Question

A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the seat of the car pushing upward on her with a force equal to five times her weight as she goes through the dip. If r = 17.0 m, how fast is the roller coaster traveling at the bottom of the dip

Answer 1

The velocity of the roller coaster at the bottom of the dip can be found using the formula v = sqrt(5gr), where r is the radius of the dip and g is the acceleration due to gravity.

Step-by-step explanation:

In order to find the speed of the roller coaster at the bottom of the dip, we can use the concept of centripetal force. The force pushing upward on the passenger is equal to five times her weight, which can be represented as 5mg, where m is the mass of the passenger and g is the acceleration due to gravity. At the bottom of the dip, the centripetal force is provided by the passenger's weight, which can be represented as mg. Equating these two forces, we have 5mg = mg, where g is cancelled out. Solving for v, the velocity of the roller coaster at the bottom of the dip, we get v = sqrt(5gr), where r is the radius of the dip.

Substituting the given values, with r = 17.0 m, the velocity of the roller coaster at the bottom of the dip is v = sqrt(5 * 9.8 * 17.0) = 25.17 m/s.

Answer 2

16.3 m/s

Step-by-step explanation:

Let m = mass of the passenger,

v = speed of the roller coaster

Forces on the passenger are

1. Force by seat = 2.18 mg in upward direction.

2. Weight mg in downward direction

Therefore, net upward force = 2.18 mg - mg = 1.18 mg

Also, net upward force = centripetal force = mv^2/r

Therefore,

1.18 mg = mv^2/r

Dividing by m,

1.18 g = v^2/r

Or v^2 = 1.18 gr

Or v = sqrt(1.18 gr)

= sqrt(1.18 * 9.81 * 22.9)

= sqrt(265)

= 16.3 m/s