A simple circle problem

Question

There is a big circle of radius 20cm and a smaller circle 100 cm away from it of radius 5cm now imagine these two to be 2 tires connected by a chain , where the bigger one completes one rotation how many rotation will small complete??

Any idea how to Solve this??

When you say "$100$cm away," do you mean the distance between the centers, or the distance between the actual circles? — Thomas Andrews, Jul 23 '13 at 18:16
How is the chain connected? How is the distance measured (center-to-center or border-to-border)? What have you done so far? — AlexR, Jul 23 '13 at 18:16
@ThomasAndrews i think 100 cm would be = when taken by radius or from bottom...yeah..possible, a taunt.May be related to RPM — joey rohan, Jul 23 '13 at 18:19
If this is a classical "linked wheels" problem, the things to ask yourself are how do the circumferences of the two wheels compare, and how far will the chain (or belt) move when the large wheel turns one completely? (And, yes, if the two wheels are connected by the chain or belt without any slippage, the distance the wheels are separated by is irrelevant. In fact, the two wheels could touch each other at a tangent point without any chain at all and the answer would be the same.) — colormegone, Jul 23 '13 at 18:20
@AlexR i tried up taking their RPM's and dividing...Cannot get it though — joey rohan, Jul 23 '13 at 18:20
@RecklessReckoner If they are connected, then the smaller one will turn only after 20*5 turns of the 1st one.Then how to relate it with a single round? — joey rohan, Jul 23 '13 at 18:24
Don't the two wheels and the connecting chain all turn together at the same rate? For the large tire to turn once means a point on its circumference has moved $ \ 2 \cdot \pi \cdot 20 \ $ cm., so every point on the connecting chain and on the smaller tire will have moved by the same amount. How many times will the small tire have turned if a point on its circumference has moved by $ \ 2 \cdot \pi \cdot 20 \ $ cm. ? — colormegone, Jul 23 '13 at 18:45
A point on the circumference of the large time has traveled $ \ 40 \pi \ $ cm. after one rotation. So must everything else in the system connected to that tire. So the small tire must turn four times for a point on its edge to circle by the same distance. (This illustrates a general principle in the design of mechanical linkages and gear trains...) — colormegone, Jul 23 '13 at 18:56
"100 cm away from it" This information is almost useless.. don't worry about it ;) ... 20cm(some variables)/5cm(some variables) ... some variables cancel out.. 20/5 = 4... 4 spins — Albert Renshaw, Jul 23 '13 at 19:08

score 1 · Answer 1 · answered Jul 23 '13 at 18:30

1

Assuming they are connected in a $1:1$ means of distance (i.e. a fixed chain on a cog), you have to get the circumference via $$s = 2\pi r$$ where $r$ is the radius of the cog. The number is then computed by $${\rm rot}_{\rm small} \cdot s_{\rm small} = {\rm rot}_{\rm big} \cdot s_{\rm big}$$

answered Jul 23 '13 at 18:30

AlexR

24,905

so the rot by small wheel is simply 20/5? – joey rohan Jul 23 '13 at 18:37
Yes: if everything is moving together and the small tire has one-quarter the diameter (and thus one-quarter the circumference) of the large time, then it will rotate four times as rapidly. – colormegone Jul 23 '13 at 18:47
@RecklessReckoner so here r should be the radius of circles right? Why the cog? – joey rohan Jul 23 '13 at 18:57
The cog as an example of lossles transfer. A rubber band would have friction loss etc, wich would not guarantee the $1:1$ tranmission rate – AlexR Jul 23 '13 at 19:29

score 1 · Accepted Answer · answered Jul 23 '13 at 19:01

When the chain moves by some amount $s$ (measured somewhere between the two wheels) then $$r_{\rm big}\cdot \phi_{\rm big}=s=r_{\rm small}\cdot \phi_{\rm small}\ ,$$ where $\phi_{\rm big}$ and $r_{\rm big}$ are the turning angle and the radius of the big wheel; and similarly for the small wheel. It follows that $$\phi_{\rm small}={r_{\rm big}\over r_{\rm small}}\cdot \phi_{\rm big}=4\phi_{\rm big}\ .$$ Therefore the small wheel will complete four full turns when the big wheel completes one.

A simple circle problem

2 Answers2