# Roller coaster, ball attached to coaster by a string calculate various tension

1. Mar 2, 2009

### yoyolala

1. The problem statement, all variables and given/known data

http://img57.imageshack.us/my.php?image=fullscreencapture322009.jpg

Part of the track of an amusement park roller coaster is shaped as shown below. A safety bar is oriented length-wise along the top of each car. In one roller coaster car, a small 0.10 kilogram ball is suspended from this bar by a short length of light, inextensible string.

The car is then accelerated horizontally, goes up a 30° incline, goes down a 30° incline, and then goes around a vertical circular loop of radius 25 meters. For each of the four situations described in parts (B) to (E), do all three of the following. In each situation, assume that the ball has stopped swinging back and forth. 1) Determine the horizontal component Th of the tension in the string in newtons and record your answer in the space provided. 2) Determine the vertical component Tv of the tension in the string in newtons and record your answer in the space provided. 3) Show on the adjacent diagram the approximate direction of the string with respect to the vertical. The dashed line shows the vertical in each situation.

[[[[ I'm having trouble on parts C and D ]]]]] which look like...

http://img10.imageshack.us/my.php?image=fullscreencapture322009.jpg

2. Relevant equations

I'm pretty sure I got all the other parts right and I guess I'm just confused on parts c and d

The answers are supposed to be:

(c) T horiz = 0N
T vert = 1N
(d) T horiz = .43 N
T vert = .75 N

3. The attempt at a solution

I drew FBD for parts C and D and my explanations are based off of those and I'm thinking maybe I didn't draw them right...

In part C:

for the horizontal component, I got the tension to be 0N because it says it is pulled up the incline with a constant speed therefore a=0.
F=ma a=0
Tsin(theta)=0
T=0 N

And for the vertical component
,
F=ma a=0
Tcos(theta)-mg=0
solve for T where m=.1 kg and theta=30 degrees
T= 1.13 N which isn't right

In part D:

totally got this wrong...

Moves down the incline with a=5.0 m/s^2

For the horizontal component of the tension:
F=ma
Tsin(theta)=ma
T= (ma) / sintheta where a=5, and theta=30
T=1.0N??? Not right

For the vertical component of the tension:

F=ma
Tcos(theta)-mg=ma
T=(ma+mg)/(costheta) where a=5, and theta=30
T= 1.71 N ??

Help...?