Saturday, February 11, 2006

Weightlessness in a Ferris Wheel

Centripetal and Normal Forces

How many Revolutions per Minute would a 20m Diameter Ferris Wheel have to make in order that the occupants are weightless at the top of the wheel? How much do the occupants weigh at the bottom of the wheel? Describe what happens at the top if the rotation rate is further increased?

If the Ferris Wheel is in constant, uniform, rotational motion the net force on a body on the rim of the wheel must be the correct centripetal force, Fc, always pointed towards the center of the wheel. The actual forces acting on the occupants are gravity (downwards) and some sort of normal force between the body and the structure of the wheel.

Lets assume the occupant is standing in a carriage that is free to rotate so that the ground is always down (at least until it all becomes weightless) - then the normal force between the occupant and the carriage is pointed directly upwards. Thus (taking upwards to be positive):


But, we know that Fnet must be equal to Fc, where


where R is the radius of the wheel and the negative is because the centripetal force points downwards atthe top of the wheel.




Now, at the exact speed required for weightlessness, the normal force must go to zero (the condition for weightlessness):



Now, we have the tangential velocity required for weightlessness. In order to find the equivalent "revolutions per minute" we first work out the time for a single revolution, T:

    T=2 Pi R/v

    T=2 Pi Sqrt(R/g)

If we are working in SI units, the units of T are seconds per revolution. Then the number of revolutions in 1 minute, N, is:

    N=(60 seconds/minute)/(T seconds/revolution)


Substituing in our values, we see that:

    N=9.5 rpm

Now, at the bottom of the wheel, the Normal force is still pointed upwards and graivty is pointed downwards, but the centripetal force that we need to create has to point upwards. Thus there is no way to create weightlessness here! However, we can calculate the weight we experience at the bottom, because:

from force addition and


from the requirements for uniform circular motion

thus, at the bottom. one feels twice ones normal weight! If the rpm was slowly increased past the value worked out above, we would find that at (and around) the top of the wheel, there would be no forces immediately able to create the required centripetal force and the uniform motion would break down. The occupants would "be shot off in a tangential direction" and move like a projectile under gravity. If the speed was just a little above gravity, the occupants would rise off the ground and land a little later. Also, if the carriage was free to rotate all the way around its connection to the wheel, it would tend to flip upside down during the "less than weightless" phases of motion. If the velocity was high enough, the carriage would flip and then continue in (a new) uniform circular motion - the centripetal force provided by the new downwards normal force provided by the strong connection betweent he carriage and the wheel - and the occupant would also probably find them selves pressing against the floor - which is now the cieling in the upside down carriage.

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