Answer:
The linear speed of the car is approximately 27.30 m/s
Explanation:
The question parameters are;
The mass of the person on the rollercoaster, m = 57.0 kg
The radius of the rollercoaster track, r = 42.7 m
The normal force felt by the person, F = 995 N
The centripetal force acting on the person keep the circular motion is given by the following equation;
[tex]Centripetal \, force \ F_c = \dfrac{m \times v^2}{r}[/tex]
Where;
v = The linear velocity of motion = The linear speed of the car
The centrifugal force, F, is the force normal force felt by the person and is equal to the centripetal force, therefore, we have;
[tex]Centripetal \, force \ F_c = Centrifugal \, force \ F = \dfrac{m \times v^2}{r}[/tex]
From which we have;
[tex]F = 995 = \dfrac{57 \times v^2}{42.7}[/tex]
[tex]\therefore v = \sqrt{\dfrac{995 \times 42.7}{57} } \approx 745.38[/tex]
The linear speed of the car = v ≈ 27.30 m/s
The angular speed of the car, ω = v/r ≈ 27.30/42.7 ≈ 0.639 rad/s