## A Paradox of Galileo

Going by the popularity of the little physics problem I posted last week, I thought some of you might be interested in this little conundrum which dates back to the 17th Century (to Galileo Galilei, in fact). It’s not a problem to which there’s a snappy answer, so no poll this time, but I think it’s quite a good one to think about – and please try to resist the temptation to google it!

The above figure shows a large circular wheel, which rolls from left to right, without slipping, on a flat surface, along a straight line from P to Q, making exactly one revolution as it does so. The distance PQ is thus equal to the circumference of the wheel.

Now, consider the small circle, firmly fixed to the larger circle with the two centres coincident. The small circle also makes one complete revolution as the wheel rolls, and it travels from R to S. Similarly, therefore, the distance RS must be the circumference of the small circle.

Since RS is clearly equal to PQ it follows that the circumferences of the large wheel and the small circle must be equal!

Since the radii of the large and small circles are different, this conclusion is clearly false so what’s wrong with the argument?

June 1, 2011 at 12:45 pm

It’s a lot older than Galileo isn’t it?

June 1, 2011 at 12:49 pm

I don’t know. I have a reference to it in Galileo’s

Dialogues Concerning Two New Sciences(1638), but I’d be very interested to know if it has a longer provenance…June 1, 2011 at 12:55 pm

Don’t click on the below if you want to work out the answer

>> http://mathworld.wolfram.com/AristotlesWheelParadox.html

June 1, 2011 at 1:31 pm

Oh right. Aristotle. I’ve heard of him. Something about a bottle.

June 1, 2011 at 2:00 pm

Yes, one of the few incidences of a double rhyme in Cockney rhyming slang.

June 1, 2011 at 12:49 pm

The speeds of points R & P /as they rotate) are different so, although RS = PQ and a definite amount of time, t, has passed, since the speeds are different, it cannot follow that the circumferences are equal.

June 1, 2011 at 12:56 pm

If you pretend the inner circle is rolling along the line RS, then I don’t think the inner circle could be described as rolling without slipping. So the argument must fall over at that point.

Is that right?

June 1, 2011 at 1:45 pm

Here’s a big hint. What’s the difference between a cycloid and a trochoid?

June 1, 2011 at 2:41 pm

Okay, let’s be pedantic!

A cycloid is a type of trochoid. It is the path of a point on the exterior of a ball that rolls without slipping. Apparently (and yes I had to look this up) it’s another trochoid – a curtate cycloid- that describes the path of an exterior point on the inner ball.

If our trochoid has a distance “b” from the center of the outer circle of radius “a”, the rate of change of the horizontal coordinate of the trochoid is proportional to “a – b cos(phi)”, where “phi” is the angle between the horizontal and the location on trochoid.

If a=b (a cycloid), then the rate of change is zero for a particular phi (i.e. the point of the ball in contact with the ground has zero velocity, which is what we mean by saying there is no slipping).

If b < a (for our curtate cycloid), there is no phi for which this is the case. Hence the inner ball does not follow a cycloid (or as I said before, it can't be described as rolling without slipping – the point on the trochoid in contact with the 'floor' always has a non-zero velocity).

So I think that's right.

June 1, 2011 at 2:45 pm

>Here’s a big hint. What’s the difference between a cycloid and a trochoid?

Isn’t it that the point at the bottom of a cycloid is stationary, where as in a general trochoid it is not (i.e. sliding)?

June 1, 2011 at 2:51 pm

Alternatively, think about what happens if you shrink the small circle…

June 1, 2011 at 1:06 pm

The problem is that the statement “the distance RS must be the circumference of the small circle” is false, even though it’s phrased as some sort of obvious implication of the preceding information. Effectively the small circle is “skidding” along while it rolls, a bit like a snooker ball when first it.

Hence I think this problem is not due to Galileo, but to Steve Davis.

June 1, 2011 at 1:07 pm

How is the small circle attached to the large circle? If it is mud that holds them together, then the problem with the argument is that the small circle would fly off in a vertical direction when point R meets point P.

