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Suppose I have a track that starts at a height h, goes down (not a vertical fall, but gradually in a slope), enters a circular loop of radius r (the loop is 'laying' on the ground so the top of the loop is at a height 2r from the ground) and then continues at ground level.

- Calculate the minimum height h so that the mass remains on the track at all times (assume no friction).
- Let 2h be the release height. What is the normal force on m at the bottom of the loop (before entering it), at the top of the loop, and when the block exits the loop onto the flat section.
- Show that the difference in the apparent weight of the mass at the top of the loop and the bottom of the loop is 6mg, and that this answer does not depend on the size of the loop or the speed of the mass as long as the speed is above the minimum required to go through the loop.

Quite a quesiton huh? Anyways, I need confirmation for my answers to a and b, and I need some help with c.

- As I figure it, as long as the mass can make it through 3/4 of the loop, it should make it through the entire loop (i.e. if the velocity of the mass is zero when it goes through 3/4 of the loop, then gravity should help push it through the entire loop at that point). Using energy methods, I can equate the total mechanical energy at the top of the track (at a height h) with the total mechanical energy when it's gone through 3/4 of the loop. So,

[tex]mgh = mgr \Rightarrow h = r[/tex] - For the normal force N
_{1}on m before entering the loop:

[tex]mv^2/r = N_1 - mg \Rightarrow N_1 = mv^2/r + mg[/tex][tex]2mgh = mv^2/2 \Rightarrow mv^2 = 4mgh[/tex][tex]N_1 = mv^2/r + mg = 4mgh/r + mg[/tex]For the normal force N_{2}on m at the top of the loop:

[tex]mv^2/r = N_2 + mg \Rightarrow N_2 = mv^2/r - mg[/tex][tex]2mgh = mv^2/2 + 2mgr \Rightarrow mv^2 = 4mg(h - r)[/tex][tex]N_2 = mv^2/r + mg = 4mg(h - r)/r + mg[/tex]For the normal force N_{3}when the mass exists the loop:

[tex]0 = N_3 - mg \Rightarrow N_3 = mg[/tex] - Suppose I have a spring scale used for measuring the weight of the mass when it's at the top of the loop. Let w be the force that the spring measures (i.e. the force in which the spring pulls up on the mass). Applying Newton's 2nd law yields:

[tex]mv^2/r = N_2 + mg - w \Rightarrow w = N_2 + mg - mv^2/r[/tex][tex]mv^2 = 4mg(h - r) \Rightarrow w = N_2 + mg - 4mg(h - r)/r[/tex]This is as far as I can go. Maybe this approach is wrong. Any tips?

e(ho0n3