Question

You design a roller coaster with a loop-the-loop with a 12 m radius and no seatbelts.

A) What is the minimum speed the roller coaster must have at the top of the loop for the passengers to not fall out?
B) If your car goes twice this fast at the top of the loop what will a scale (measuring kg ) read with a 100 kg passenger sitting on top of it?

Answer 1

The minimum speed of the roller coaster at the top of the loop is approximately 13.86 m/s. If the car goes twice this fast, the scale will read 1960 N.

Step-by-step explanation:

To find the minimum speed of the roller coaster at the top of the loop, we need to consider the forces acting on the passengers. At the top of the loop, the downward acceleration due to gravity must be equal to the centripetal acceleration.

We can use the equation g = v^2 / r, where g is the acceleration due to gravity (9.8 m/s^2), v is the speed, and r is the radius of the loop. Rearranging the equation, we have v = sqrt(g * r).

Substituting the given values, v = sqrt(9.8 * 12) = 13.86 m/s. Therefore, the minimum speed the roller coaster must have at the top of the loop is approximately 13.86 m/s.

If the car goes twice this fast at the top of the loop, the scale will read twice the normal force acting on the passenger. The normal force is equal to the gravitational force acting on the passenger, which is given by F = m * g, where m is the mass of the passenger and g is the acceleration due to gravity. In this case, the scale will read 2 * (100 kg * 9.8 m/s^2) = 1960 N.