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November 29, 2014

November 29, 2014

Posted by **Greg** on Saturday, December 10, 2011 at 11:28am.

A former amusement park attraction consisted of a huge vertical cylinder rotating rapidly around its axis. The rotation speed should be sufficient for the people found inside to be stuck to the wall when the floor was removed. what should be the minimum speed of rotation for this to happen? The coefficient of static friction between the people and the wall of the cylinder is 0.300 and the cylinders radius is 4 meters.

Solution:

F = mv^2\r

fs = mv^2\r

us(mg) = mv^2\g

us = v^2\g

0.300 * 9,8 m\s^2 = v^2\4m.

2.94 m\s^2 * 4m = v^2

square root (11.76 m^2\s^2) = v

v = 3.43 m\s

Thank you

- Physics (verification)2 -
**drwls**, Saturday, December 10, 2011 at 1:55pmSome of your equations are wrong, even dimensionally. For example, us is dimensionless but v^2/g has dimensions of length.

The equation should be

(M V^2/r)*us = M g

V = sqrt(r*g/us) = 11.43 m/s

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