Question

A small mass particle is projected with an initial velocity $v_0$ tangent to the horizontal rim of smooth hemispherical bowl at a radius $r_0$ from the vertical centre line, as shown at point $A$. As the particle slide past point $B$, a distance $h$ below $A$ and a distance $r$ from the vertical centre line, its velocity $v$ makes an angle $\theta$ with the horizontal tangent to the bowl through $B$. Determine $\theta$.

Answer 1

Downward motion

$h=\frac{1}{2}g{t}^2$

$\Rightarrow t=\sqrt{\frac{2h}{g}}$

(Refer to Image)

Lateral motion

$r-r_o=V_o\sqrt{\frac{2h}{g}}$

Velocity= $V_o\hat{t}+gt\hat{d}$ $\hat{t}=\text{tangent};\hat{d}=\text{downward}$

$=V-V_o\sqrt{\frac{g}{2h}}+g\sqrt{\frac{2h}{g}}$

$tan\theta =\frac{gt}{V_o}=\frac{g\sqrt{\frac{2h}{g}}\times \sqrt{\frac{g}{2h}}}{V-V_o}$

$=\frac{g}{V-V_o}$