Rotating a Smaller Circle outside a Larger one!

Geometry Level 5

Two circular discs have radii 8 and 28. The larger disc is fixed, while the smaller disc rolls around the outside of the larger disc. In their original positions, point $A$ on the smaller disc co-incides with the point $B$ on the larger disc. The least number of rotations that the small disc makes about its centre before $A$ and $B$ again co-incide is:

7 3.5 12 9 7.5 10.5 6 4.5

3 solutions

Helen Bergstrom
Aug 9, 2015

The circumference of circle B is 2 pi 28 and the circumference of circle A is circumference 2 pi 8. After two revolutions around the larger circle, the smaller circle will have rolled a distance 2 pi 56, totaling 7 rotations. If this distance were traveled along a straight line, the answer would simply be 7 rotations.

But this distance is not traveled along a straight line, it's traveled around a circle -- twice. Imagine that the outer circle "slid", keeping the indicated point in contact with the inner circle. The outer circle would now rotate twice, even though it is sliding and not "rolling" relative to the inner circle.

To get the total number of rotations of the outer circle, we need to sum these two sources of rotation: rolling a distance of 2 pi 56 = 7 rotations, , and rotating twice around the inner circle = 2 rotations. The total is 9 rotations.

@Helen Bergstrom We liked that solution in the comment, and so we have converted it into an actual solution. If you subscribe to this post, you will receive notifications about future comments.

Calvin Lin Staff - 5 years, 10 months ago

Satyajit Mohanty
Aug 6, 2015

When two circles are tangent (like the smaller and the larger circles), then the line joining their centres passes through the point of tangency.

Thus the distance between their centres is the sum of their radii, or $8+28=36$ .

Therefore the centre of the smaller circle is always at a distance of 36 from the centre of the larger circle. Hence the centre of the smaller circles travels around a circle of radius 36.

The centre of the smaller circle moves 2 times around this circle, so travels a total distance of $2(2\pi (36)) = 144 \pi$ .

Circumference of the small circle = $2\pi(8) = 16 \pi$ .

Therefore total number of rotations = $\dfrac{144\pi}{16\pi} = 9$ .

Why does "the center of the circle moves X" and "the circumference is Y" imply that the number of rotations is $X / Y$ ?

Calvin Lin Staff - 5 years, 10 months ago

$\dfrac{X}{Y} = \dfrac{\text{Distance moved by centre}}{\text{Unit rotation distance}}$

Kishore S. Shenoy - 5 years, 9 months ago

Ujjwal Rane
Feb 13, 2016

First let us consider both centers to be fixed.

A and B reunite when both the circles complete whole number of revolutions simultaneously. The ratio of their radii is 28 : 8, giving an LCM = 56 Thus the reunion occurs after 56/28 = 2 and 56/8 = 7 revolutions of the large and small circles.

Now if we stand on the large circle, it will appear stationary. And its two rotations will get added to the apparent motion of the smaller circle giving 7 + 2 = 9 revolutions. This is pretty much like the apparent 'rotation of the sun about the earth' 365 times a year.

To put it algebraically, the rotations of the two circles are opposite (so +7 and -2). The change in observer's position amounts to adding (+2) to both. Making their revolutions (7+2 = 9) and (-2 + 2 = 0)

Rotating a Smaller Circle outside a Larger one!

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