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3 years ago

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A circular-motion addict of mass 83.0 kg rides a Ferris wheel around in a vertical circle of radius 13.0 m at a constant speed o

f 6.10 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?

1 answer:

Svetllana [295]3 years ago

5 0

Answer:

(A) Time period T = 6.28 SEC

(B) At highest point fore is - 575.83 N

(B) At lowest point force is 1050.97 N

Explanation:

We have given that mass m = 83 kg

Radius r = 13 m

Speed v = 6.10 m/sec

(A) Time period of the motion is given by $T=\frac{2\pi r}{v}=\frac{2\times 3.14\times 13}{6.10}=6.28sec$

(b) Net force is given by $F_{NET}=\frac{mv^2}{r}=\frac{83\times 6.10^2}{13}=237.571N$

Force due to gravity $F_{gravity}=mg=83\times 9.8=813.4N$

At highest point $F_{NORMAL}=F_{NET}-F_{GRAVITY}=237.57-813.4=-575.83$

(B) At lowest point $F_{NORMAL}=F_{NET}+F_{GRAVITY}=237.57+813.4=1050.97N$

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For the magnetic field we have the next equation:

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It is the same as E. Here we just need to find B(max).

We can use this equation:

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$B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}$

$B_{max}=1.17*10^{-8}T$

Putting this in (3), finally we will have:

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