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Homework Statement
In one roller coaster car, a small 0.10 kilogram ball is suspended from a safety bar by a short length of light, inextensible string. The car is then accelerated horizontally, goes up a 30 degree incline, goes down a 30 degree incline, and then goes around a vertical circular loop of radius 25 meters. For each of the four situations described in parts (b) and (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 of the string in Newtons.
2. Determine the vertical component Tv of the tension in the string in Newtons.
(b) The car is at point B (flat surface) moving horizontally with an acceleration of 5.0 m/s squared.
(c) The car is at point C and is being pulled up the 30 degree incline with a constant speed of 30 m/s.
(d) The car is at point D moving down to 30 degree incline with an acceleration of 5.0 m/s squared.
(e) The car is a tpoint E moving upside down with an instantaneous speed of 25 m/s and no tangential acceleration at the top of the vertical loop of radius 25 meters.
Homework Equations
F = ma
At equilibrium, F1 + F2 + F3 = 0
A squared plus B squared = C squared
The Attempt at a Solution
(b)
Fg = Tv
Fg = ma
Fg = 0.10 * 9.8 m/s^2
Fg = 0.98 N
Tv = 0.98 N
Fh = 0.10 kg * 5.0 m/s^2
Fh = 0.5 N
Fh = Th
Th= 0.5 N
(c) This is where it gets complicated because there's an angle involved.
See: http://img52.imageshack.us/img52/9761/physics0001.jpg
Here there's no acceleration, so there's equilibrium. Therefore, the only tension is vertical against gravity. Right?
Tv= 0.98 N
Th = 0 N
(d) Here, there's an acceleration and an angle. I solved it as if there was no angle- does the angle matter?
Tv= 0.98 N
Th= 0.5 N
(e) No acceleration and no angle, but the car is on a circle.
Tv= 0.98 N
Th = 0N
I can't believe that the angle has absolutely nothing to do with anything. Where have I gone wrong?
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