Perplex City - We Love Puzzles

Perhaps they mean the speed at which 1 rotates about 2 is twice the speed at which the free spinning cog 1 rotates?

Shouldn't they both rotate about at the same speed if they are connected gears. Unless, as you say Cog 1 is free spinning but then that would mean it could stay perfectly still as well.

It depends what you are measuring the rotation relative to. Relative to each other, they will be the same angular velocity but in opposite directions. Relative to the viewer, as A moves over B it must rotate more than B as it travels. Consider the situation where B stays fixed and A moves over B. B doesn't rotate, but A certainly will.

Cog 2 driving at v
Cog 1 rotating about B at 2v but "free-wheeling"

I've tried cog 1 rotating about A at twice the speed of cog 2 rotating about B and cog 1 rotating about B at twice the speed of cog 2 rotating about B.
I've also tried to get cog 1 to the opposite side of where it is touching cog 2 at the start (cog 1 ends up somewhere between the indicated start and end points on the card). But nothing seems to produce a correct answer.

The trick is, that for every turn the fixed cog makes, the other one rotates about it 180 degrees. THAT's what it means by the speed of one is half that of the other. Terribly worded card...

You need to look at how much Gear A rotates not from the reference frame of the gear itself, but you looking at the whole contraption. If you look at one spot on Gear A as it moves around, you will see it rotate 360 degrees by the time it reaches the other side of Gear B. However, if you were just looking at it from Gear A's perspective, it would have only gone 180 degrees.

Since Gear B moves half as fast as Gear A in the opposite direction, it will have added another 90 degrees to the total travel distance. This would be have the rotation amount, of 180 degrees.

So the total will then be 360 + 180 = 540 degrees.

Fixed cog rotates through 180, which moves the other cog through 90 degrees to the 12 o'clock position. Fixed position cog rotates through another 180 degrees, the other one moves through the final 90 degrees to the 3 o'clock position.

The fixed cog has spun 360, the other one has been 'forced' via the teeth of the fixed cog through 180.

360 + 180 = 540

First just consider B doesn't rotate and A just moves over to the other side of B. This means that A does a 360 degree cw rotation.

Now we are told that B rotates at half the speed of A, so by the end B will have rotated ccw (rB) half the degrees of A's total rotation cw (rA). Because they are the same size B's ccw rotation will be added cw to A's rotation caused by the arm moving.

rA = 360+rB (A's total = arm rotation + same as B)
rB = rA/2 (B's total rotation = half A's total)

So solve the simultaneous equations:
rA = 360+rA/2 --> 2rA - rA = 720 --> rA = 720 ==> rB = 360

But the question asks for A's rotation around axis 1, which has also rotated 180 cw, so we need to subtract that 180 degrees off the final answer for rA.

Answer = rA - 180 = 720 - 180 = 540.