{y}^2 + \dot{z}^2\right)$

The potential energy of the particle is: $\mathcal{V} = mgy$

System Lagrangian: $\mathcal{L} = \mathcal{T}-\mathcal{V}$

Lagrange's equations read:

$\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{y}}\right) - \frac{\partial \mathcal{L}}{\partial y}=0$ $\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{z}}\right) - \frac{\partial \mathcal{L}}{\partial z}=0$

$\ddot{y} = -\frac{5g}{28}$ $\ddot{z} = 0$

Using the constraint equation:

$y = \frac{x}{\sqrt{3}}$ $\implies \ddot{y} = \frac{\ddot{x}}{\sqrt{3}}$

Therefore:

$\ddot{x} =-\frac{5\sqrt{3}g}{28}$

Solving these three equations gives (Only showing $y$ and $z$ )

$y = v_{oy}t - \frac{5gt^2}{56}$ $z = v_{oz}t$

The Y coordinate of the particle becomes zero at $t=0$ and:

$t = \frac{28u}{5\sqrt{2}g}$ At this instant, the Z-coordinate is:

$z = \frac{28}{10}\frac{u^2}{g}$ $\implies R_2 = 2.8 R_1$