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The first known diagram of the puzzle, from Claude Dechales Cursus seu Mundus Mathematicus (1674).<\/p>\nEarlier today I set you a puzzle that Isaac Newton posed in 1679 in a latter to Robert Hooke. I mentioned that the answer is counter-intuitive, and also gave you the hint that it might have something to do with figure skating. Here\u2019s the problem again, with its solution.<\/p>\n
The tower drop<\/p>\n
The first known diagram of the puzzle, from Claude Dechales Cursus seu Mundus Mathematicus (1674).<\/p>\n
An object is dropped from a high tower. Bearing in mind the rotation of the Earth itself, where will the object land?<\/p>\n
slightly to the west<\/p>\n
slightly to the east<\/p>\n
exactly at the bottom<\/p>\n
Solution 2. slightly to the east.<\/p>\n
The Earth rotates from west to east.<\/p>\n
Our first thought might be that the object falls slightly to the west, on the grounds that as the object is falling the Earth will rotate slightly to the east.<\/p>\n
But such an argument ignores the fact that the object is originally \u2013 like the tower itself \u2013 rotating with the Earth. Our second thought might be that it lands exactly at the bottom of the tower.<\/p>\n
Again wrong! The object lands slightly to the east because as it falls it spins about its axis slightly faster, by essentially the same mechanism that makes a spinning ice-skater spin faster when they pull in their arms. It is in this way, then, that the object lands slightly to the east.<\/p>\n
Illustration from Wonders in Motion by David Acheson<\/p>\n
The effect is small, but not absurdly so. If the Eiffel tower were on the equator the deflection to the east would be about 11cm.<\/p>\n
In more technical detail, the result follows from conservation of angular momentum, which is a measure of the momentum of an object rotating about an axis.<\/p>\n
When an object is falling from a tower, gravity is pushing it towards the centre of mass of the Earth. Yet the object is also spinning in a circle around the centre of the Earth.<\/p>\n
As the object falls its angular momentum stays the same. Which means that the expression wr2 \u2013 where w is the object\u2019s angular velocity and r is its distance to the rotational axis (in this case, the centre of the Earth) \u2013 stays the same.<\/p>\n
Since the value of r decreases during the fall, the value of w must increase, which means that the object spins faster. If it spins faster than the Earth is spinning it will land slightly to the east of the position directly underneath the tower.<\/p>\n
Thanks to David Acheson for today\u2019s puzzle. He is the author of a new book, Wonders in Motion, The Story of Dynamics, which among other things explains the background to Newton\u2019s puzzle in more detail.<\/p>\n
I hope you had fun. I\u2019ll be back in two weeks.<\/p>\n
Wonders of Motion by David Acheson is on sale at the Guardian Bookshop.<\/a><\/p>\n