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 six times her weight as she goes through the dip. If r = 20.0 m, how fast is the roller coaster traveling at the bottom of the dip?
6 M g = M(V^2/R + g)
Cancel out the mass (M) and solve for V.
5 g = V^2/R
V = sqrt(5gR)
To determine the speed of the roller coaster at the bottom of the dip, we can use the concept of centripetal force. At the bottom of the dip, the net force acting on the passenger is the sum of two forces: the weight force (mg) acting downwards, and the normal force (N) acting upwards.
The normal force is the force that the seat of the car exerts on the passenger to keep her in circular motion. It can be calculated by multiplying the passenger's weight (mg) by the acceleration due to gravity (g). In this case, the normal force is 6 times the weight, so N = 6mg.
At the bottom of the dip, the net force acting on the passenger is equal to the centripetal force required to keep her in circular motion, which is given by the equation:
Fc = mv^2 / r
Here, m is the mass of the passenger and v is the velocity of the roller coaster.
Since the passenger's weight (mg) is equal to the difference between the normal force (6mg) and the weight force (mg), we can write:
6mg - mg = mv^2 / r
Simplifying the equation, we get:
5mg = mv^2 / r
Now we can solve for v:
v^2 = 5rg
v = sqrt(5rg)
Plugging in the values given in the question, with r = 20.0 m:
v = sqrt(5 * 9.8 * 20.0)
v = sqrt(980.0)
v = 31.3 m/s
Therefore, the roller coaster is traveling at a speed of 31.3 m/s at the bottom of the dip.