Question

The radius of curvature of a loop-the-loop for a roller coaster is 12.4 m. at the top of the loop (with the car inside the loop), the force that the seat exerts on a passenger of mass m is 0.80mg. find the speed of the roller coaster at the top of the loop.

Answer 1

14.79 m/s At the top of the loop, there's 2 opposing forces. The centripetal force that's attempting to push the roller coaster away and the gravitational attraction. These 2 forces are in opposite directions and their sum is 0.80 mg where m = mass and g = gravitational attraction. So let's calculate the amount of centripetal force we need. 0.80 = F - 1.00 1.80 = F So we need to have a centripetal force that's 1.8 times the local gravitational attraction which is 9.8 m/s^2. So 1.8 * 9.8 m/s^2 = 17.64 m/s^2 The formula for centripetal force is F = mv^2/r where F = force m = mass v = velocity r = radius We can eliminate mass from the equation since the same mass is being affected by both the centripetal force and gravity. So: F = v^2/r 17.64 m/s^2 = v^2/12.4 m 218.736 m^2/s^2 = v^2 14.78972616 m/s = v So the velocity at the top of the loop (rounded to 2 decimal places) is 14.79 m/s.