This is a gem of a problem pointed to me by an avid puzzler and an IITD friend (now in Philippines) – Girish Tutakne. Thanks Girish. Even though I know the answer, it still baffles me!
Suppose we have 2 coins, with radii 2 cm and 6 cm respectively. The coin with the 2 cm radius is rotated on the circumference of the bigger coin. How many times will it rotate before it comes back to it’s starting point?
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Alok – since it was a puzzle, I knew it could not be 6/2 = 3 🙂 I thought that actually it is the center of the small coin which is moving in a circle with a radius equal to 6+2 = 8 and the movement is accomplished by the small coin of radius 2. So my answer is 8/2 = 4.
Had me stumped for a long time. Thinking about it the whole day! had some practical results which I was looking for a conceptual validation, which I got from a Japanese economics professor´s blog! So, it is 4, but for different reasons than mentioned by Prakhar. Dont want to give the reason, so that people have the pleasure of thinking through this interesting problem!
Curious to know how you did it. Please do drop me a note on my e-mail firstname.lastname@example.org
Beautiful problem. Answer is 4. Hint: Consider both rotation and revolution of the smaller coin.
Beautiful problem. Answer is 4.
Hint: Consider both rotation and revolution of the small coin.
There will be two motions of outer(smaller) coin of radius 2 cm:
a) Rotational motion due to no slip condition at point of contact.
2*pi*6 = n1*(2*pi*2) => n1 = 3.
b) Rotational motion due to revolution about larger coin.
n2 = 1
so, total rotation is n1+n2 = 4
Really good puzzle!
Actually another way to look at it is this – the number of rotations of a circle around its center is directly proportional to the linear distance travelled by the center of the circle. Since the linear distance travelled by the center of the smaller coin is equal to 2*pi*8, the number of revolutions will be 4.