A Spring Physics Problem
It’s been a while since I posted anything in the Cute Problems category, so since Spring is in the air I thought I’d post a physics problem which involves springing into the air…
Two identical fleas, each of which has mass m, sit at opposite ends of a straight uniform rigid hair of mass M, which is lying flat and at rest on a smooth frictionless table. If the two fleas make simultaneous jumps with the same speed and angle of take-off relative to the hair as they view it, under what circumstances can they change ends in one jump without colliding in mid air?
UPDATE Monday 10th March: No complete answers yet, so let’s try this slightly easier version:
Two identical fleas, each of which has mass m, sit at opposite ends of a straight uniform rigid hair of mass M, which is lying flat and at rest on a smooth frictionless table. Show that, by making simultaneous jumps with the same speed and angle of take-off relative to the hair as they view it, the two fleas can change ends without colliding in mid-air as long as 6m>M.
Answers via the comments box please..
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In the Dark 
March 9, 2014 at 2:39 pm
Is the hair rotating?
March 9, 2014 at 2:58 pm
No, not to begin with…
March 9, 2014 at 2:41 pm
They can’t exchange ends – if they are both at the same angle, they are both heading in the same direction!
March 9, 2014 at 2:43 pm
are they?
March 9, 2014 at 2:54 pm
Yes, unless the two angles are in a different coordinate system. My hair, incidentally, is on a table, as you didn’t specify where it was.
March 9, 2014 at 2:56 pm
Yes, I forgot to specify the position of the hair.
The two flies do the same jump in terms of speed and angle, so what each flea sees is the same.
March 9, 2014 at 2:49 pm
The orientation and velocity of the hair was not specified; could this be significant? Assuming it’s not in a vacuum & standard gravity?
March 9, 2014 at 2:52 pm
An inclined hair
March 9, 2014 at 2:57 pm
No, I’ve clarified the problem – the hair is horizontal, lying on a smooth table.
March 9, 2014 at 3:16 pm
Too late – get it right first time or fail! Seriously though, one of my fleas is at the end of the hair and jumping off the hair away from the hair, so it doesn’t matter what the hair does in terms of rotation, the flea will never return to the hair. It is moving away from the hair.
March 9, 2014 at 3:18 pm
Well, then you’ve failed 😉
March 9, 2014 at 3:24 pm
Not at all – you are suffering from the classic academic’s fallacy that a problem only has one solution. That’s what my fleas did. 😉
March 9, 2014 at 3:02 pm
OK, so if they both jump sideways and impart a spin to the hair around its central point, the hair would rotate under them and they land on the other ends.
March 9, 2014 at 3:19 pm
Doesn’t have to be completely sideways, just with a sideways component….
March 9, 2014 at 3:23 pm
Ah yes, otherwise they’d go off at tangents and miss the landing
March 9, 2014 at 3:02 pm
an even number of jumps
March 9, 2014 at 3:08 pm
One jump
March 9, 2014 at 3:06 pm
What I win?
March 9, 2014 at 3:14 pm
A vertical jump, and I rotate the hair
March 9, 2014 at 4:59 pm
The fleas should jump at an angle A to the long axis of the hair (so that they jump off opposite sides of the hair). By conservation of (angular) momentum, the hair will then be set rotating in the opposite sense. This brings the far end of the hair round into the path of each falling flea. By geometry, the angle H the hair rotates by must be equal to 2A.
(You can conserve angular momentum to find another relation between H and A in terms of m, M, and L (hair length), and eliminate one of the angles to get A in terms of m,M,L, but it’s not very pretty.
March 9, 2014 at 5:32 pm
You’re on the right lines. Say the fleas jump at angles (pi/2-theta/2) to the initial direction of the hair; then the impulsive couple generated must make the hair rotate by an angle (pi-theta) if they are to land at opposite ends from where they take off. You need to consider the launch speed v, the angle of take-off alpha, and the length of the hair in order to derive a necessary condition for the operation to be possible…
March 10, 2014 at 9:22 am
If the hair is lying North-South and not on the equator?
March 10, 2014 at 12:49 pm
I don’t understand: no solver!
For the conservation of the angular momentum, there is an evaluation of the constant angular velocity of the hair (for the initial impulse of the fleas); if I don’t make mistake \omega = 12m v/ (M L^2).
So that the other two velocities (vertical and along the hair) are arbitraries: if the fleas know the ballistic, they don’t miss the target, and for each tangential velocity there are infinite solutions.
It is a fun problem.
March 10, 2014 at 12:56 pm
It helps to focus on the time the fleas spend in the air…
March 10, 2014 at 4:10 pm
The time of flight is 2 s/g where the s is the velocity along the gravity.
The distance that the flea travel is L cos[12 m v t / (M L^2)], for the constraint in the circular motion.
The distance that the flea travel for the ballistic trajectory is t \sqrt{u^2+v^2}.
These two distance must be the same: L cos[12 m v t / (M L^2)] = t \sqrt{u^2+v^2} so that g L cos[24 m v s/ (g M L^2)] = 2 \sqrt{(u^2+v^2) s} or cos( A v s) = B \sqrt{u^2+v^2} so that the for each tangential velocity v, there are infinite solutions (singular solution when v or s are zeros).
There a mix of kinematics, dynamic, conservation laws … poor fleas.
March 10, 2014 at 4:28 pm
Fleas are well-known circus performers. I don’t think they’d do the maths but rather practice until they got it right.
March 10, 2014 at 10:55 pm
Let L be half the length of the hair, v be the horizontal component of each flea’s velocity, T be the time of flight, and theta be the angle between the direction of v and the hair (in the horizontal plane). The angular momentum imparted during the jumps is 2mvL sin(theta), and the hair’s moment of inertia is ML^2/3, so the hair acquires angular speed omega = 6(m/M)(v/L) sin(theta).
In order for the flea to land on the hair, the hair must turn through an angle (2 theta + n pi) for some integer n. (A picture would help here.) I’m going to leave off the + n pi part, because unless the fleas are very massive they can’t make the hair turn more more than the minimum amount. You can put that back in and see what happens if you want. So
6(m/M)(v/L) sin(theta) T = 2 theta.
The horizontal distance each flea travels is
v T = 2L cos(theta).
Eliminating T gives
6(m/M)(2 sin(theta) cos(theta)) = 2 theta
where I deliberately didn’t cancel some 2’s because I wanted to make the substitution u = 2 theta to get
6(m/M) sin(u) = u.
If 6(m/M)<1, then the left side is strictly less than the right side. If it's greater than 1, then there's a solution for positive u. (Taking derivatives, we see that the left side is greater than the right for sufficiently small u, but clearly left is less than right for sufficiently large u.)
I guess there are infinitely many choices for the initial velocity vector: the above equations can be solved for v T, but you can choose any value you like for v (which is just the horizontal velocity component) and then choose the initial vertical velocity to get the corresponding T.
March 11, 2014 at 7:23 am
Bravo! So you see there is no solution unless 6m>M…so in terms of the first wording of the problem that’s the circumstance that makes it possible.
March 11, 2014 at 10:26 pm
I’m sure this problem would be easier if it involved frogs jumping on a stick floating on a lake, instead of fleas on a hair. 😉