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You go to an amusement park with your friend Betty, who wants to ride the 80-m-diameter Ferris wheel. She starts the ride at the lowest point of a wheel that, as you face it, rotates counterclockwise.

Part A

What is the magnitude of her displacement vector when the wheel has rotated by an angle of 40 ∘?

Express your answer with the appropriate units.

Part B

What is the direction of her displacement vector when the wheel has rotated by an angle of 40 ∘, measured counterclockwise from the horizontal?

Respuesta :

MechEngineer

Answer:

A. 27.4 m

B. 0.35 rad or 20 degrees

Explanation:

A Ferris wheel with 80m in diameter would have a radius of 40m

Let the center of the wheel be the origin, so the vector (let call it a) from the the center of the wheel to Betty's starting (lowest) point is

[tex]\vec{a} = <0, -40>[/tex]

When the wheel rotated by 40 degrees, Betty would have traveled a distance of:

horizontally [tex]rsin\theta = 40sin40^0 = 25.71 m[/tex]

vertically [tex]r - rcos\theta = 40 - 40cos40^0 = 9.36 m[/tex]

So her displacement vector is <25.71, 9.36>

A. Magnitude of this vector is

[tex]\sqrt{25.71^2 + 9.36^2} = 27.4 m[/tex]

B. Direction of this vector is

[tex]tan\alpha = 9.36 / 25.71 = 0.364 [/tex]

[tex]\alpha = tan^{-1}0.364 = 0.35 rad [/tex] or 20 degrees

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