A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius 11.0 m. A passenger feels the seat of the car pushing upward on her with a force equal to three times her weight at the bottom of the dip. What is her speed as she goes
through the dip?
To calculate the passenger's speed as she goes through the dip, we can use the concept of centripetal force.
1. Start by considering the forces acting on the passenger at the bottom of the dip. There are two forces to consider: the gravitational force (weight) and the normal force (the seat pushing upward). Given that the normal force is three times her weight, we can write the equation:
n = 3w
2. Use the relationship between the normal force and the gravitational force:
n = mg
3w = mg
w = mg/3
3. Next, consider the forces causing the centripetal acceleration in the vertical circular motion. At the bottom of the dip, the net force acting on the passenger is the sum of the gravitational force (weight) and the normal force:
net force = w + n
net force = w + 3w
net force = 4w
4. Since the net force is responsible for the centripetal acceleration, we can equate it to the centripetal force:
net force = centripetal force
4w = m * (v^2 / r) (centripetal force = m * (v^2 / r), where m is mass, v is velocity, and r is radius)
5. Substituting the value of the weight (w = mg/3) into the equation:
4 (mg/3) = m * (v^2 / r)
6. Simplify the equation by canceling out the mass (m) common to both sides:
4g/3 = v^2 / r
7. Rearrange the equation to solve for the velocity (v):
v^2 = (4g/3) * r
v = sqrt((4g/3) * r)
8. Finally, substitute the given value for the radius (r = 11.0 m) and the acceleration due to gravity (g ≈ 9.8 m/s^2):
v = sqrt((4 * 9.8 / 3) * 11.0)
v ≈ 14.6 m/s
Therefore, the passenger's speed as she goes through the dip is approximately 14.6 m/s.