- #1
rollorcoaster
- 3
- 0
A small diameter (2.00 mm), solid steel ball rolls from rest, without slipping, down a track and into a loop-the-loop of 1.50 meters diameter. Between the starting point on the track and the top of the loop the ball converts 10.0% of its initial mechanical energy into other forms of energy. From how high above the ground must the ball be released in order to just make it through the top of the loop? Consider the diameter of the ball to be small compared to the diameter of the loop but don't forget to consider rotational kinetic energy! KE = .5I(omega)^2 + .5mv^2
i of sphere 2/5MR^2
d = 1.5m
mgh - mgd = [1/5mR^2 (V/R)^2 + .5mV^2](.9)
v = [2g(h - d)]^(1/2)
gh - gd = [2/5g(h - d)+ g(h - d)](.9)
h - d = 7/5(h - d)(.9)
h = [(-7/5)(.9)d + d]/(1-(7/5)(.9))
h = .975
when the hight of a track before the loop needs to be 5/2 larger without a loss term and rotational kinetic energy ??
i of sphere 2/5MR^2
d = 1.5m
mgh - mgd = [1/5mR^2 (V/R)^2 + .5mV^2](.9)
v = [2g(h - d)]^(1/2)
gh - gd = [2/5g(h - d)+ g(h - d)](.9)
h - d = 7/5(h - d)(.9)
h = [(-7/5)(.9)d + d]/(1-(7/5)(.9))
h = .975
when the hight of a track before the loop needs to be 5/2 larger without a loss term and rotational kinetic energy ??
Last edited: