Solving for Speed on a Roller Coaster's Top of Hill

[SilverAu]

Homework Statement

A roller coaster car is going over the top of a 18-m- radius circular rise. At the top of the hill, the passengers "feel light," with an apparent weight only 60% of their true weight. How fast is the coaster moving?

Homework Equations

I think F=mv^2/ (r)
F=ma

The Attempt at a Solution

Attempt:
F=mv^2/r
(.6)g=v^2/r
(9.8*.6)r=v^2
5.88*18=v^2
v=10.3 m/s My answer is said to be wrong

 

[Mark44]

This is not advanced physics, so moving to the intro physics section.

 

[nasu]

Homework Statement

A roller coaster car is going over the top of a 18-m- radius circular rise. At the top of the hill, the passengers "feel light," with an apparent weight only 60% of their true weight. How fast is the coaster moving?

Homework Equations

I think F=mv^2/ (r)
F=ma

The Attempt at a Solution

Attempt:
F=mv^2/r
(.6)g=v^2/r
(9.8*.6)r=v^2
5.88*18=v^2
v=10.3 m/s My answer is said to be wrong

You did not calculate what the question asks.
It's not the centripetal force which is 60% of the weight.
Draw a diagram with all the forces.

 

[SilverAu]

OK well if I am solving for the velocity of the moving coaster, how would I do that because I've drawn the diagram with forces and I don't see it.

 

[nasu]

Write Newton's second law for circular motion. The acceleration is the centripetal acceleration.
What is the expression for the net force at the top of the hill?

 

[HallsofIvy]

If they only felt 0.6 of their weight then the centripetal force on them is 1- 0,6= 0.4 of their weight.

 

[SilverAu]

Yeah I actually just figured it out, that is the only thing I messed up... Should be
F=mv^2/r
(.4)g=v^2/r
(9.8*.4)r=v^2
3.92*18=v^2
v=8.4m/s
Thanks for helping you guys