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Imagine two circles one of radius "r" another "2r" . Let the

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Imagine two circles one of radius "r" another "2r" . Let the [#permalink]  23 Jul 2003, 04:51
Imagine two circles one of radius "r" another "2r" . Let the bigger circle be stationay and the smaller circle moves on the bigger circle with out " SLIP". In one complete revolution of the smaller circle on the larger one :

How many rotations does the smaller circle go through around its center in completing the revolution.
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geometry.( some circles in life!!) [#permalink]  23 Jul 2003, 05:28
Brainless wrote:
4

Explain...

I think it should be 2
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The answer is 2. Their circumferences are in the ratio of 1:2.
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C/c=pi*4*r/pi*2*r=2

2 revolutions
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Re: geometry.( some circles in life!!) [#permalink]  24 Jul 2003, 08:57
abhaypathania wrote:
Imagine two circles one of radius "r" another "2r" . Let the bigger circle be stationay and the smaller circle moves on the bigger circle with out " SLIP". In one complete revolution of the smaller circle on the larger one :

How many rotations does the smaller circle go through around its center in completing the revolution.

2 is correct : But this it becomes tricky when I would ask how many revolutions around the CENTER of the BIGGER circle.
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it is given -- one revolution
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Re: geometry.( some circles in life!!) [#permalink]  25 Jul 2003, 00:13
abhaypathania wrote:
Imagine two circles one of radius "r" another "2r" . Let the bigger circle be stationay and the smaller circle moves on the bigger circle with out " SLIP". In one complete revolution of the smaller circle on the larger one :

How many rotations does the smaller circle go through around its center in completing the revolution.

Be careful gang! This problem is a LOT trickier than it looks!!!!

I submit that the answer is THREE! <evil grin>

Any takers?

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AkamaiBrah
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MBA, Anderson School of Management, UCLA, Class of 1993

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Re: geometry.( some circles in life!!) [#permalink]  25 Jul 2003, 00:20
abhaypathania wrote:
abhaypathania wrote:
Imagine two circles one of radius "r" another "2r" . Let the bigger circle be stationay and the smaller circle moves on the bigger circle with out " SLIP". In one complete revolution of the smaller circle on the larger one :

How many rotations does the smaller circle go through around its center in completing the revolution.

2 is correct : But this it becomes tricky when I would ask how many revolutions around the CENTER of the BIGGER circle.

I respectfully disagree with ALL of you. The correct answer is THREE! Think about it....
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AkamaiBrah
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The center of the smaller circle draws a new circle, its radius being 3r.
If we imagine a new circle as a line (its lenght is 6r*pi), and if we fix a point on the smaller circle, then the number of the point's revolutions will be proportional to the circumference of the smaller circle.

Cnew=2*3*pi*r=6r*pi
Csmall=2r*pi

Thus, 3?

Is my reasoning correct?
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but in this case, the number of revolutions around the center of the bigger circle is 2.
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stolyar wrote:
The center of the smaller circle draws a new circle, its radius being 3r.
If we imagine a new circle as a line (its lenght is 6r*pi), and if we fix a point on the smaller circle, then the number of the point's revolutions will be proportional to the circumference of the smaller circle.

Cnew=2*3*pi*r=6r*pi
Csmall=2r*pi

Thus, 3?

Is my reasoning correct?

Stolyar, the smaller circle's centre will draw a new circle with a radius of r and not 3r.
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prashant wrote:
stolyar wrote:
The center of the smaller circle draws a new circle, its radius being 3r.
If we imagine a new circle as a line (its lenght is 6r*pi), and if we fix a point on the smaller circle, then the number of the point's revolutions will be proportional to the circumference of the smaller circle.

Cnew=2*3*pi*r=6r*pi
Csmall=2r*pi

Thus, 3?

Is my reasoning correct?

Stolyar, the smaller circle's centre will draw a new circle with a radius of r and not 3r.

Try this:

Take two quarters. Both have radius r. Suppose you keep one quarter stationary while the other quarter rotates around it one time (say start at 12 o'clock position, roll clockwise, then end at 12 o'clock position). According to logic seen so far, the outer quarter will only make ONE revolution about its center (that's the question asked, right?). Put the outer quarter in the heads position with George facing left and "LIBERTY" on top. Let's define one revolution of the outer quarter as the most obvious one: when George facing left and "liberty" is on top again. Now roll the outer quarter carefully around the stationary quarter, being careful not to let it slip. Note when the quarter is at about the 6 o'clock position, GEORGE IS FACING LEFT and LIBERTY IS ON TOP. As far as I am concerned, and by any reasonable definition, the outer quarter has completed one full revolution about ITS center. Keep going and teh quarter will make yet another revolution about its center and end up at teh 12 o'clock position with George once again facing left.

HOW DID THIS HAPPEN? Well, the quarter made ONE revolution RELATIVE TO THE SURFACE of the inner quarter, but the in following the surface of the inner quarter, forced it to make an additional revolution.

