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Weekly Brain Teaser | Consider a vertical wheel of radius 10 cm

fight it out

Every week we feature a Brain Teaser or a thought-provoking question.

If you have one that you would like to share with fellow engineers, please email it to info@engineeringdaily.net.

So, here is this week’s,

Consider a vertical wheel of radius 10 cm. Now suppose a smaller wheel of radius 2 cm, is made to roll around the larger wheel in the same vertical plane while the larger wheel remains fixed.

What is the total number of rotations the smaller wheel makes when its center makes one complete rotation about the larger wheel?

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Posted by Doug Fred E.I. on Apr 8th, 2009 and filed under Discussions. You can follow any responses to this entry through the RSS 2.0. You can leave a response by filling following comment form or trackback to this entry from your site

21 Responses for “Weekly Brain Teaser | Consider a vertical wheel of radius 10 cm”

  1. Mike Sprayberry says:
    July 10, 2009 at 12:16 pm

    Here is the magic trick to this puzzle no one has pointed out yet:

    Imagine you open up the large wheel into a straight line. Easy math tells us the small wheel will roll 5 times around if it travels the length of the straight line. Well if the line is curved back up into a circle (wheel), then the line is the same length and the small wheel has not changed, so why does it take 6 (instead of 5) rotations to go around? The answer if 5 + 1 really. The 5 rotations along the line PLUS one more big rotation of the small wheel around the big wheel.

    Another way to look at is assume the big wheel was a single point. The outer wheel would not “roll” one bit. A mark on the outer wheel where it touches single point would stay put. Yet the outer wheel still rotates 1 time around the point. 0 + 1 times.

    This works for any two wheels. Do the simple math like the inner wheel is a straight line and then add 1 rotation. This interesting single extra rotation does not change no matter the size of either wheel. It is always one more rotation.

    Reply
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