Question

You go to an amusement park with your friend Betty, who wants to ride the 70-m-diameter Ferris wheel. She starts the ride at the lowest point of a wheel that, as you face it, rotates counterclockwise.
Part AWhat 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 BWhat is the direction of her displacement vector when the wheel has rotated by an angle of 40 ∘, measured counterclockwise from the horizontal?

Answer 1

Answer:

Part a)

$d = 23.94 m$

Part b)

$\theta = 110 degree$

Explanation:

Part a)

As we know that initially the position vector is r

then the same magnitude position vector is rotated by 40 degree angle

so displacement magnitude is the magnitude of change in position vector

so it is given as

$d = \sqrt{r_1^2 + r_2^2 + 2r_1r_2cos(180-\theta)}$

$r_1 = r_2 = 35 m$

$d = \sqrt{35^2 + 35^2 -2(35)(35)cos40}$

$d = 23.94 m$

Part b)

now we need to find the direction of the displacement vector

so let say it makes an angle with x axis so we have

$tan\theta = \frac{rsin\phi}{r - rcos\phi}$

$tan\theta = \frac{sin 40}{1 - cos40}$

$\theta = 110 degree$