A roller coaster track is designed so that the car travels upside down on a certain portion of the track as shown in the figure below. What is the minimum speed the roller coaster can have without falling from the track? Assume the track has a radius of curvature of 20 m.
The force of the centripetal acceleration must be greater than or equal to the force of gravity, or
m * v^2 / r >= mg
v^2 / 20 >= 9.8
solve for v
To determine the minimum speed the roller coaster can have without falling from the track, we can use the concept of centripetal force. The centripetal force is the force that keeps an object moving in a circular path, and it is given by the equation:
F = (m * v^2) / r
where F is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of curvature.
In this case, the roller coaster car is traveling upside down, which means the centripetal force must be provided by the normal force (the force exerted by the track on the car) to keep the car from falling off the track. The normal force acts toward the center of the circular path.
At the minimum speed, the normal force is zero, implying that the roller coaster car is just about to lose contact with the track. Therefore, the centripetal force is equal to the weight of the car (mg), where g is the acceleration due to gravity (9.8 m/s^2).
Setting the centripetal force equal to the weight:
(m * v^2) / r = m * g
Canceling out the mass terms:
v^2 / r = g
Solving for v:
v = sqrt(r * g)
Plugging in the given values:
v = sqrt(20 m * 9.8 m/s^2)
v = sqrt(196 m^2/s^2)
v ≈ 14 m/s
Therefore, the minimum speed the roller coaster can have without falling from the track is approximately 14 m/s.