Design a Roller coaster project help

Help for the Roller Coaster Project!

Name __________________________________________ Science ______

The Standard Roller Coaster Assignment (20 points)

Carefully design and draw a roller coaster to scale on graph paper. Use a compass and a ruler to draw neat lines. Do not exceed a starting height of 75 meters. (5 points)

Choose a scale that makes sense to you. For example, one square = 5 meters in height. Identify your scale in a key at the bottom of your drawing. (5 points)

Calculate the velocity at three different locations. Assume the initial velocity is zero. ("d" is the dropping distance from the top first hill!) Use one of the following equations. (5 points)

Velocity =

Show your work for each calculation on your graph or on a separate sheet of paper. Be meticulous. (5 points)

Example problem 1: Assume that your roller coaster car has an initial velocity of 6 m/s and it drops for 10 meters. What will its final velocity be?

Velocity	=
	= (6)² + 2 x 9.8 x 10
	= 36 + 196
	= 232
	Get the square root
	= 15.2 m/s

Extra credit challenge # 1: G-forces at the Dip of a Hill (15 points)
Calculate the centripetal acceleration and G-forces at the dip of a hill. (5 points) Make sure that you do not exceed 5 G-forces. (5 points) Show your work for each calculation on your graph or on a separate sheet of paper. Be meticulous. (5 points)
C.A. =	r =	Gs felt at top of a loop Subtract a G due to gravity
Gs = C.A. divided by 9.8m/s²		Gs felt at bottom of a loop Add a G due to gravity

Example problem 2: Let's assume that the bottom of the first dip is 10 meters from the top of the ride. Let's limit the number of Gs to 3, which is roughly equivalent to a centripetal acceleration of 30 m/s². When gravity is taken into account, our people will feel 4 Gs. We will not exceed 5 Gs. That is good.

To solve this problem, we need to determine our radius.

radius	=
	= 2 x 9.8 x 10/ 30
	= 196/30
	= 6.5 meters

Extra credit challenge # 2: Calculate a parabolic hill

that will allow your car to free fall! (20 points)

Calculate the velocity at the top of a hill:

Velocity =

(5 points)

Use D=VT and D=1/2gt² to calculate the position of the coaster car at 1 second, 2 seconds, 3 seconds, etc…. Be meticulous. (15 points)

Draw the other side of the hill to match.

Example problem 3: The car is traveling uphill. What velocity will it have when it is 4 meters below the starting hill?

Velocity	=
	= (6)² + 2 x 9.8 x 4
	= 36 + 78.4
	= 114.4
	Calculate the square root of 114.4
	= 10.7 m/s

Now, if it is traveling at 10.7 m/second, how far will it travel horizontally at 1 second, 2 seconds, and at 3 seconds?

Time	D	= VT
1 second:		= 10.7 x 1	= 10.7 m
2 seconds:		= 10.7 x 2	= 21.4 m
3 seconds:		= 10.7 x 3	= 32.1 m

How far will it fall from the top of the hill?

Time	D	= 0.5gt²
1 second:		= 0.5 x 9.8 x 1²
		= 4.9 x 1	= 4.9 m

2 seconds:		= 0.5 x 9.8 x 2²
		= 4.9 x 4	= 19.6 m

3 seconds:		= 0.5 x 9.8 x 3²
		= 4.9 x 9	= 44.1 m

Use the calculated information to plot the path of the coaster.

Time	Horizontal Distance (m)	Vertical Distance (m)
1 second	= 10.7 m	= 4.9 m
2 seconds	= 21.4 m	= 19.6 m
3 seconds	= 32.1 m	= 44.1 m

Extra credit challenge # 3: Include a Clothoid Loop (20 points)
Calculate the centripetal acceleration and G-forces at the top of the loop. (5 points) Do not exceed 2 G-forces at the top of the loop. (5 points) Calculate the centripetal acceleration and G-forces at the side of the loop. (5 points)
		4) Show your meticulous work. (5 points)
C.A. =	r =	Gs felt at top of a loop Subtract a G due to gravity
Gs = C.A. divided by 9.8m/s²		Gs felt at bottom of a loop Add a G due to gravity

Example problem 4: Let's assume that our coaster loop is 16 meters down from the top of the lift hill.

We know that we cannot exceed 2 G-forces. However, since gravity is acting against the Gs, we can pump the centripetal acceleration up to 29 m/s².

Radius	= 2gd/C.A.
	= 2 x 9.8 x 16/29
	= 313.6/29
	= 10.8 meters

A radius of 10.8 meters will give a centripetal acceleration of 29 m/s².

Gs = Centripetal acceleration/9.8m/s²= 29/9.8 = 3 Gs

Since gravity acts in the other direction, the riders will feel 2 Gs. Perform similar calculations to discover the radius you need at the side of the loop.

Extra credit challenge # 4: Calculate the Optimum Banking Angle and Gs on a horizontal turn:

Banking angle	G-forces
Tan	Click this link to see calculations
How to calculate Gs on a banked turn

Example problem 5: What must the curve's angle be for a roller coaster car to travel around a curve of radius 30 m at 20 m/s?

Tangent (theta)	= V²/rg
	= 20²/30 x 9.8
	= 400/294
	= 1.36
Click to see trigonometric tables
	= 53 degrees

Extra problems:

1. What must the curve's angle be for a roller coaster car to travel around a curve of radius 30 m at 20 m/s?

2. How many g's are felt by a rider as he travels around the banked curve in the previous problem?

3. A car is to make it around a banked curve. The radius is 15.35 m and the car will travel at 30 m/s. What is the optimum banking angle of the curve?

4. A car is to make it around a banked curve. The radius is 15.35 m and the car will travel at 30 m/s. This roller coaster is on the moon where the acceleration due to gravity is 1.67 m/s2. What is the optimum banking angle of the curve?

5. A rider is to make it around a curve of radius of 24.28 m so that the rider will feel 2.50 g's. What is the angle of the banked curve?

6. A rider is to make it around a curve of radius of 31.15 m so that the rider will feel 1.64 g's. How fast must the rider be traveling?

A rider is to make it around a curve of radius of 51.15 m so that the rider will feel 4.52 g's. How fast must the rider be traveling?

ANSWERS

1 53.68° 2 1.69 g's 3 80.51° 4 88.37°

5 66.42° 6 19.92 m/s (52.69°) 7 47.01 m/s (77.22°)