Show that a roller coaster with a circular vertical loop. The difference in your apparent weight at the top of the circular loop and the bottom of the circular loop is 6 g's-that is, six times your weight. Ignore friction. Show also that as long as your speed is above the minimum needed, this answer doesn't depend on the size of the loop or how fast your go through it. My working for the first half of the problem, the 6g's part is as follows Radius if loop is R Height from which it is released is h The speed at bottom of the loop is determined by the conservation of mechanical energy 1/2 mvbottom2=2mgh Apparent weight at the bottom of the loop is obtained by the below: mvBot2= FNorm. Bot.-mg Therefore apparent weight at bottom is FNorm. Bot.=mvbot2/R+mg FNorm. Bot.=2mgh/R+mg (using result obtained via conservation of energy) To find speed at top of the loop we have from Conservation of Energy 1/2 mvtop2+mg(2R)=mgh mvtop2=2mg(h-2R) Therefore using the above the apparent weight at the top of the loop is mvTop2/R = FNorm. Top.+mg Therefore Apparent weight is : FNorm. Top. = (2mg(h-2R))/R - mg Hence FNorm. Bot. - FNorm. Top. = 2mgh/R + mg - [((2mg(h-2R))/R - mg)]= 2mgh/R + mg - 2mgh/R + 4mg + mg= 6mg That's how I proved the first part of the problem. Can anyone please tell me how to complete the second part of the problem; namely proving that as long as your speed is above the minimum needed, the answer doesn't depend on the size of the loop or how fast your go through it. This part of the problem is really bugging me and I've tried heaps of ways but can't come up with the answer.