Consider the Ferris wheel above. It 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 are horizontally in line with the axle, moving upwards.
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.
STOP CHEATING EMAD!!!
TASK 1:
I. To form an equation that models the displacement of the carriage from the axle for one revolution, we need to consider the geometry of the Ferris wheel. Since the carriage is horizontally in line with the axle and moves upwards, we can represent the displacement as a function of the angle (in radians) that the wheel has rotated.
Let θ be the angle in radians that the wheel has rotated, and let d(θ) represent the displacement of the carriage from the axle. The radius of the wheel is 35m, and the height of the axle is 38m.
Assuming the carriage starts at the bottom and moves upwards, we can use trigonometry to determine the displacement. As the wheel rotates, the angle θ changes, and the vertical displacement can be calculated using the sine function:
d(θ) = 38 + 35sin(θ)
This equation models the displacement of the carriage from the axle for one revolution. The assumption made here is that the carriage starts at the bottom, and the effect of this assumption is that the equation only holds for one revolution.
II. To sketch this equation, we can use technology such as graphing calculators or online graphing tools. Plotting the graph will show the vertical displacement of the carriage at different angles (θ). The points of interest on this graph are the maximum and minimum values of the displacement, which correspond to the top and bottom positions of the carriage. Additionally, the x-intercepts represent the angles at which the carriage is in line with the axle.
III. For the vertical velocity of the carriage:
a. The shadow appears stationary when the vertical velocity is zero.
b. The shadow travels the fastest upwards when the vertical velocity is maximum.
c. The shadow travels the fastest downwards when the vertical velocity is minimum.
IV. The manual sketch of the vertical velocity of the carriage should indicate the general shape of the curve. It doesn't need to be accurate but should show that the velocity is positive when the shadow is moving upwards and negative when it is moving downwards.
V. The period of the vertical velocity function is the time it takes for one complete revolution, which is 5 minutes in this case. The amplitude represents the maximum value of the vertical velocity, which depends on the radius of the wheel. The possible equation for this curve can be determined by taking the derivative of the displacement function.
VI. To compare the derivative of the displacement function with the vertical velocity function, we can graphically plot both functions and observe their similarities. Verbally, the derivative of the displacement function gives the rate of change of the displacement, while the vertical velocity function gives the actual velocity of the carriage. So the derivative can be considered as the rate of change of velocity.
TASK 2:
1. To find the equation that models the height of the car over time, we need to consider the relationship between the height and the angle of rotation. Since one revolution takes 5 minutes, which is equivalent to 2π radians, and the height changes as the angle changes, we can express the height as a function of time.
Let t be the time in minutes, and let h(t) represent the height of the carriage. Since one revolution takes 5 minutes, we can define the period as T = 5 minutes.
Using the equation for displacement from Task 1, we can substitute θ with ωt, where ω is the angular velocity defined as ω = 2π/T:
h(t) = d(ωt) = 38 + 35sin(ωt)
2.
a. To find the height of the carriage after 2 minutes into the ride, we can substitute t = 2 into the equation h(t) and solve for h(2). The direction of motion can be determined by observing whether the height is increasing or decreasing. If h(2) is greater than the initial height, the carriage is moving upwards. Otherwise, it is moving downwards.
b. To find the times during one revolution when the carriage is 20m off the ground, we can set h(t) = 20 and solve for t. The solutions will give us the times when the height of the carriage is 20m.
3. To find the equation that models the height of the car if we start our analysis one minute into the ride, we can use the same equation as in part 1 but substitute t with t - 1, which represents starting one minute into the ride.
h(t) = 38 + 35sin(ω(t - 1))
4. To find the equation representing the height if the carriage takes 10 minutes for one revolution, we need to adjust the period of the function. The new period is T = 10 minutes, which gives us a new angular velocity ω = 2π/T. Substituting this value into the equation from Task 2.1 will yield the desired equation.
5. Assuming that the sun is directly above the carriage, the shadow of the carriage will be directly beneath it. The equation that models the displacement of the car's shadow from the shadow of the axle can be formed by subtracting the height of the carriage from the height of the axle, and then taking the absolute value. The strengths of this model are that it represents the vertical displacement accurately. However, limitations include the assumption of a flat ground and that the terrain is level.