I-have-two-circles-side-by-side-Circle-1-has-radius-r-Circle-2-has-radius-1-3-r-If-I-roll-circle-2-around-circle-1-until-it-reaches-the-begining-how-many-times-will-it-roll-

Question Number 5546 by FilupSmith last updated on 19/May/16

I have two circles side by side .

Circle 1 has radius r .

Circle 2 has radius \frac{1}{3} r .

If I roll circle 2 around circle 1 until

it reaches the begining,

how many times will it roll ?

Commented by FilupSmith last updated on 19/May/16

Correct if wrong but i think i have

my answer .

Mapping a point on the circle, it produces

a sine wave (if the circle is stationary)

the length of a sine wave y = \sin (x)

is equal to π

L e n g t h = \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x = \int_{0}^{2 π} \sqrt{1 + \cos x} d x = π

from what i have gathered this means id

that by the time it reaches 2 π, t h e c u r v e

l e n g t h i s π .

similarly, once the spining coin makes it

half way around the other coin (π d i s t a n c e),

it rotates 2 π = 360^{o}

s o r r y f o r b a d o r i n c o r r e c t s o l u t i o n

Commented by Yozzii last updated on 20/May/16

T h e i n t e g r a l e x p r e s s i o n f o r t h e a r c

l e n g t h o f t h e s i n e c u r v e c a n n o t b e

e x p r e s s e d i n t e r m s o f e l e m e n t a r y f u n c t i o n s .

T h i s i s b e c a u s e n o s u c h r e s u l t c a n b e

t h e a n t i d e r i v a t i v e o f \sqrt{1 + {c o s}^{2} x} .

F o r t h e c l o s e d i n t e r v a l [0, 2 π] t h e a r c

l e n g t h \int_{0}^{2 π} \sqrt{1 + {c o s}^{2} x} d x \approx 7.6 \neq π .

Commented by FilupSmith last updated on 19/May/16

Commented by Yozzii last updated on 19/May/16

2 π r \times \frac{1}{2 π r / 3} = 3 t i m e s

Commented by Yozzii last updated on 19/May/16

T h e m o b i l e p o i n t, s a y P, o n C 2 w i l l m o v e

c o v e r i n g a d i s t a n c e o f 2 π r w h i c h i s

t h e c i r c u m f e r e n c e o f C 1 . B u t, i n o n e

r e v o l u t i o n o f C 2 a b o u t i t s c e n t r e

P c o v e r s a d i s t a n c e o f o n l y \frac{2 π r}{3} .

S o, f o r t h e t o t a l d i s t a n c e 2 π r t o b e

c o v e r e d b y P, C 2 w i l l m a k e \frac{2 π r}{2 π r / 3} = 3

c o m p l e t e r e v o l u t i o n s t o r e t u r n t o B .

G e n e r a l l y, f o r c i r c l e s C 1 a n d C 2 h a v i n g

r a d i i r_{1} a n d r_{2} r e s p e c t i v e l y, i f C 1 i s

s t a t i o n a r y a n d C 2 i s m a d e t o r o l l

a l o n g t h e c i r c u m f e r e n c e o f C 1, t h e n

C 2 m a k e s \frac{2 π r_{1}}{2 π r_{2}} = \frac{r_{1}}{r_{2}} r e v o l u t i o n s a b o u t

i t s c e n t r e .

Commented by FilupSmith last updated on 19/May/16

I thought so, too . What I found Is that

if i have two identical coins, coin 1 will

rotate TWICE as if does a full cycle .

I dont know why

Commented by Yozzii last updated on 19/May/16

S o t h e m o b i l e c o i n r e v o l v e d t h r o u g h

720 ° w h i l e b o t h c o i n s a r e i d e n t i c a l ?

Commented by FilupSmith last updated on 19/May/16

correct! try it yourself!

Commented by FilupSmith last updated on 19/May/16

I {can}^{'} t understand why this happenes!

Commented by FilupSmith last updated on 20/May/16

Oh, i see, i reaslised my mistake .

So is there a way to determine its length ?

Commented by Yozzii last updated on 20/May/16

Commented by Yozzii last updated on 20/May/16

Commented by Yozzii last updated on 20/May/16

I a m y e t t o r e a c h s u c h a l e v e l o f

C a l c u l u s a n d A n a l y s i s t o e x p l a i n

t h e e l l i p t i c i n t e g r a l . I a l s o c a n n o t

s a y t h a t t h e r e {i s n}^{'} t a w a y t o

f i n d t h e e x a c t a r c l e n g t h o f t h e s i n e

c u r v e i n t e r m s o f e l e m e n t a r y f u n c t i o n s .

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