A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop is 12 m, with what minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top? Assume the rider is not strapped to the car.
Question
A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop is 12 m, with what minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top? Assume the rider is not strapped to the car.
Solution
To solve this problem, we need to use the physics concept of centripetal force. When the roller coaster is at the top of the loop, the force of gravity and the normal force (the force exerted by a surface that supports the weight of an object resting on it) are both acting downwards. The normal force becomes zero when the car is just about to fall, which is the minimum speed we need to find.
The centripetal force needed to keep the car moving in a circle is provided by the weight of the car when the normal force is zero. So, we have:
mg = mv^2/r
where: m is the mass of the car (which will cancel out), g is the acceleration due to gravity (9.8 m/s^2), v is the speed of the car, and r is the radius of the loop (12 m).
Rearranging the equation to solve for v gives:
v = sqrt(g*r)
Substituting the given values gives:
v = sqrt(9.8 m/s^2 * 12 m) = sqrt(117.6 m^2/s^2) = 10.85 m/s
So, the car must traverse the loop with a minimum speed of 10.85 m/s to ensure that the rider does not fall out while upside down at the top.
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