Tuesday, September 21, 2010

My Eyes Are Buggin’

I had an interesting conversation with one of my physics students recently.  She had a dance class right before she came to lab.


Her: I hate dancing with glasses.

Me: Because they fall off when you’re doing the turn-y things?

Her: Yes, it’s so much easier with contacts.

Me: You know, you can calculate how fast you’d have to turn to get the contacts to pop out.


How fast would a dancer have to spin to get her contacts to pop out?

Contact lenses weigh about 0.05 g and are about 1.0 cm2 in area.  They are held on mostly by suction.  Suction occurs because there is a partial vacuum created under the contact as it's pulled away from the eye.  To pop out, a contact needs to overcome air pressure, which for our atmosphere is about 100 kPa.  As the dancer spins, her contacts move in a circle with roughly a 10 cm radius.  The force to keep an object in uniform circular motion is given by

force = (mass) · (velocity)2/ (radius).

The maximum this force can be before the contact pops out is determine by atmospheric pressure

force = (pressure) · (area).

Solving for the velocity, we get

velocity = [(pressure) · (area) · (radius) / (mass)]1/2
= [(100 kPa) · (1 cm2) · (10 cm) / (0.05 g)]1/2
= 140 m/s.

A dancer would have to rotate about 1400 per second to get her contacts to pop out.


2 comments:

  1. Something tells me her eyes wouldn't be far behind.

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  2. What would the number be if it was surface tension of the thin liquid layer between the contact and the surface of the eye? ;)

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