Hi, this is the problem I was given, I am not too sure whereto start here. Can y

Question

Hi, this is the problem I was given, I am not too sure whereto start here. Can you please help? Thanks. Deb
Part of a roller-coaster ride involves coasting down an incline andentering a loop 7.00 m in diameter.For safety considerations, the roller coaster's speed at the top ofthe loop must be such that the force of the seat on a rider isequal in magnitude to the rider's weight. From what height abovethe bottom of the loop must the roller coaster descend to satisfythis requirement? Hi, this is the problem I was given, I am not too sure whereto start here. Can you please help? Thanks. Deb
Part of a roller-coaster ride involves coasting down an incline andentering a loop 7.00 m in diameter.For safety considerations, the roller coaster's speed at the top ofthe loop must be such that the force of the seat on a rider isequal in magnitude to the rider's weight. From what height abovethe bottom of the loop must the roller coaster descend to satisfythis requirement?

Explanation / Answer

First determine the required speed at the top of theloop. Centripetal acceleration is: ac = V2 /R         where V isthe tagential velocity and R is the radius of the circle Using newton's second law, F=ma: mac = Fn +mg         whereFn is the normal force Fn = mac - mg We are given Fn = mg mac = 2mg V = (2Rg) = (2*9.81m/s2*3.5m) =8.28m/s Then use energy concepts to determine the height. Initial energy: Ue = mgh Final energy: Ef = mg*2R + 1/2mV2 Conservation of energy: mgh = mg*2R + 1/2mV2 h = 2R + V2 / (2g) = 2*3.5m + (8.28m/s)2/ (2*9.81m/s2) = 10.49m

PLEASE CHECK MY MATH AND RATE
Conservation of energy: mgh = mg*2R + 1/2mV2 h = 2R + V2 / (2g) = 2*3.5m + (8.28m/s)2/ (2*9.81m/s2) = 10.49m

PLEASE CHECK MY MATH AND RATE

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