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Sagot :
To prevent cars from falling, the radius at the top of the circle should be
small such cars inverted at the top remain attached during motion.
Correct response;
The radius of the coaster can be C) 20 m
Method by which the above option is selected
Mass of roller coaster car, m = 500 kg
Speed at the top of the circle, v = 20 m/s
Required:
The maximum radius of the circular path the roller coaster car.
Solution:
[tex]\displaystyle Centrifugal \ force, \, F_{c} = \mathbf{\frac{m \cdot v^2}{r}}[/tex]
Where;
r = The radius of the circular path.
Weight of the roller coaster car = m·g = The centripetal force
Where;
g = Acceleration due to gravity = 9.81 m/s²
At equilibrium, we have;
Centrifugal force = Centripetal force
[tex]\displaystyle \frac{m \cdot v^2}{r} = \mathbf{ m \cdot g}[/tex]
Therefore;
[tex]\displaystyle r = \mathbf{ \frac{v^2}{g}}[/tex]
Which gives;
[tex]\displaystyle r = \frac{20^2}{9.81} \approx 40.77[/tex]
The maximum radius for safety of a roller coaster, r ≈ 40.77 meters
[tex]\displaystyle Range \ of \ radius \ of \ the \ circle = \frac{40.77}{4} \leq Radius \ of \ circle \leq 40.77[/tex]
Which gives;
Range of the radius of the circle = 10.2 ≤ Radius of circle ≤ 40.77
The correct option for safety considerations is therefore;
- C) 20 m
The possible question options are;
A) 5 m B) 10 m C) 20 m D) 40 m E) 80 m
Learn more about centripetal force here:
https://brainly.com/question/12674230
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