Question

A 57.0 kg person in a

Answer 1

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;

$Centripetal \, force \ F_c = \dfrac{m \times v^2}{r}$

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;

$Centripetal \, force \ F_c = Centrifugal \, force \ F = \dfrac{m \times v^2}{r}$

From which we have;

$F = 995 = \dfrac{57 \times v^2}{42.7}$

$\therefore v = \sqrt{\dfrac{995 \times 42.7}{57} } \approx 745.38$

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