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3 years ago

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A circular-motion addict of mass 82.0 kg rides a Ferris wheel around in a vertical circle of radius 10.0 m at a constant speed o

f 7.10 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:

alisha [4.7K]3 years ago

8 0

Answer:

(a) Time period is 8.85 s

(b) Normal force at the highest point is 390.2 N

(c) Normal force at the lowest point is 1217 N

Explanation:

Given:

Mass of person, m = 82.0 kg

Radius of the wheel, r = 10.0 m

Speed of the wheel, v = 7.10 m/s

(a) Time period of the circular motion is determine by the relation:

$T=\frac{2\pi r}{v}$

Substitute the suitable values in the above equation.

$T=\frac{2\pi\times10 }{7.10}$

T = 8.85 s

(b) The normal force ( F ) at the highest point of the circular path is given by the relation:

F = F₁ - F₂ ....(1)

Here F₁ is gravitational downward force acting on the person and F₂ is the centripetal force.

Gravitational Force, F₁ = mg

Here g is acceleration due to Earth's gravity.

Centripetal force, F₂ = mv²/r

Thus, the equation (1) becomes:

$F=mg-\frac{mv^{2} }{r}$

Substitute the suitable values in the above equation.

$F=82\times9.8-\frac{82\times7.10^{2} }{10}$

F = 390.2 N

(c) The normal force ( F ) at the lowest point of the circular path is given by the relation:

F = F₁ + F₂ ....(2)

Here F₁ is gravitational downward force acting on the person and F₂ is the centripetal force.

Thus the equation (2) becomes:

$F=mg+\frac{mv^{2} }{r}$

$F=82\times9.8+\frac{82\times7.10^{2} }{10}$

F = 1217 N

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