My holiday book giveaway is still going strong until January 1, and you can still get your name in there to try to win the daily random drawing, but I figured it was high time for another challenge question. The first person to post a comment on this post with the correct answer will win a copy of the Math Guide. The usual contest rules apply.

Charles has painted a diameter on a wheel with radius of x centimeters, and is holding the wheel in place on an inclined plane so that the painted line is parallel to the ground as shown in the figure above. If he releases the wheel and it begins to roll without slipping, how many centimeters, in terms of x, will it roll before the painted diameter is perpendicular to the inclined plane for the first time?

As I’ve mentioned on the last few challenge questions, some of my older posts have been getting spammed so I’ve had to add some security to the comment system. If your comment doesn’t appear right away, don’t panic. That’s normal. I still receive an email for every comment as it comes in, so I will still know who got the answer first. You don’t need to resubmit 50 times. 🙂

Good luck!

UPDATE: Many congratulations to Muna for getting it first. Solution below the cut.

The key to this question is really just figuring out how many degrees of rotation the wheel will make for the painted diameter to be parallel to the ramp. To figure that out, note that right now, it’s at a 30º angle from the ramp:

It’s going to roll downhill, obviously, so the first 30º it rolls will make the painted line parallel to the ramp:

From here, it just needs to roll 90º further to make the line perpendicular to the ramp. Total revolution: 30º + 90º = 120º.

Of course, 120º is 1/3 of 360º, so the wheel is going to make only 1/3 of a revolution.

Circles, as you probably know, roll a distance of one circumference when they make one complete revolution. Since this wheel only makes 1/3 of a revolution, it’s only going to roll a distance of one circumference.

Comments (9)

Whoops, thought x was the diameter. Muna’s got it. After pi/6 the painted diameter is parallel to the incline, then after another pi/2 it will be perpendicular to the incline. The arc length is (pi/6+pi/2)*x=2x*pi/3.

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