**Roller Coaster Physics: Understanding Loop Dynamics**As part of understanding the physics behind roller-coaster rides, consider the scenario where the coaster coasts down an incline and enters a loop that is 8.20 meters in diameter. To ensure safety, the speed of the roller coaster at the top of the loop should be such that the force exerted by the seat on a rider is equal in magnitude to the rider's weight. **Problem Statement:**From what height above the bottom of the loop must the roller coaster descend to meet this requirement?**Calculation Section:**- Diameter of the loop: 8.20 meters**Interactive Component:**Would you like assistance or further explanation? Click the "Read It" button for additional help and resources.---**Visual Aids:**There are no graphs or diagrams in this text. If needed, schematic representations of the roller coaster loop or free-body diagrams can be provided in educational resources to aid in understanding the forces in play.

Given, the diameter of the loop is 8.2 m. So the radius of the loop, R=4.1m Let the mass of the rider is M, so the centripetal acceleration of the roller coaster will be, ac=v2R And from the question, force on the rider due to the seat is equal to its weight. But here two forces on the rider one is gravity and the other is normal, so the resultant force must be equal to centripetal force, that is, mv2R=2mgv2=2Rg Nw at the top of the loop the potential energy of roller coaster is, P=mg2R=2mgR And the kinetic energy is, K=12mv2=mRg Let the height of the roller coaster for satisfying the safety condition is, h So potential energy at this height is U=mgh and kinetic energy is 0. Using energy conservation mgh=P+Kmgh=mRg+2mgR So, h will be, h=R+2R=4.1m+8.2m=12.3mAnswer: The height of the roller coaster for satisfying the safety condition is 12.3 m.

Answered: Part of a roller-coaster ride involves…