At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered" against the wall. Suppose the radius of the room is 3.37 m and the speed of the wall is 19.0 m/s when the floor falls away. The source of the centripetal force on the riders is the normal force provided by the wall. (a) How much centripetal force acts on a 52.9 kg rider? (b) What is the minimum coefficient of static friction that must exist between the rider's back and the wall, if the rider is to remain in place when the floor drops away?
At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered" against the wall. Suppose the radius of the room is 3.37 m and the speed of the wall is 19.0 m/s when the floor falls away. The source of the centripetal force on the riders is the normal force provided by the wall. (a) How much centripetal force acts on a 52.9 kg rider? (b) What is the minimum coefficient of static friction that must exist between the rider's back and the wall, if the rider is to remain in place when the floor drops away?
At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered" against the wall. Suppose the radius of the room is 3.37 m and the speed of the wall is 19.0 m/s when the floor falls away. The source of the centripetal force on the riders is the normal force provided by the wall. (a) How much centripetal force acts on a 52.9 kg rider? (b) What is the minimum coefficient of static friction that must exist between the rider's back and the wall, if the rider is to remain in place when the floor drops away?
At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered" against the wall. Suppose the radius of the room is 3.37 m and the speed of the wall is 19.0 m/s when the floor falls away. The source of the centripetal force on the riders is the normal force provided by the wall. (a) How much centripetal force acts on a 52.9 kg rider? (b) What is the minimum coefficient of static friction that must exist between the rider's back and the wall, if the rider is to remain in place when the floor drops away?
Transcribed Image Text:The image depicts a diagram of a "rotating cylindrical ride" often found in amusement parks, illustrating the concept of centripetal force and circular motion.
In the diagram:
1. **Cylindrical Structure**: A large vertical cylinder is shown with a partial cutaway to display its interior.
2. **Person Inside**: Inside the cylinder, there is a person standing against the wall with their arms extended. The person represents an object experiencing the effects of centripetal force.
3. **Rotation Details**:
- A dashed line extends vertically through the center of the cylinder, indicating the axis of rotation.
- A curved arrow at the top of the cylinder shows the direction of rotation around this vertical axis.
4. **Centripetal Force and Velocity**:
- The letter "r" next to a radial line inside the cylinder signifies the radius from the axis of rotation to the wall.
- The red arrow pointing along the interior wall in the direction of motion is labeled "v", representing the tangential velocity of the person at the cylinder's edge.
This diagram helps explain how the centripetal force keeps the person pressed against the cylinder wall as it rotates, preventing them from falling. The forces at play include gravity and the normal force exerted by the wall, with friction typically providing the necessary centripetal force to maintain circular motion.
Definition Definition Force on a body along the radial direction. Centripetal force is responsible for the circular motion of a body. The magnitude of centripetal force is given by F C = m v 2 r m = mass of the body in the circular motion v = tangential velocity of the body r = radius of the circular path
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.