So, this problem is about energy. The question is: what must the minimum height (in R) of a height be, for a ball to go down it, through a loop the loop (with radius R), and then through a rough surface 6R's long with u= 0.5.
gravitational potential energy= mgh
kinetic energy= 1/2mv^2
work of friction= umgs
The Attempt at a Solution
Hopefully this is right: I did initial potential energy= mgh= 1/2mv^2 (kinetic needed to get past the top of the loop) + umgs (work of friction of the surface) + mgh (potential energy at top of loop). Also, I got v^2=Rg, since using the centripetal force equation mv^2/R, I made it equal to the normal force, mg, and v^2=Rg.
The mg's cancel out, so you're left with h=1/2R + (0.5)6R + 2R, which equals 5.5R. Is this right??