• anonymous
i don't get this , A special electronic sensor is embedded in the seat of a car that takes riders around a circular loop-the-loop ride at an amusement park. The sensor measures the magnitude of the normal force that the seat exerts on a rider. The loop-the-loop ride is in the vertical plane and its radius is 18 m. Sitting on the seat before the ride starts, a rider is level and stationary, and the electronic sensor reads 745 N. At the top of the loop, the rider is upside down and moving, and the sensor reads 395 N. What is the speed of the rider at the top of the loop?
  • Stacey Warren - Expert
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  • schrodinger
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  • matt101
You just need to think about the forces acting on the rider in each situation. In the first part, the rider is level and stationary, meaning he is only experiencing the force of gravity (balanced by the normal force). If the force of gravity is 745 N, you can calculate the rider's mass (which will be important for the second part) using F=ma: 745=m(9.8), so m=76 kg. In the second part, the rider is upside down and moving. Importantly, he's moving in a circle (since it's a loop-the-loop). That means the net force on the rider is a centripetal force, in order to keep him moving in a circle. Since at the top of the loop-the-loop, the net force (centripetal force) is in the direction of the gravitational force (straight down for gravitational force=radially inward for centripetal force in this position), we know the gravitational force must outweigh the normal force. That means we can set up the equation as follows: \[F_{net}=F_c=F_G-F_N\]\[ma_c=mg-F_N\]\[\frac{mv^2}{r}=mg-F_N\] We calculated mass from the first part, and both the radius and normal force are given in the question. Plug in the values and solve for v! Let me know if that makes sense!

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