I hate this problem....these type of problems involve deriving and I hate doing that.
In Figure 11-32, a solid brass ball of mass m and radius r will roll without slipping along the loop-the-loop track when released from rest along the straight section. For the following answers use g for the acceleration due to gravity, and m, r, and R, as appropriate, where all quantities are in SI units.
![[http://www.webassign.net/hrw/hrw7_11-32.gif]](http://www.webassign.net/hrw/hrw7_11-32.gif "http://www.webassign.net/hrw/hrw7_11-32.gif")
(a) From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop? (The radius of the loop-the-loop is R. Assume R r.)
I started off by using law of conservation of energy.
Ui = Kf + Uf
Then inserted in the kinetic energy of rolling
mgh = .5 Icom omega^2 + .5 Mv^2 + mg(2R)
mgh = .5 (2/5mr^2)(v/r)^2 + .5 Mv^2 + mg2R
the Icom equation is for a sphere and the v/r is equilivalent to omega or angular speed.
canceling m and r
gh = 7/10v^2 + 2gR
h = (7/10v^2)/g + 2R
Why is this not right then?
(b) If the marble is released from height 6R above the bottom of the track, what is the magnitude of the horizontal component of the force acting on it at point Q?
In Figure 11-32, a solid brass ball of mass m and radius r will roll without slipping along the loop-the-loop track when released from rest along the straight section. For the following answers use g for the acceleration due to gravity, and m, r, and R, as appropriate, where all quantities are in SI units.
(a) From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop? (The radius of the loop-the-loop is R. Assume R r.)
I started off by using law of conservation of energy.
Ui = Kf + Uf
Then inserted in the kinetic energy of rolling
mgh = .5 Icom omega^2 + .5 Mv^2 + mg(2R)
mgh = .5 (2/5mr^2)(v/r)^2 + .5 Mv^2 + mg2R
the Icom equation is for a sphere and the v/r is equilivalent to omega or angular speed.
canceling m and r
gh = 7/10v^2 + 2gR
h = (7/10v^2)/g + 2R
Why is this not right then?
(b) If the marble is released from height 6R above the bottom of the track, what is the magnitude of the horizontal component of the force acting on it at point Q?
![[http://www.cubcarson.com/mediac/400_0/media/bunny~with~pancake~head.jpg]](http://www.cubcarson.com/mediac/400_0/media/bunny~with~pancake~head.jpg "http://www.cubcarson.com/mediac/400_0/media/bunny~with~pancake~head.jpg")