# I Centripetal Force, Friction, Loop de Loop, and Normal Forces

Tags:
1. Oct 9, 2016

### Abhishek Jain

1. How is the force of friction of the tires pointed inwards to cause centripetal acceleration? If the tire is rotated when trying to turn, the car is still traveling with a velocity perpendicular to the circle and force of friction would still be counteracting the direction of movement?

2. How can gravity = centripetal acceleration at the top of loop de loop? They are pointed in same direction downward? Also i have been told that normal force is zero when centripetal acceleration = gravity. If normal force is zero, the car wouldn’t be in contact with the road then and would fall? Second, I was told that normal force increases as centripetal acceleration increases. However, isn’t normal force pointing downward toward the center counteracting the car pushing on the road upward? But also if gravity is pointing downward, how is the car even exerting force on the road upward? To me it seems all the forces are pointed in the downward direction at the top of the loop.

2. Oct 9, 2016

### A.T.

No, static friction can have a different direction.

Draw a free body diagram.

3. Oct 9, 2016

### PhanthomJay

centripetal force is a center seeking force and is not a force of its own, rather, it is the net force acting toward the center. For the car in a circle on a level road, the centripetal force is the net force toward the center, and the only force toward the center is friction. The car has a tendency to move outward, so friction is opposite the pending relative motion, thus, inward.
gravityfirce is one of the forces causing centripetal acceleration , and the other is the normal force. At the top, Net force must always be down in the direction of the downward inward acceleration.
when gravity is the only force causing the acceleration, the normal force is a hair above zero at the critical speed . It doesn't fall unless the speed at the top is less than the critical speed.
yes, normal force increases and points down in same direction as gravity force, increasing the net force and centripetal acceleration downward, and the speed.

4. Oct 9, 2016

### Abhishek Jain

These are the diagrams I had?

So as acceleration increases in the direction of the center, the velocity increases perpendicular to the circle to make up for the increased acceleration and in turn making sure the car doesn't just fall to the center correct? I know they are inter related by the formula ac = v2/r. I am just trying to understand it visually. Basically your magnitude of speed accounts for the accounts for the increased downward acceleration?

Also I am having trouble understand what is exactly keeping the car in contact with the road in the loop de loop. If all the forces are pointed downward and in reality it is falling but also moving at a speed forward, is it that velocity keeping it on the road?

#### Attached Files:

File size:
108.1 KB
Views:
153
5. Oct 10, 2016

### A.T.

The curvature of the loop is stronger than the curvature of the free fall parabola it would take without the loop, so the road is in it's way.

6. Oct 10, 2016

### PhanthomJay

[QUOTE="Abhishek Jain,

in circular motion, the direction of the velocity is tangent to the circular path and gives rise to the centripetal acceleration because of its change in direction As the magnitude of the velocity increases, both the tangential and centripetal acceleration increases.
The car will not fall off the track as long as there is a normal force pushing on it.
Normal forces are."pushing" forces. At the top of the loop you are upside down beneath the track. The speed increases this force of the seat which is pushing downward on you. So by Newton's third law you are pushing up on the seat. This is the contact force. The contact force keeps you pressed to the seat. So at the top, the centripetal force is mg + N downward,. and per newton 2, mg + N =mv^2/r, that is, N =mv^2/r -mg. Therefore v^2/r must be greater than g or else N would be negative, that is to say, v must be greater than the sq.rt of rg. This is the min speed required at the top to keep it moving in the curved path without falling.

With an ordinary roller coaster, not a loop de loop, but rather, one where at the top you are above the track, not below it, then the normal force pushes upward on you, opposite the gravity direction. In this case, mg - N = mv^2/r, that is, N = mg-mv^2/r . Therefore, v^2/r must be less than g, or v must be less than the sq.rt of rg, the max speed to prevent the car from flying off the track (although due to safety features, this speed can be exceeded since the safety bar and wheels beneath the track can provide the force necessary to keep you in circular motion without flying out).

Confusing, isn't it? We could get into inertial forces that will likely serve to compound the confusion.