Help for the Roller Coaster Project!
Name __________________________________________ Science ______
The Standard Roller Coaster Assignment (20 points)
Velocity = 
Velocity = 
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: Gforces at the Dip of a Hill (15 points) 

(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/s^{2} 
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/s^{2}. 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) 

Velocity = (5 points) 



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 
= 0.5gt^{2} 

1 second: 

= 0.5 x 9.8 x 1^{2} 



= 4.9 x 1 
= 4.9 m 




2 seconds: 

= 0.5 x 9.8 x 2^{2} 



= 4.9 x 4 
= 19.6 m 




3 seconds: 

= 0.5 x 9.8 x 3^{2} 



= 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) 

(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/s^{2} 
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 Gforces. However, since gravity is acting against the Gs, we can pump the centripetal acceleration up to 29 m/s^{2}.
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/s^{2}.
Gs = Centripetal acceleration/9.8m/s^{2 }= 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 
Gforces 
Tan 

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) 
= V^{2}/rg 

= 20^{2 }/30 x 9.8 

= 400/294 

= 1.36 

= 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?
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°)