What is the normal acting force?

Question

i need help with this: Judy (62 kg) is riding a roller coaster with a large loop. At the top of the loop, Judy is traveling 35 m/s. The loop has a radius of curvature of 23 m. From Judy's frame of reference, what is the normal force acting on her at the top of the loop?

Answer 1

I assume the car is on the under side of the loop at the top, and not on the top side of the loop at the top.  If it were on top, it would be going too fast to stay on the track, unless there were some grabbing mechanism which is not described.  If that were true, the normal force of the track on the car would be zero.

 

So I presume it is on the under side of the top of the loop.  In that case, with the circumstances described in the problem, we expect two forces acting, both directed down:  the force of gravity, and the normal force of contact of the track on the car.

 

The result of these two forces is the centripetal force.  An important point to note about centripetal force is that it is NOT a separate force from the other forces in the problem; it is the RESULT of the other forces in the problem.  It is analogous to having a big, heavy wooden box on a concrete floor.  If you push it one way with a force F_p, and then have friction (f) oppose that motion, you could write (via Newton's 2nd law) that F_p - f = ma (taking the direction of F_p to be positive).  What you would *not* say about the "ma" term is that it is a separate force; it is the result of the other forces.  Centripetal force is the same way.  It is not as if there are a bunch of forces and then the centripetal force; the centripetal force is created by the other forces.  The resultant force (in the radial direction) on any object moving in a circle must be the centripetal force.

 

The expression for the centripetal force is m(v²)/r

 

At the top of the loop, the centripetal force must point down, toward the center.  Since both of the forces (normal force, N, and gravitational force, mg) also point down, let's call down positive.  Then we can set up Newton's 2nd law in the vertical direction (the only direction which matters here):

 

F_net = N + mg = m(v²)/r

 

You can use this to solve for the normal force, since all other quantities are given.

 

This may be a bit troubling.  How can all the forces be down and yet the object not fall?  In this case, the key is in having velocity around the circle (in what we call the tangential direction).  If the car were stationary at the top, of course it would fall under the influence of gravity alone.  

 

But think of the car going through the top of the loop and then descending one of the sides.  Once it starts descending, how does it vertical position compare to when it is at the top?  It is lower!  The downward forces DID draw the car downward.  But because it was moving tangentially, by the time it started going down, the car was on a different part of the circle and subject to different forces, so it can keep going in the circle.  That is why it can be that way at the top, because it has (tangential) velocity.

 

This is the same reason satellites can remain in circular orbits even though the only force on them is gravity (pointing toward the ground).  As long as they move fast enough in the tangential direction around the Earth, then by the time gravity pulls them a meter toward the Earth, the curvature of the Earth will have caused the ground to fall away by 1 m under them.  That is why satellites need a certain speed to attain orbit.  If they stopped, they would fall out of the sky (and this is one way to draw orbiting spacecraft back to the Earth).

 

I hope this helps!  If you have any further questions or concerns, please let me know.

Answer 2

Make a force diagram. You have two forces going down, and the normal force going up. If there were no normal force going up, Judy would be pulled down into the loop! The two forces are F = mg, 62 x 9.8  and F = mass x v²/r, or 62 x 35²/23. Remember the centripetal acceleration is straight down from the top, 90 degrees down from the velocity which is tangent to the top of the curve. Add up the two forces going down.  Pick a direction for the force up and the two forces down. I tend to call down negative and up positive, but follow the teacher's directions or just define your terms. The normal force is equal and opposite to the total of the down forces, since Judy is not flying up or down (Newton's laws).