the ferris wheel problem
problem statement:
On a ferris wheel whose center is 65 feet from the ground and whose radius is 50 feet, a diver and his assistant spin counterclockwise starting at “3:00”. A tub of water is constantly moving on a track that is parallel to the ferris wheel’s plane. The track starts 240 feet to the left of the center of the ferris wheel. The divers assistant must determine the exact moment to release the diver from a platform suspended from a moving ferris wheel so that he lands in the tub of water.
Process:
Create a ferris wheel: To start off the project, we were told that we had to model what was being asked of us. We were put into groups and had to create a ferris wheel, track, and cart out of popsicle sticks. This would give us a visual as to what we are working with.
Horizontal position: 50sin(90-9t)=x
Vertical position: 50sin(9t)[+ or - 65 depending on whether the section of the circle the diver is in is above(+) or below(-) the diameter of the ferris wheel]
50: radius of ferris wheel
9t: the number of degrees that the ferris wheel moves per second (9 degrees) multiplied by the time the ferris wheel moves
(+) or (-) 65: if the quadrant of the ferris wheel in which the diver is located above the diameter, which is 65 feet from the ground, then you would add 65 to the height of the diver above the diameter of the ferris wheel. However, if the diver is located in a quadrant that is below the diameter of the ferris wheel, you would subtract the height that you got from 65 feet.
Position of the cart at any time (& in relation to the ferris wheel):
How fast the diver is falling: time position formula: 50cos9t
Finding the formula: -240+15t+15[(57+50sin9t)/16] = 50cos9t
Formula breakdown
-240+15t : Position of the cart
-240 feet - how far away cart was at start, made negative since it
has yet to reach the ferris wheel
15t - The cart is going 15 feet per second
15[(57+50sin9t)/16] : How far the cart goes as the diver falls
15 - height ferris wheel is off the ground
57 - The cart is 8 feet in height, therefore 65-8 = 57
50sin9t - vertical position formula
50cos9t : horizontal position of the diver
Horizontal position: 50sin(90-9t)=x
Vertical position: 50sin(9t)[+ or - 65 depending on whether the section of the circle the diver is in is above(+) or below(-) the diameter of the ferris wheel]
50: radius of ferris wheel
9t: the number of degrees that the ferris wheel moves per second (9 degrees) multiplied by the time the ferris wheel moves
(+) or (-) 65: if the quadrant of the ferris wheel in which the diver is located above the diameter, which is 65 feet from the ground, then you would add 65 to the height of the diver above the diameter of the ferris wheel. However, if the diver is located in a quadrant that is below the diameter of the ferris wheel, you would subtract the height that you got from 65 feet.
Position of the cart at any time (& in relation to the ferris wheel):
How fast the diver is falling: time position formula: 50cos9t
Finding the formula: -240+15t+15[(57+50sin9t)/16] = 50cos9t
Formula breakdown
-240+15t : Position of the cart
-240 feet - how far away cart was at start, made negative since it
has yet to reach the ferris wheel
15t - The cart is going 15 feet per second
15[(57+50sin9t)/16] : How far the cart goes as the diver falls
15 - height ferris wheel is off the ground
57 - The cart is 8 feet in height, therefore 65-8 = 57
50sin9t - vertical position formula
50cos9t : horizontal position of the diver
solution:
The solution to this problem is 12.28 seconds. At 12.28 seconds the diver needs to be released in order for the diver to fall directly inside the moving cart filled with water. At 13 seconds, the diver and the cart are directly on top of each other, but we need to take in consideration that the cart is still moving as the diver is falling towards the ground.
evaluation:
Overall, I though that this problem was very fun. We used a lot of different math skills to solve the entire problem, an I think that in the end, everyone (including myself) learned something new. This problem, in my opinion, was educationally worthwhile mostly for that reason. Also, I think that everyone got the chance to work with and learn from every one of our classmates. If I had to change this problem to make it better, I would probably just make sure that we worked on this problem for about two weeks straight instead of twice a week for several weeks. That way, all of the information and formulas that we learn stays fresh in our minds. But even so, I thought that this problem was very fun and challenging and allowed us to use a lot of critical thinking in combination with math skills.
Self-evaluation:
Based on the work that I did for this POW, I would grade myself an A to an A+ because I contributed everything that I said I would and I think that as a team, we did a really good job at equally distributing the workload.