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I was sitting down yesterday and saw my phone vibrate on a side, and it moved about a centimetre per vibration. I wondered why it moves, and thought perhaps that the side it was on had a slight slope, because it tended to move in the same direction every time. So I then though, what if the side was perfectly flat? Well, then I considered the fact that the phone's centre of mass would make it move in a biased direction (is this correct?).

So I then thought, what if the phone's centre of mass was in the exact centre, and the slop was perfectly flat?

Would the phone still move?

And if so, what factor would make it move in a particular direction?

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4 Answers 4

up vote 3 down vote accepted

So I then thought, what if the phone's centre of mass was in the exact centre, and the slope was perfectly flat?

Close, though what you need is not for the phone's centre of mass to be in the centre. You need the force produced by the vibrating element to pass through the centre of mass (and be normal to the surface(.


In the example I've drawn above, the force produced by the vibration passes through the COM so there is no net torque on the phone and it will rise straight up even though the vibration isn't coming from the centre. If you move the force to the left of the COM it will apply a clockwise torque to the phone. The friction between the right edge of the phone and the table will pull the phone to the right. Likewise, moving the force to the right of the COM will tend to pull the phone to the left.

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This is a really helpful answer. Could the effect be calculated by simple Principle of Moments calculation? Also, how can I draw diagrams like that and get them onto answers/questions here? –  Olly Price Apr 29 '13 at 18:09
@OllyPrice: I use the Drawing app in Google Drive to create my drawings. I screenshot the drawings and save them as a GIF, then insert the GIF into my posts. Google Drawing is rather basic, but it's fine for the sort of diagrams I need for this site. –  John Rennie Apr 30 '13 at 5:43

If the phone was placed onto a perfectly flat, frictionless surface, then there would be no net movement after it vibrates. This is due to conservation of momentum. The phone's center of mass does not matter in this case.

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Well, you should qualify that no net angular momentum is induced by the vibration ( it could be conserved by radiation em and acoustic). If there is, it will rotate. IMO also if the vibration devise in the phone is asymmetric, again there will be a motion from a net effect of radiation leaving. So that symmetry should be in the assumptions. –  anna v Apr 18 '13 at 4:17

Since vibration comes from a rotary motor with off-center weight.

Now as the weight ceases to move upwards, the phone is lifted off the surface a little. Now the weight moves sideways, and pulls the phone in opposite direction as it moves - the phone moves almost frictionlessly as it's lifted off the surface by the prior sudden jerk. Then it drops, and finally so drops the phone. Now the return movement is with the phone lying flat on the surface with much higher friction as it pushes with its whole weight.

In essence, the mechanism is based on increasing and decreasing friction by shaking the phone up and down, and applying lateral force in one direction when the friction is low, and in opposite when the friction is high. The result is horizontal movement in direction where the force acts during the low friction phase.

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No it won't move. Simple prove is tide. You can consider moon as a motor with asymmetric center of gravity. So it makes water moving back and further, however water doesn't move in any direction constantly.

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