Question

A Ferris wheel of radius 100 feet is rotating at a constant angular speed ω rad/sec counterclockwise. Using a stopwatch, the rider finds it takes 4 seconds to go from the lowest point on the ride to a point Q, which is level with the top of a 44 ft pole. Assume the lowest point of the ride is 3 feet above ground level. Let q(t)=(x(t),y(t)) be the coordinates of the rider at time t seconds; i. e., the parametric equations. assuming the rider begins at the lowest point on the wheel, then the parametric equations will have the form: x(t)=rcos(ωt-ω/2) and y(t)=rsin(ωt -ω/2), where r,ω can be determined from the information given. provide answers below accurate to 3 decimal places. (note: we have imposed a coordinate system so that the center of the ferris wheel is the origin. there are other ways to impose coordinates, leading to different parametric equations.)

Find r, ω during the first revolution of the wheel, find the times when the rider's height above the ground is 80 feet.

Answer 1

Answer:

First time t = 5.699 s

Second time t = 21.049 s

Step-by-step explanation:

Find the change in angle for 4 seconds.

cos(Δθ) = (100 + 3 - 44 ft) / (100 ft)

Δθ = 0.9397375 rad

Calculate ω.

ω = Δθ/Δt

ω = (0.9397375 rad) / (4 s)

ω = 0.2349 rad/s

Parametric equation for y:

y(t) = 3 ft + (100 ft) sin(0.0.2349t - π/2)

Find first 2 times for y = 80 ft.

80 = 103 + 100 sin(0.2349t - π/2)

-0.23 = sin(0.2349t - π/2)

-0.232078 + 2πn = 0.2349t - π/2 or (π + 0.232078) + 2πn = 0.2349t - π/2

First time:

-0.232078 = 0.2349t - π/2

1.338718 = 0.2349t

t = 5.699 s

Second time:

π + 0.232078 = 0.2349t - π/2

4.944467 = 0.2349t

t = 21.049 s

Answer 2

The vertical and horizontal motion of the Ferris wheel are combined to

give the locus of a point on the rotating wheel.

• a) r = 100 feet

• b) ω ≈ 0.235 rad/s

• c) First time ≈ 5.697 sec, Second time ≈ 21.041 sec

Reasons:

The given parameters are;

Radius of the Ferris wheel = 100 feet

Time it takes the rider to go from the lowest point to the point Q which is

the top of 44 ft. pole = 4 seconds

The coordinate of the rider's location, Q(t) = (x(t), y(t))

x(t) = r·cos(ω·t - π/2)

y(t) = r·sin(ω·t - π/2)

The center of the Ferris wheel = The origin

a) The radius r = 100 feet

b) The angular velocity is given by the arccos of the ratio of the distance of

the center of the Ferris wheel to the radius of Ferris wheel as follows;

$cos (\theta) = \dfrac{100 + 3 - 44}{100} = \dfrac{59}{100}$

$The \ angle \ of \ rotation \ \theta = arccos \left(\dfrac{59}{100} \right)$

$The \ angular \ velocity, \, \omega =\dfrac{\theta}{t} = \dfrac{ arccos \left(\dfrac{59}{100} \right)}{4} \approx 0.235$

The angular velocity, ω ≈ 0.235 rad/s

c) When the height 80 feet, we have;

$\theta = arccos \left \left(\dfrac{100 + 3 - 80}{100} \right) = arccos(0.23)$

$Time, \ t = \dfrac{\theta}{\omega}$

Which gives;

$t_1 = \dfrac{arccos \left(0.23 \right)}{0.235} \approx 5.697$

The first time, t₁ ≈ 5.697 seconds

The second time, is given by t₂ = t₁ + Δt

Where;

$\Delta t = \dfrac{2 \cdot \pi - 2 \times arccos(0.23)}{0.235} = 15.344$

Which gives;

t₂ = 5.697 + 15.344 = 21.041

The second time the Ferris wheel is at the height 80 feet, t₂ ≈ 21.041 s.

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