June 1, 2011 at 1:21 pm

>> Effectively the small circle is “skidding” along while it rolls

But which way does it skid?? 🙂

June 1, 2011 at 1:29 pm

You don’t get any marks for skidding.

June 1, 2011 at 1:54 pm

Shame on you.

June 1, 2011 at 2:57 pm

The difference is that the two points on the wheel are moving at different speeds. The reason why PQ = 2*pi*OP (O is the origin of the wheel) is that the time taken to make one rotation equals the time to move from P to Q at a constant velocity v which is equal to the tangential velocity of P. At R the tangential velocity is lower than the translational velocity of the wheel so this no longer holds.

June 1, 2011 at 5:05 pm

The point on the circumference is grounded, but the other one is pi in the sky.

June 1, 2011 at 5:25 pm

Hmm…. I’m curious about the construction of this wheel system. If they’re the same piece, then they would both be moving with the same rotational velocity. Because they have the same rotational velocity, there’s no slipping. However, the difference in radius means that the outer wheel has a higher tangential velocity, making this example appear paradoxical.

So, and correct me if I’m wrong, the problem with the argument lies in this part: Similarly, therefore, the distance RS must be the circumference of the small circle.

The phrasing makes it seem obvious, but the problem, I think, lies in the fact that the small circle is not the one rolling. The distance traveled depends solely upon the outer circumference and the velocity of the wheel.

June 2, 2011 at 1:07 am

[…] A Paradox of Galileo Going by the popularity of the little physics problem I posted last week, I thought some of you might be interested in […] […]

June 2, 2011 at 1:29 am

Maybe this will make visualization easier:

Imagine the two circles are gears, still firmly attached. For convenience, let the smaller circle be 1/2 the diameter of the larger (this argument will be similar for any ratio of diameters).

Now imagine two more gears, both the same size as the larger circle. One gear meshes with the larger circle, the other with the smaller. Obviously, they can’t both be on the same shaft, so they can rotate at different speeds.

Since the first gear is the same size as the larger circle, it rotates at the same speed as the larger circle.

But the second gear, meshing with the smaller circle, will be a “lower” gear, and will rotate at 1/2 the speed as the first gear.

So after one revolution of the two original circles, a point on the circumference of the first gear will have travelled the distance PQ. But a similar point on the second gear will have travelled 1/2 as far.

So, since the two circles are concentric, when the larger circle moves forward the distance PQ, so must the smaller circle. But only 1/2 that distance (in this example) is due to the smaller circle’s rolling, and the rest must result from slipping. Galileo is spoofing us.

June 2, 2011 at 9:48 pm

The actual distance traveled along the arc is less for the smaller circle by a factor equal to the ratio of the radii. As the arc is shallower for the inner circle it means that it travels a larger transverse distance. Make sense?

June 5, 2011 at 6:38 am

Well, obviously it’s rolling on the large circle, not the small one. Imagine if it were rolling on the hub instead, then allow the hub to shrink to an arbitrarily small diameter. As it approaches zero, RS also approaches zero, yet the diameter of the large circle hasn’t changed at all. To top it off, the speed of rotation would approach infinity! So there isn’t any slippage; you’re just measuring something other than the diameters when trying to compare the two circles, tho with my limited knowledge I don’t know what it’s called. Neat problem tho…

June 9, 2011 at 8:12 pm

Imagine a wheel with a fixed drive shaft. Construct a shelf parallel to the ground with its upper surface tangential to the underside of the drive shaft (in other words line RS).

Now roll the wheel. Clearly, the drive shaft will roll but not enough to travel from R to S. The rest comes from slipping. Not sure how to correctly describe the direction of slippage – I’d call a “braking” slippage as the drive shaft or inner circle is “constrained” from rolling more rapidly – a kind of partly rolling skid.

Building on SteveT above, shrinking the inner circle/shaft reduces the amount of roll and increases the amount of slippage or skidding.

These are fun – I can think about them visually and physically but should learn some of the math.

February 13, 2014 at 4:20 am

The Line RS wouldn’t be straight since it has excess length that’s not rolled up by the smaller circle. It would be sagging.