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Question Number 138696 by mr W last updated on 16/Apr/21

Commented by mr W last updated on 16/Apr/21

$a s m a l l m a s s i s r e l e a s e d o n t h e$ $i n t e r i o r s u r f a c e o f a h o l l o w c o n e$ $a t t h e p o s i t i o n a s s h o w n . f i n d t h e$ $t i m e t h e m a s s n e e d s t o r e a c h t h e t i p$ $o f t h e c o n e . t h e r e i s n o f r i c t i o n .$
	Answered by mr W last updated on 18/Apr/21

	Commented by ajfour last updated on 28/Apr/21

	$L e t a p o i n t o n a x i s o f c o n e b e$ $A (ρ \cos ϕ, 0, ρ \sin ϕ)$ $e q . o f p l a n e n o r m a l t o a x i s a n d$ $t h r o u g h A i s$ $(\bar{r} - i ρ \cos ϕ - k ρ \sin ϕ) \cdot (i \cos ϕ + k \sin ϕ) = 0$ $\cos ϕ (x - ρ \cos ϕ) + \sin ϕ (z - ρ \sin ϕ) = 0$ $\Rightarrow x \cos ϕ + z \sin ϕ = ρ$ ${(x - ρ \cos ϕ)}^{2} + {(y - ρ \sin ϕ)}^{2} + z^{2} = r^{2}$ $x^{2} + y^{2} + z^{2} = ρ^{2} + r^{2}$ $l e t y = r \cos θ, x = ρ \cos ϕ - r \sin θ \sin ϕ$ $z = ρ \sin ϕ + r \sin θ \cos ϕ$ $\frac{r}{ρ} = \tan α = m = \frac{R}{h}$ $a_{s} = \frac{- g z}{\sqrt{ρ^{2} + r^{2}}}$ $a_{s} = \frac{- (g \cos α) z}{ρ}$ $s = \frac{ρ}{\cos^{2} α}$ $d s = d ρ (1 + m^{2})$ $\frac{v_{s} {d v}_{s}}{d s} = - \frac{(g \cos α) z}{ρ}$ $v_{s} {d v}_{s} = - \frac{(g \cos α) (1 + m^{2}) z d ρ}{ρ}$ $(d x) \cos ϕ + (d z) \sin ϕ = d ρ$ $d x = (d ρ) \cos ϕ - (m \sin ϕ) d (ρ \sin θ)$ $d ρ - (d z) \sin ϕ = (d ρ) \cos^{2} ϕ$ $- m \sin ϕ \cos ϕ [(d ρ) \sin θ + ρ \cos θ d θ]$
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