Centripetal force on a top down loop

AI Thread Summary
The discussion revolves around calculating the speed of a rider at the top of a vertical loop in an amusement park ride, using the normal force readings from a sensor. Initially, the rider experiences a normal force of 950 N when stationary, which corresponds to a mass of 96 kg. At the top of the loop, the normal force drops to 300 N, prompting the need to analyze the forces acting on the rider, including gravitational force. The net centripetal force is determined by the sum of the normal force and gravitational force, leading to the equation for speed. The final calculation yields a speed of approximately 15 m/s at the top of the loop, emphasizing the importance of understanding the forces involved in circular motion.
onyxorca
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Homework Statement



A special electronic sensor is embedded in the seat of a car that takes riders around a circular loop-the-loop at an amusement park. The sensor measures the magnitude of the normal force that the seat exerts on a rider. The loop-the-loop ride is in the vertical plane and its radius is 18 m. Sitting on the seat before the ride starts, a rider is level and stationary, and the electronic sensor reads 950 N. At the top of the loop, the rider is upside-down and moving, and the sensor reads 300 N. What is the speed of the rider at the top of the loop?

Homework Equations



a=v^2/r

The Attempt at a Solution



950/9.8=96 kg

300=790-96(v^2/18)

?
 
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onyxorca said:

Homework Statement



A special electronic sensor is embedded in the seat of a car that takes riders around a circular loop-the-loop at an amusement park. The sensor measures the magnitude of the normal force that the seat exerts on a rider. The loop-the-loop ride is in the vertical plane and its radius is 18 m. Sitting on the seat before the ride starts, a rider is level and stationary, and the electronic sensor reads 950 N. At the top of the loop, the rider is upside-down and moving, and the sensor reads 300 N. What is the speed of the rider at the top of the loop?

Homework Equations



a=v^2/r

The Attempt at a Solution



950/9.8=96 kg

300=790-96(v^2/18)

?
I'm not sure where your 790 figure is coming from. You must look at the forces acting on the rider at the top of the loop. What are the 2 forces? In which direction do they act? What's the net force and the direction of the net force? In which direction is the centripetal acceleration? Then use Newton's 2nd law.
 
Fn=Fc-Fg

Fn=mv^2/r-mg

300=mv^2/r-950

mv^2/r=300+950

v=sqrt((300+950)r/m)

=15.2350 m/s ?
 
onyxorca said:
Fn=Fc-Fg

Fn=mv^2/r-mg

300=mv^2/r-950

mv^2/r=300+950

v=sqrt((300+950)r/m)

=15.2350 m/s ?
Looks good, nice work! you should round off your answer to maybe v =15m/s
 
I thought at the top it would be Fn+Fg=mv^2/r
and then Fg is the only force acting because of centripetal acceleration and it is free fall for an instant... so Fn=0?
 
Last edited:
bonz said:
I thought at the top it would be Fn+Fg=mv^2/r
That is corrrect, both the normal and gravity forces act downward toward the center of the circle, so their sum provides the net centripetal force
and then Fg is the only force acting because of centripetal acceleration and it is free fall for an instant... so Fn=0?
It is given that the nornal force is 300 N at the top of the loop, so why are you setting it equal to 0?? Setting the normal force equal to 0 will give you the absolute minimum speed required to keep the coaster and rider moving in a circle, but that is not what is being asked in this problem.
 
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