Final answer:
The minimum speed at the top of a loop-the-loop to avoid falling is derived by equating the centripetal force to the gravitational force, leading to the formula v = sqrt(gr).
Step-by-step explanation:
To find the minimum speed at which a car must travel at the top of a loop-the-loop to avoid falling, we use the concept of centripetal force. At this position, the only force providing the centripetal acceleration is the car's weight. Therefore, the condition for minimum speed is when the centripetal force is exactly equal to the gravitational force acting on the car at the top of the loop.
Using the formula for centripetal force Fc = mv2/r, where m is the mass of the car, v is the speed of the car at the top, and r is the radius of the loop, we can equate it to the gravitational force Fg = mg, where g is the acceleration due to gravity.
The radius of the loop is half of the diameter, so in this case, it is 45m. At the minimum speed v, we have: mv2/r = mg. Solving for v gives us v = sqrt(gr). Using g = 9.8 m/s2 and r = 45 m, we find v = sqrt(9.8 * 45), which gives us the minimum speed the car must have to maintain contact with the track at the top of the loop.