Question-61994

Question Number 61994 by ajfour last updated on 13/Jun/19

Commented by ajfour last updated on 13/Jun/19

F i n d t h e \boldsymbol r o f l a r g e s t s i z e s p h e r e

t h a t c a n r e s t a g a i n s t t h e r i m o f

c y l i n d e r o n o n e s i d e a n d t h e

s t i c k o n t h e o t h e r s i d e .

Answered by mr W last updated on 13/Jun/19

\frac{r + h}{r} = \frac{\sqrt{{(2 R)}^{2} + h^{2}}}{2 R}

\Rightarrow r_{m a x} = \frac{h}{\sqrt{1 + {(\frac{h}{2 R})}^{2}} - 1}

Commented by ajfour last updated on 13/Jun/19

n o t s u r p r i s i n g, s e e m s n a t u r a l S i r .

I j u s t v i e w e d y o u r c o n c l u s i o n s i r .

Commented by mr W last updated on 13/Jun/19

Commented by mr W last updated on 13/Jun/19

t h i s i s m a y b e n o t t h e l a r g e s t p o s s i b l e

s p h e r e, s i n c e t h e s t i c k (b l u e l i n e) c a n

a l s o b e i n o t h e r p o s i t i o n s . i t o n l y

n e e d s t o t o u c h t h e s p h e r e .

Commented by mr W last updated on 13/Jun/19

Commented by ajfour last updated on 13/Jun/19

O^{'} y e s, u n d e r s t o o d s i r, s p h e r e

c a n b e b a l a n c e d a t o n e p o i n t e v e n .

T h a n k s .

Commented by mr W last updated on 13/Jun/19

p o s i t i o n o f s t i c k i s θ . i t t o u c h e s t h e

s p h e r e a t p o i n t P .

C (- R, 0, h + r)

A (- R \cos θ, - R \sin θ, 0)

B (R \cos θ, R \sin θ, h)

e q n . o f l i n e A B (s t i c k) :

\frac{x + R \cos θ}{2 R \cos θ} = \frac{y + R \sin θ}{2 R \sin θ} = \frac{z}{h} = k

\Rightarrow x = R \cos θ (2 k - 1)

\Rightarrow y = R \sin θ (2 k - 1)

\Rightarrow z = h k

d i s t a n c e f r o m p o i n t C t o l i n e A B = d :

{D = d^{2} = {[R \cos θ (2 k - 1) + R]}^{2} + [R \sin θ (2 k - 1))]}^{2} + {[h k - h - r]}^{2}

= 2 R^{2} [2 k^{2} - (2 k - 1) (1 - \cos θ)] + {(h k - h - r)}^{2}

= 4 R^{2} [k^{2} - (2 k - 1) \sin^{2} \frac{θ}{2}] + {(h k - h - r)}^{2}

\frac{d D}{d k} = 0

4 R^{2} (k - \sin^{2} \frac{θ}{2}) + (h k - h - r) h = 0

\Rightarrow k = \frac{(h + r) h + 4 R^{2} \sin^{2} \frac{θ}{2}}{4 R^{2} + h^{2}} = \frac{1 + 4 {(\frac{R}{h})}^{2} \sin^{2} \frac{θ}{2} + \frac{r}{h}}{1 + 4 {(\frac{R}{h})}^{2}}

{(\frac{r}{h})}^{2} = 4 {(\frac{R}{h})}^{2} [k^{2} - (2 k - 1) \sin^{2} \frac{θ}{2}] + {(k - 1 - \frac{r}{h})}^{2}

l e t ρ = \frac{R}{h}, λ = \frac{r}{h}

\Rightarrow k = 1 + \frac{λ - 4 ρ^{2} \cos^{2} \frac{θ}{2}}{1 + 4 ρ^{2}}

0 = 4 ρ^{2} [{(\frac{λ}{1 + 4 ρ^{2}} - \frac{4 ρ^{2} \cos^{2} \frac{θ}{2}}{1 + 4 ρ^{2}})}^{2} - 2 (\frac{λ}{1 + 4 ρ^{2}} - \frac{4 ρ^{2} \cos^{2} \frac{θ}{2}}{1 + 4 ρ^{2}}) \cos^{2} \frac{θ}{2} - \cos^{2} \frac{θ}{2}] + (\frac{λ}{1 + 4 ρ^{2}} - \frac{4 ρ^{2} \cos^{2} \frac{θ}{2}}{1 + 4 ρ^{2}} - 2 λ) (\frac{λ}{1 + 4 ρ^{2}} - \frac{4 ρ^{2} \cos^{2} \frac{θ}{2}}{1 + 4 ρ^{2}})

\Rightarrow m a x . λ i s w h e n θ = 0

Commented by mr W last updated on 13/Jun/19

Commented by mr W last updated on 13/Jun/19

y e s s i r!

o p t i m u m i n t h e n a t u r e i s a l w a y s i n

s y m m e t r y .

Facebook Tweet Pin