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I dropped a Tic-Tac:

enter image description here

(no worries, it happened before), and I noticed as it bounced on the floor that it would first jump 20 cm high, and at the next bounce for instance 50 cm high. Shouldn't it always have less energy with each bounce, and therefore bounce less high each time?

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You should stick to Ritter Sport. – NikolajK Jun 15 '12 at 10:31
@Nick - Nice one! Maybe I should. :-) – stevenvh Jun 15 '12 at 11:41

OK, I think I know. I didn't pay special attention, but it's possible that at the lower bounce it had a higher horizontal speed, so that potential + kinetic energy still decrease.

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In this paper, the calculations are detailed. The problem is close to the one you have tried to solve. There are two impact coefficients that need to be known, since they varie with velocity, the problem can not be solved easily. But you were right in assuming the rotational kinetic energy is transferred into linear kinetic energy : – Shaktyai Aug 14 '12 at 15:27

I suspect that on the first bounce a lot of energy went into making the tictac rotate, then on the second bounce the tictac stopped rotating and the energy went back into linear motion.

I don't have any tictacs to hand, but if you'd like to weigh one of yours we could work out how fast it would have to rotate to absorb 30cm worth of bounce potential energy.

Response to comment: actually I didn't need the mass, I just need the length. Since I forgot to ask for the length of a tictac let's assume it's 1cm.

If you look at the moment of inertia of a beam of length L is:

$$I = \frac{mL^2}{12}$$

Strictly speaking the tictac is better modelled as a cylinder, but treating it as a beam is a reasonable approximation and makes the calculation easier. Anyhow, let's suppose you drop the tictac from a height $h$ so it's kinetic energy when it hits the ground is $mgh$. The kinetic energy of a rotating body is $0.5Iw^2$, so if all the kinetic energy is converted into rotation we get:

$$ \frac{1}{2}Iw^2 = mgh $$


$$ w = \sqrt{\frac{2mgh}{I}} $$

Substitute $I = mL^2/12$ and we get:

$$ w = \sqrt{\frac{24gh}{L^2}} $$

Now feed in $d$ = 1 metre and $L$ = 0.01 metres and we get $w$ is about 1,500 radians per second, which is a frequency of rotation of about 250Hz.

OK, I think it's pretty unlikely that the tictac is rotating 250 times a second, so I think your explanation is far more likely i.e. that you're converting vertical motion into horizontal motion. In my defence, you didn't say that the tictac was bouncing along horizontally. If you've ever watch a rugby ball or American football ball bouncing along the ground you'll know that the height of the bounces is very irregular.

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Thanks. I took the trouble of counting how many there are in an 18 g box, and there's 37 of them, so one weighs 0.5 g. Length is 11 mm, I think you need that for rotational energy. – stevenvh Jun 15 '12 at 9:37
I've updated my answer - as you'll see, all I've done is prove I was wrong :-) – John Rennie Jun 15 '12 at 9:43
@ Stevenvh I do not understand your reasoning. You just have proved the tic tac does not convert all its potential energy into rotational kinetic energy when it first touchs the ground. The problem is to determine the repartition of rotational and translational kinetic energy after a rebound. – Shaktyai Aug 14 '12 at 14:14

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