A ferris wheel has an axle standing 38m off the ground and a radius of 35m. The wheel takes 5 minutes to complete one revolution. The wheel moves in clockwise motion and you are sitting in one of the carriages which is horizontally in line with the axle, moving upwards.
Form an equation that models the displacement of the carriage from the axle one revolution.
The height of the axle does not matter.
Let the axle be at (0,0)
Assuming that at t=0 the carriage is horizontally in line with the axle,
x(0) = 35
y(0) = 0
The amplitude is 35.
The period is 5, so we are looking at
y(t) = 35sin(2pi/5 t)
x(t) = 35cos(2pi/5 t)
x^2 + y^2 = 35^2
TASK 1:
Consider that it is sunset with the sun’s rays parallel to the rim of the wheel, so that the wheel’s shadow is falling on to the vertical wall of the building.
The car’s shadow moves up and down the wall. You may like to use a wheel and torch to experience what this may look like.
I. Form an equation that models the displacement of the carriage from the axle one revolution. (B maps) (State any assumptions that you have made and the effects they have on the overall result)(ASSUMPTIONS AND EFFECTS MAPS A OR B)
II. Sketch this equation with the aid of technology. State the points of interest on this graph. (C KAPS)
III. Consider vertical velocity of the carriage.
When does the shadow appear to be :
a. Stationary
b. Travelling the fastest upwards
c. Travelling the fastest downwards (C KAPS)
IV. Manually sketch the vertical velocity of the carriage. (This does not need to be accurate, a sketch indicating the general shape of the curve is sufficient)(C KAPS)
V. State the period, amplitude and possible equation for the curve. (C KAPS)
VI. Calculate the derivative of the displacement function and compare this to the vertical velocity function graphically and verbally. (COMPARISON MAPS C)
TASK 2: KAPS AND MAPS
1. Find the equation that models the height of the car over time. (B MAPS)
2. Use this equation to find:
a. The height of the carriage after 2 minutes into the ride and the direction of its motion. (KAPS)
b. The times (minutes and seconds) during one revolution, when the carriage is 20m off the ground.(KAPS)
3. Find the equation that models the height of the car, if we start our analysis one minute into the ride. (B MAPS)
4. Using the original equation (starting at time zero), find an equation to represent height, if the carriage now takes 10 minutes for one revolution. (B KAPS)
5. Assuming that the sun is now directly above your carriage so that the shadow of the carriage you are in is directly beneath you. Write the equation that models the displacement that the car’s shadow is from the shadow of the axle. (STATE THE STRENGTHS AND LIMITATIONS OF THE DEVELOPED MODEL) (A LEVEL)
TASK 3: KAPS AND MAPS
1. The equations for height and velocity from previous tasks were expressed using degrees. Using fractions, rewrite them using radians. (KAPS)
2. Find the maximum upward velocity of the wheel and the time this occurs within one revolution. (KAPS/MAPS)
3. Using the ‘rule’ for speed, calculate the constant speed of your carriage and then compare this with the maximum upward velocity.( KAPS) (COMPARISON MAPS C)
Throughout this entire EMPS, there are opportunities to justify your responses and the reasonableness of them. Data is modeled regularly and you have an opportunity to state the strengths and limitations of each of them. Ensure your responses are logical, coherent and concise, to ensure that you have the opportunity to receive a high grade. Follow the criteria to achieve.
To form an equation that models the displacement of the carriage from the axle, we need to consider the vertical and horizontal components separately.
First, let's focus on the vertical component. Since the carriage is moving upwards, the vertical component of the displacement is the height gained during one revolution. We know that the radius of the Ferris wheel is 35m and the axle is 38m off the ground. This means that the highest point the carriage reaches during one revolution is 35m + 38m = 73m above the ground.
To calculate the vertical displacement, we need to find the difference between the initial height (38m) and the highest point reached (73m). So the equation for the vertical component of the displacement, denoted as "Dv", would be:
Dv = 73m - 38m = 35m
Next, let's consider the horizontal component. Since the carriage is in line with the axle and moving upwards, there is no horizontal displacement. Therefore, the equation for the horizontal component of the displacement, denoted as "Dh", is:
Dh = 0m
Combining the vertical and horizontal components, the equation that models the displacement of the carriage from the axle during one revolution is:
D = √(Dv^2 + Dh^2)
D = √(35m^2 + 0m^2)
D = √1225m^2
D = 35m
So, the equation for the displacement of the carriage from the axle during one revolution is D = 35m.