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In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 268 kg and moves with speed v = 16.24 m/s. The loop-the-loop has a radius of R = 10.5 m.

What is the magnitude of the normal force on the care when it is at the bottom of the circle? (But as the car is accelerating upward.)

2) What is the magnitude of the normal force on the car when it is at the side of the circle (moving vertically upward)?

3) What is the magnitude of the normal force on the car when it is at the top of the circle?

Answer :

taskmasters
Before we go through the questions, we need to calculate and determine some values first.

r = 11.5 m 
m = 280 kg 
Centripetal force = m x v^2/r = 280 x (17.1^2/11.5) = 7119.55 N 

1) What is the magnitude of the normal force on the care when it is at the bottom of the circle. 

Centripetal force + mg = 7119.55 + (280 x 9.8) = 9863.55 N 

2) What is the magnitude of the normal force on the car when it is at the side of the circle. 

Centripetal force = 7119.55 N 


3) What is the magnitude of the normal force on the car when it is at the top of the circle. 

Centripetal force - mg = 7119.55 - (280 x 9.8) = 4375.55 N 

4) What is the minimum speed of the car so that it stays in contact with the track at the top of the loop. 

√(gr) 

√(9.8 x 11.5) = 10.62 m/s

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