Question

# A circular-motion addict of mass 80 kg rides a Ferris wheel around in a vertical circle of radius 12 m at a constant speed of 5.5 m/s. (a) What is the period of the motion? What is the magnitude of the normal force on the addict from the seat when both go through (b) the highest point of the circular path and (c) the lowest point?

Solution

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To solve this problem we will use the fact that the circumference depends on the radius and calculate the time needed for one full revolution. Then, we will note that at the top the centripetal force and the gravitational force have the same direction whereas at the very bottom their direction is opposite. Let's do it.

a) The period needed for one full revolution can be found from

$T=\frac{L}{v}$

where he path covered by the rider $L$ is given as $L=2\pi R$
so we have that

$T=\frac{2\pi R}{v}=\frac{2\pi 12}{5.5}$

$T=13.7\textrm{s}$

b) At the highest point the normal force from the sit on the rider is gonna be given as the difference of the magnitudes of the centripetal force $F_c=mv^2/R$ and the gravitational force $F_g=mg$

$F_n=mg-\frac{mv^2}{R}=80\times 9.8-\frac{80\times 5.5^2}{12}$

which gives that

$F_n=582\textrm{N}$

c) At the lowest point the normal force from the sit on the rider is gonna be given as the sum of the magnitudes of the centripetal force $F_c=mv^2/R$ and the gravitational force $F_g=mg$

$F_n=\frac{mv^2}{R}+mg=\frac{80\times 5.5^2}{12}+80\times 9.8$

which gives that

$F_n=986\textrm{N}$

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