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 = 22.0 m, how fast is the roller coaster traveling at the bottom of the dip?
To find the speed of the roller coaster at the bottom of the dip, we can use the concept of centripetal force.
Centripetal force is the force that keeps an object moving in a circular path. In this case, the centripetal force is the force exerted by the roller coaster on the passenger, which is equal to six times her weight.
We can start by finding the weight of the passenger. The weight of an object is given by the product of its mass and the acceleration due to gravity (9.8 m/s^2 on Earth). Let's represent the weight of the passenger as W.
W = m * g
Now, we know that the force exerted by the roller coaster on the passenger during the dip is six times her weight. Therefore, the centripetal force (F_c) can be expressed as:
F_c = 6 * W
The centripetal force is also related to the speed (v) and the radius (r) of the circular path. It can be calculated using the formula:
F_c = (m * v^2) / r
We can equate these two expressions for the centripetal force:
6 * W = (m * v^2) / r
Since we are looking for the speed (v) at the bottom of the dip, we'll assume the mass of the passenger cancels out in the equation. So, we can simplify the equation to:
6 * g = v^2 / r
Now, we can solve for the speed (v) by rearranging the equation and isolating v:
v^2 = 6 * g * r
v = √(6 * g * r)
Substituting the values, where g = 9.8 m/s^2 and r = 22.0 m, we can calculate the speed (v):
v = √(6 * 9.8 * 22.0) m/s
v ≈ 24.2 m/s
Therefore, the roller coaster is traveling at a speed of approximately 24.2 m/s at the bottom of the dip.