Confused about weightlessness and feeling an increased apparent weight.
Ok, so picture this, you're riding on a roller coaster and constantly going around a circular loop. At the top of this loop, since gravity acts down, and the reaction force acts to the centre (so down), you would feel no weight (weightless) since the forces cancel out. On the other hand, if you are at the bottom of the loop, you have weight acting downwards, and a reaction force acting upwards, so would feel an increased apparent weight as these vectors add up.
Oh god I'm so confused !
OK. There are two possible scenarios and they're both quite different. Let's consider the first case, in which the rollercoaster is above the tracks (non-inverted, facing up) at the top of the loop. The reaction force acts UPWARDS. As the net force must be down, towards the centre of the circle, Newton's second law gives mv^2/r = mg - N (if we let down be positive and the normal reaction is now negative). For circular motion to exist at that radius and speed, N>=0, so N = mg-mv^2/r>=0. Or, g>=v^2/r. This makes sense as if the rollercoaster moves too fast, you can imagine the people flying off the track. The limiting case, N=0, is when the weight force supplies all of the acceleration and the normal reaction force isn't needed to maintain a circular path of that radius and speed.
The second case is when the rollercoaster is under the tracks. Now, both the normal reaction and weight force act down, so mv^2/r = mg + N, or N = mv^2/r - mg. Here, if the speed isn't high enough, v^/r < g and N<0, which again is physically meaningless; the rollercoaster just falls from the sky. The limiting case here is again when v^2/r = g, again when the weight force is just able to provide all of the acceleration.
Now let's look at the bottom of the loop. Here, the normal reaction acts upwards and the weight force acts downwards. The net force is upwards, so N>mg and mv^2/r = N - mg, or N = mg + mv^2/r. This is why N>mg. The normal reaction force has to both counter gravity and push the rollercoaster in a circular path.
With these questions, draw the normal reaction force and its direction in the direction it would make sense to act in. If the rollercoaster is on top of the tracks, it only makes sense that the reaction force is acting upwards.
EDIT:
And again!!
So a car at a height of 22 metres has a GP of 55 000. When he goes down to ground level, his KE would be 55000 as energy is not lost. This car then goes to a loop, and the egp at the top of the loop is 40 000. If it wants me to find the speed at that loop, why can't I just use the 40 000?
I'm assuming it b/c kinetik energy is 0 at that height. So hence, I would figure out the amount of energy (and hence velocity) lost at that point of time, and that being 15 000 joules. I then use these 15 000 joules to figure out the velcoity..
As you can see, I've confused the shit out of myself in these two questions. PB, Lxznl, Honghyo and others please help!! 
There are several problems with this question or the way you've worded it. Firstly, potential energy values themselves have no real meaning. You need to define a zero point somewhere as really, only changes in potential energy have any meaning. I am assuming you mean that the potential energy is zero at ground level.
The next issue is with the bit about KE being 55000 J. Now, this is only valid if at the top of the loop the car was stationary; otherwise, you must add 55000 J to the initial KE at the top.
To do this question, you would use the gravitational potential energy initially (55000 J) and relate it to the energy at the top of the loop (40000 J) to find the increase in KE. Use that to find the speed.
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