Centripetal force on a top down loop

[onyxorca]

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)

?

 

[PhanthomJay]

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.

 

[onyxorca]

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 ?

 

[PhanthomJay]

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

 

[bonz]

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?

 

[PhanthomJay]

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.