# A car in a vertical loop

1. May 7, 2017

### Saracen Rue

1. The problem statement, all variables and given/known data

2. Relevant equations
$E_{\operatorname{total}}=E_{kin}+E_g$
$F_{net}=F_N+F_g$
$a=\frac{v^2}{r}$
3. The attempt at a solution
I've already technically completed question 4 part c, getting an answer of 8168 N downwards (which is what the answers say). However, this is implying that the net force, the gravitational force and the normal force are all acting in the same downward direction. This doesn't make sense to me because it doesn't include a force acting upon the car which is keeping it connected to the track. There has to be a force keeping the car in contact with the track, otherwise it would just fall off. Is there an extra force which would be acting in this situation which has been exempted from the question to prevent the question form becoming too complicated? An explanation for this would be much appreciated.

Thank you all for you time.

2. May 7, 2017

### Staff: Mentor

The car and the track exert forces on each other. (The normal force.) If there were no normal force acting downward at point Z the car would go flying off. (The gravitational force alone is not enough to force the car around the bend.)

If the car wasn't going fast enough, then it would lose contact with the track. That's part 5. (But here it's going plenty fast enough.)

The only forces acting on the car at point Z are gravity and the normal force. They both act down.

3. May 7, 2017

### Saracen Rue

The part that I'm really struggling to understand is how two downward acting forces are keeping the car in contact with the track. I understand that the car itself would have to be exerting a force onto the track in order for it to remain in contact with the track, but what is generating this force? And would this force have to be equal in magnitude to combination of the gravitational and normal forces?

4. May 7, 2017

### Staff: Mentor

Realize that the car is moving and accelerating at the top of the loop. The acceleration -- and thus the net force needed -- is downward. No upward force acts on the car!

The car is pressed against the track because the track is forcing the car to change direction. (Otherwise the car would just shoot off into space.) The faster the car is moving at point Z, the greater the force the track needs to exert on the car to change its direction. (And that force that the track exerts on the car is the normal force.)

And realize -- from Newton's 3rd law -- that the car and track exert equal and opposite forces on each other.
There is no mysterious third force!

5. May 7, 2017

6. May 8, 2017

### CWatters

Velocity has components speed an direction. When moving in a circle the direction is changing so the velocity is changing. That means there is an acceleration towards the centre of the circle. Therefore the net force must be towards the centre of the circle.

The force required to move in a circle is given by mv^2/r. If the net force is greater than this the radius of the circle will tend to reduce (eg it will fall down at the top). If the net force is less than this the radius will tend to increase. However the track prevents it increasing. The normal force increases until the net force matches that required to move in a circle equal to the radius of the track.

7. May 8, 2017

### Saracen Rue

I think I'm starting to get it now :) Also, because the velocity is constantly tangential to the circle and momentum is given by $P=m\cdot v$, that would mean that relative to the car, the car will always be having a forward momentum. However, the centripetal acceleration is also dependent on velocity. While the acceleration (and thus net force) is always acting towards the centre of the circle, the velocity/momentum of the car is always acting tangential to the circle which will keep the car moving so long as the velocity is above a certain value