Thus for the circle of radius r rolling around the circle of radiur 2r. The circle will make 2 complete revolutions RELATIVE TO THE SURFACE OF THE INNER CIRCLE, and just following the path of the surface of the inner circle will take the outer circle through one additional complete revolution, thus it will make a total of 3 revolutions. (Cut 2 circles out of paper, draw a reference arrow, and try it yourself if you are not convinced!!!).
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geometry.( some circles in life!!) [#permalink]  25 Jul 2003, 02:27
Fair try Akamaibrah. I see your point.

But could you please give us a more mathematical explanation rather than physical visualisation based explanation. I'm still struggling to solve it without using coins and paper cuttings

Thanks!
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Re: geometry.( some circles in life!!) [#permalink]  25 Jul 2003, 02:36
prashant wrote:
Fair try Akamaibrah. I see your point.

But could you please give us a more mathematical explanation rather than physical visualisation based explanation. I'm still struggling to solve it without using coins and paper cuttings

Thanks!

I've got it!

Stolyar - your reasoning is correct.

In the case where the two coins have the same radius r, the centre of the outer coin will draw out a circle of radius 2r as it goes around the circumference of the inner coin. Therefore, the distance travelled by the centre of the outer coin to come back to its starting position is 2*pi*2r which implies that the outer coin would have completed (2*pi*2r)/(2*pi*r) = 2 revolutions

With the same logic, in abhay's this case, the centre of the outer circle with radius r draws a new circle of radius 3r, as Stolyar mentioned. Therefore, the revolutions done by smaller circle, using the same logic, would be 3.

I feel quite dumb....this wasn't so tough.....

but anyway...c'est la vie
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Re: geometry.( some circles in life!!) [#permalink]  25 Jul 2003, 02:39
prashant wrote:
Fair try Akamaibrah. I see your point.

But could you please give us a more mathematical explanation rather than physical visualisation based explanation. I'm still struggling to solve it without using coins and paper cuttings

Thanks!

If you were to "unravel" the circumference of the inner circle into a straight line, the outer coin would clearly rotate 2 times since the circumference of the inner circle is twice that of the outer circle. However, since the actual path is also ONE complete circle, just following the path adds another circle to the number of rotations (imagine a ant walking along the edge of a coin held upright. Just by following his nose completely around the quarter, he completes one full "somersault" without knowing it).

Hence, the number of complete rotations is

[(CircumferenceInnerCircle)/(CircumferenceOUterCircle)] + 1.

Since Circumference is proportionate to Radius, the equation is also:

R_inner/R_outer + 1

and in this case

2r/r + 1 = 2 + 1 = THREE.

Stolyar's explanation is a little hard to understand (for me anyway), but I think he means this. Let's say you poke a hole in the middle of the outer circle and put in an axle. As the outer circle of radius r goes around the circle of radius 2r, the axle travels a total distance of 2*Pi*3r.

If you took that same "circle" (it is now, i suppose, a wheel") and rolled it along a path of length 2*Pi*3r, it would make 3 revolutions. That really is exactly equivalent to my equation.

The radius of the axle's path is 2*Pi*(R_inner + R_outer). Tne number of rotations is this length divided by the circumference of the outer circle which is 2 * Pi * R_outer. Hence:

Rotations = 2*Pi * (R_inner + R_outer) / (2 * Pi * R_outer)
= (R_inner + R_outer) / R_outer
= R_inner/R_outer + R_outer/R_outer
= R_inner/R_outer + 1 which is the same result that I derived "logically" (I don't trust the math unless I can visualize it in a way that makes sense).

The main thing is that we all agree that the answer is THREE.

HTH!
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Vice President, Midtown NYC Investment Bank, Structured Finance IT
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MBA, Anderson School of Management, UCLA, Class of 1993

Last edited by AkamaiBrah on 25 Jul 2003, 03:03, edited 2 times in total.
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Re: geometry.( some circles in life!!) [#permalink]  25 Jul 2003, 02:47
AkamaiBrah wrote:
prashant wrote:
Fair try Akamaibrah. I see your point.

But could you please give us a more mathematical explanation rather than physical visualisation based explanation. I'm still struggling to solve it without using coins and paper cuttings

Thanks!

If you were to "unravel" the circumference of the inner circle into a straight line, the outer coin would clearly rotate 2 times since the circumference of the inner circle is twice that of the outer circle. However, since the actual path is also ONE complete circle, just following the path adds another circle to the number of rotations (imagine a ant walking along the edge of a coin held upright. Just by following his nose completely around the quarter, he completes one full "somersault" without knowing it).

Hence, the number of complete rotations is

[(CircumferenceInnerCircle)/(CircumferenceOUterCircle)] + 1.

Since Circumference is proportionate to Radius, the equation is also:

R_inner/R_outer + 1

and in this case

2r/r + 1 = 2 + 1 = THREE.

And the ant example works very well !

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Bingo...

Three it is.

the explaination given is correct too. The other way u could solve it is by making the radius of the inner circle to zero... then the outter cirlce would rotate about a point... ( ONCE ) and still would have travelled zero distance about the point...the other way to resolve the mistry is to make diagrams and then evaluate rotation angle ("thita') about the point of rotation of the outter circle.
Absolute Revolutions are always calculated based on the point of rotation rather then the center of rotation. where the point of rotation in this is the point of contact between the bigger and the smaller circle.
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